Search: boubaker
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A135936
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Irregular triangle read by rows: row n gives coefficients of Boubaker polynomial B_n(x) in order of decreasing exponents (another version).
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+20
4
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1, 1, 1, 2, 1, 1, 1, 0, -2, 1, -1, -3, 1, -2, -3, 2, 1, -3, -2, 5, 1, -4, 0, 8, -2, 1, -5, 3, 10, -7, 1, -6, 7, 10, -15, 2, 1, -7, 12, 7, -25, 9, 1, -8, 18, 0, -35, 24, -2, 1, -9, 25, -12, -42, 49, -11, 1, -10, 33, -30, -42, 84, -35, 2, 1, -11, 42, -55, -30, 126, -84, 13, 1, -12, 52, -88, 0, 168, -168, 48, -2, 1, -13, 63, -130, 55, 198, -294
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0,4
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FORMULA
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T(n,m) = (-1)^m * (binomial(n-m, m) - 3*binomial(n-m-1, m-1)). (End)
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EXAMPLE
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The Boubaker polynomials B_0(x), B_1(x), B_2(x), ... are:
1
x
x^2 + 2
x^3 + x
x^4 - 2
x^5 - x^3 - 3*x
x^6 - 2*x^4 - 3*x^2 + 2
x^7 - 3*x^5 - 2*x^3 + 5*x
x^8 - 4*x^6 + 8*x^2 - 2
x^9 - 5*x^7 + 3*x^5 + 10*x^3 - 7*x
...
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MAPLE
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A135936 := proc(n, m) coeftayl( coeftayl( (1+3*t^2)/(1-x*t+t^2), t=0, n), x=0, m) ; end: for n from 0 to 25 do for m from n to 0 by -2 do printf("%d, ", A135936(n, m)) ; od; od; # R. J. Mathar, Mar 11 2008
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MATHEMATICA
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T[n_, m_] := SeriesCoefficient[SeriesCoefficient[
(1+3*t^2)/(1-x*t+t^2), {t, 0, n}], {x, 0, m}];
Table[T[n, m], {n, 0, 25}, {m, n, 0, -2}] // Flatten (* Jean-François Alcover, Mar 11 2023, after R. J. Mathar *)
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CROSSREFS
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KEYWORD
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sign,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A160242
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Triangle A(n,m) read by rows: a quarter of the Fourier coefficient [cos(m*t)] of the shifted Boubaker polynomial B_n(2*cos t)-2*cos(n*t).
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+20
0
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1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
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OFFSET
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2,2
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COMMENTS
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Starting from the polynomials B_n(x) defined in A137276 and A135929, we insert x=2*cos(t), and define the Fourier coefficients A(n,m) by B_n(2*cos t)-2*cos(n*t) = 4*sum(m=0,..,n-2) A(n,m)*cos(m*t).
A(n,m) is not an integer for n=0, so the table starts at n=1. Furthermore, A(n,m)=0 if n-m is odd, these regular zeros are skipped as usual, so effectively the first table entry appears at n=2.
Simpler definition from R. J. Mathar, Apr 15 2010: a(n)=1 if n =0 or n in A002061, otherwise a(n)=2. So this is a kind of characteristic function of the central polygonal numbers A002061.
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LINKS
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EXAMPLE
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Using T^m =cos(m*t) as a notational shortcut, the expansions start
; B_1(2 cos t)-2*cos 1 t = 0
1 ; B_2(2 cos t)-2*cos 2 t = 1
0 2 ; B_3(2 cos t)-2*cos 3 t = 2*T
1 0 2 ; B_4(2 cos t)-2*cos 4 t = 1+2*T^2
0 2 0 2 ; B_5(2 cos t)-2*cos 5 t = 2*T+2*T^3
1 0 2 0 2 ; B_6(2 cos t)-2*cos 6 t = 1+2*T^2+2*T^4
0 2 0 2 0 2 ; B_7(2 cos t)-2*cos 7 t = 2*T+2*T^3+2*T^5
1 0 2 0 2 0 2 ; B_8(2 cos t)-2*cos 8 t = 1+2*T^2+2*T^4+2*T^6
0 2 0 2 0 2 0 2 ; B_9(2 cos t)-2*cos 9 t = 2*T+2*T^3+2*T^5+2*T^7
1 0 2 0 2 0 2 0 2 ; B_10(2 cos t)-2*cos 10 t = 1+2*T^2+2*T^4+2*T^6+2*T^8
0 2 0 2 0 2 0 2 0 2 ; B_11(2 cos t)-2*cos 11 t = 2*T^3+2*T^5+2*T^7+2*T^9+2*T
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MATHEMATICA
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centralPolygonalQ[n_] := Resolve[Exists[k, k>0, n == k^2-k+1], Integers];
b[n_] := If[n == 0 || centralPolygonalQ[n], 1, 2];
a[n_] := b[n-1];
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KEYWORD
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AUTHOR
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EXTENSIONS
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Definition clarified, publication title corrected, sequence extended by R. J. Mathar, Dec 07 2009
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approved
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A007318
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Pascal's triangle read by rows: C(n,k) = binomial(n,k) = n!/(k!*(n-k)!), 0 <= k <= n.
(Formerly M0082)
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+10
2062
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1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 20, 15, 6, 1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 8, 28, 56, 70, 56, 28, 8, 1, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1
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0,5
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COMMENTS
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A. W. F. Edwards writes: "It [the triangle] was first written down long before 1654, the year in which Blaise Pascal wrote his Traité du triangle arithmétique, but it was this work that brought together all the different aspects of the numbers for the first time. In it Pascal developed the properties of the number as a piece of pure mathematics ... and then, in a series of appendices, showed how these properties were relevant to the study of the figurate numbers, to the theory of combinations, to the expansion of binomial expressions, and to the solution of an important problem in the theory of probability." (A. W. F. Edwards, Pascal's Arithmetical Triangle, Johns Hopkins University Press (2002), p. xiii)
Edwards reports that the naming of the triangle after Pascal was done first by Montmort in 1708 as the "Table de M. Pascal pour les combinaisons" and then by De Moivre in 1730 as the "Triangulum Arithmeticum PASCALANIUM". (Edwards, p. xiv)
In China, Yang Hui in 1261 listed the coefficients of (a+b)^n up to n=6, crediting the expansion to Chia Hsein's Shih-so suan-shu circa 1100. Another prominent early use was in Chu Shih-Chieh's Precious Mirror of the Four Elements in 1303. (Edwards, p. 51)
In Persia, Al-Karaji discovered the binomial triangle "some time soon after 1007", and Al-Samawal published it in the Al-bahir some time before 1180. (Edwards, p. 52)
In India, Halayuda's commentary (circa 900) on Pingala's treatise on syllabic combinations (circa 200 B.C.E.) contains a clear description of the additive computation of the triangle. (Amulya Kumar Bag, Binomial Theorem in Ancient India, p. 72)
Also in India, the multiplicative formula for C(n,k) was known to Mahavira in 850 and restated by Bhaskara in 1150. (Edwards, p. 27)
In Italy, Tartaglia published the triangle in his General trattato (1556), and Cardano published it in his Opus novum (1570). (Edwards, p. 39, 44) - Russ Cox, Mar 29 2022
Also sometimes called Omar Khayyam's triangle.
Also sometimes called Yang Hui's triangle.
C(n,k) = number of k-element subsets of an n-element set.
Row n gives coefficients in expansion of (1+x)^n.
Binomial(n+k-1,n-1) is the number of ways of placing k indistinguishable balls into n boxes (the "bars and stars" argument - see Feller).
Binomial(n-1,k-1) is the number of compositions (ordered partitions) of n with k summands.
Binomial(n+k-1,k-1) is the number of weak compositions (ordered weak partitions) of n into exactly k summands. - Juergen Will, Jan 23 2016
Binomial(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0) and (1,1). - Joerg Arndt, Jul 01 2011
If thought of as an infinite lower triangular matrix, inverse begins:
+1
-1 +1
+1 -2 +1
-1 +3 -3 +1
+1 -4 +6 -4 +1
All 2^n palindromic binomial coefficients starting after the A006516(n)-th entry are odd. - Lekraj Beedassy, May 20 2003
Binomial(n+k-1,n-1) is the number of standard tableaux of shape (n,1^k). - Emeric Deutsch, May 13 2004
Can be viewed as an array, read by antidiagonals, where the entries in the first row and column are all 1's and A(i,j) = A(i-1,j) + A(i,j-1) for all other entries. The determinant of each of its n X n subarrays starting at (0,0) is 1. - Gerald McGarvey, Aug 17 2004
Also the lower triangular readout of the exponential of a matrix whose entry {j+1,j} equals j+1 (and all other entries are zero). - Joseph Biberstine (jrbibers(AT)indiana.edu), May 26 2006
Binomial(n-3,k-1) counts the permutations in S_n which have zero occurrences of the pattern 231 and one occurrence of the pattern 132 and k descents. Binomial(n-3,k-1) also counts the permutations in S_n which have zero occurrences of the pattern 231 and one occurrence of the pattern 213 and k descents. - David Hoek (david.hok(AT)telia.com), Feb 28 2007
Consider integer lists LL of lists L of the form LL = [m#L] = [m#[k#2]] (where '#' means 'times') like LL(m=3,k=3) = [[2,2,2],[2,2,2],[2,2,2]]. The number of the integer list partitions of LL(m,k) is equal to binomial(m+k,k) if multiple partitions like [[1,1],[2],[2]] and [[2],[2],[1,1]] and [[2],[1,1],[2]] are counted only once. For the example, we find 4*5*6/3! = 20 = binomial(6,3). - Thomas Wieder, Oct 03 2007
The infinitesimal generator for Pascal's triangle and its inverse is A132440. - Tom Copeland, Nov 15 2007
Row n>=2 gives the number of k-digit (k>0) base n numbers with strictly decreasing digits; e.g., row 10 for A009995. Similarly, row n-1>=2 gives the number of k-digit (k>1) base n numbers with strictly increasing digits; see A009993 and compare A118629. - Rick L. Shepherd, Nov 25 2007
From Lee Naish (lee(AT)cs.mu.oz.au), Mar 07 2008: (Start)
Binomial(n+k-1, k) is the number of ways a sequence of length k can be partitioned into n subsequences (see the Naish link).
Binomial(n+k-1, k) is also the number of n- (or fewer) digit numbers written in radix at least k whose digits sum to k. For example, in decimal, there are binomial(3+3-1,3)=10 3-digit numbers whose digits sum to 3 (see A052217) and also binomial(4+2-1,2)=10 4-digit numbers whose digits sum to 2 (see A052216). This relationship can be used to generate the numbers of sequences A052216 to A052224 (and further sequences using radix greater than 10). (End)
Denote by sigma_k(x_1,x_2,...,x_n) the elementary symmetric polynomials. Then:
Binomial(2n+1,2k+1) = sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2(i*Pi/(2n+1)), (i=1,2,...,n).
Binomial(2n,2k+1) = 2n*sigma_{n-1-k}(x_1,x_2,...,x_{n-1}), where x_i = tan^2(i*Pi/(2n)), (i=1,2,...,n-1).
Binomial(2n,2k) = sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2((2i-1)Pi/(4n)), (i=1,2,...,n).
Binomial(2n+1,2k) = (2n+1)sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2((2i-1)Pi/(4n+2)), (i=1,2,...,n). (End)
Given matrices R and S with R(n,k) = binomial(n,k)*r(n-k) and S(n,k) = binomial(n,k)*s(n-k), then R*S = T where T(n,k) = binomial(n,k)*[r(.)+s(.)]^(n-k), umbrally. And, the e.g.f.s for the row polynomials of R, S and T are, respectively, exp(x*t)*exp[r(.)*x], exp(x*t)*exp[s(.)*x] and exp(x*t)*exp[r(.)*x]*exp[s(.)*x] = exp{[t+r(.)+s(.)]*x}. The row polynomials are essentially Appell polynomials. See A132382 for an example. - Tom Copeland, Aug 21 2008
As the rectangle R(m,n) = binomial(m+n-2,m-1), the weight array W (defined generally at A144112) of R is essentially R itself, in the sense that if row 1 and column 1 of W=A144225 are deleted, the remaining array is R. - Clark Kimberling, Sep 15 2008
If A007318 = M as an infinite lower triangular matrix, M^n gives A130595, A023531, A007318, A038207, A027465, A038231, A038243, A038255, A027466, A038279, A038291, A038303, A038315, A038327, A133371, A147716, A027467 for n=-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 respectively. - Philippe Deléham, Nov 11 2008
The coefficients of the polynomials with e.g.f. exp(x*t)*(cosh(t)+sinh(t)). - Peter Luschny, Jul 09 2009
The triangle or chess sums, see A180662 for their definitions, link Pascal's triangle with twenty different sequences, see the crossrefs. All sums come in pairs due to the symmetrical nature of this triangle. The knight sums Kn14 - Kn110 have been added. It is remarkable that all knight sums are related to the Fibonacci numbers, i.e., A000045, but none of the others. - Johannes W. Meijer, Sep 22 2010
Binomial(n,k) is also the number of ways to distribute n+1 balls into k+1 urns so that each urn gets at least one ball. See example in the example section below. - Dennis P. Walsh, Jan 29 2011
Binomial(n,k) is the number of increasing functions from {1,...,k} to {1,...,n} since there are binomial(n,k) ways to choose the k distinct, ordered elements of the range from the codomain {1,...,n}. See example in the example section below. - Dennis P. Walsh, Apr 07 2011
Binomial(n,k) is the number of subsets of {1,...,n+1} with k+1 as median element. To see this, note that Sum_{j=0..min(k,n-k)}binomial(k,j)*binomial(n-k,j) = binomial(n,k). See example in Example section below. - Dennis P. Walsh, Dec 15 2011
This is the coordinator triangle for the lattice Z^n, see Conway-Sloane, 1997. - N. J. A. Sloane, Jan 17 2012
One of three infinite families of integral factorial ratio sequences of height 1 (see Bober, Theorem 1.2). The other two are A046521 and A068555. For real r >= 0, C_r(n,k) := floor(r*n)!/(floor(r*k)!*floor(r*(n-k))!) is integral. See A211226 for the case r = 1/2. - Peter Bala, Apr 10 2012
Define a finite triangle T(m,k) with n rows such that T(m,0) = 1 is the left column, T(m,m) = binomial(n-1,m) is the right column, and the other entries are T(m,k) = T(m-1,k-1) + T(m-1,k) as in Pascal's triangle. The sum of all entries in T (there are A000217(n) elements) is 3^(n-1). - J. M. Bergot, Oct 01 2012
The lower triangular Pascal matrix serves as a representation of the operator exp(RLR) in a basis composed of a sequence of polynomials p_n(x) characterized by ladder operators defined by R p_n(x) = p_(n+1)(x) and L p_n(x) = n p_(n-1)(x). See A132440, A218272, A218234, A097805, and A038207. The transposed and padded Pascal matrices can be associated to the special linear group SL2. - Tom Copeland, Oct 25 2012
A permutation p_1...p_n of the set {1,...,n} has a descent at position i if p_i > p_(i+1). Let S(n) denote the subset of permutations p_1...p_n of {1,...,n} such that p_(i+1) - p_i <= 1 for i = 1,...,n-1. Then binomial(n,k) gives the number of permutations in S(n+1) with k descents. Alternatively, binomial(n,k) gives the number of permutations in S(n+1) with k+1 increasing runs. - Peter Bala, Mar 24 2013
Sum_{n=>0} binomial(n,k)/n! = e/k!, where e = exp(1), while allowing n < k where binomial(n,k) = 0. Also Sum_{n>=0} binomial(n+k-1,k)/n! = e * A000262(k)/k!, and for k>=1 equals e * A067764(k)/A067653(k). - Richard R. Forberg, Jan 01 2014
The square n X n submatrix (first n rows and n columns) of the Pascal matrix P(x) defined in the formulas below when multiplying on the left the Vandermonde matrix V(x_1,...,x_n) (with ones in the first row) translates the matrix to V(x_1+x,...,x_n+x) while leaving the determinant invariant. - Tom Copeland, May 19 2014
For k>=2, n>=k, k/((k/(k-1) - Sum_{n=k..m} 1/binomial(n,k))) = m!/((m-k+1)!*(k-2)!). Note: k/(k-1) is the infinite sum. See A000217, A000292, A000332 for examples. - Richard R. Forberg, Aug 12 2014
Let G_(2n) be the subgroup of the symmetric group S_(2n) defined by G_(2n) = {p in S_(2n) | p(i) = i (mod n) for i = 1,2,...,2n}. G_(2n) has order 2^n. Binomial(n,k) gives the number of permutations in G_(2n) having n + k cycles. Cf. A130534 and A246117. - Peter Bala, Aug 15 2014
C(n,k) = the number of Dyck paths of semilength n+1, with k+1 "u"'s in odd numbered positions and k+1 returns to the x axis. Example: {U = u in odd position and _ = return to x axis} binomial(3,0)=1 (Uudududd_); binomial(3,1)=3 [(Uududd_Ud_), (Ud_Uududd_), (Uudd_Uudd_)]; binomial(3,2)=3 [(Ud_Ud_Uudd_), (Uudd_Ud_Ud_), (Ud_Uudd_Ud_)]; binomial(3,3)=1 (Ud_Ud_Ud_Ud_). - Roger Ford, Nov 05 2014
The binomial coefficients binomial(n,k) give the number of individuals of the k-th generation after n population doublings. For each doubling of population, each individual's clone has its generation index incremented by 1, and thus goes to the next row. Just tally up each row from 0 to 2^n - 1 to get the binomial coefficients.
0 1 3 7 15
0: O | . | . . | . . . . | . . . . . . . . |
1: | O | O . | O . . . | O . . . . . . . |
2: | | O | O O . | O O . O . . . |
3: | | | O | O O O . |
4: | | | | O |
This is a fractal process: to get the pattern from 0 to 2^n - 1, append a shifted down (by one row) copy of the pattern from 0 to 2^(n-1) - 1 to the right of the pattern from 0 to 2^(n-1) - 1. (Inspired by the "binomial heap" data structure.)
Sequence of generation indices: 1's-counting sequence: number of 1's in binary expansion of n (or the binary weight of n) (see A000120):
{0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, ...}
Binary expansion of 0 to 15:
0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1111
(End)
T(n,k) is the number of set partitions w of [n+1] that avoid 1/2/3 with rb(w)=k. The same holds for ls(w)=k, where avoidance is in the sense of Klazar and ls,rb defined by Wachs and White.
Let {A(n)} be a set with exactly n identical elements, with {A(0)} being the empty set E. Let {A(n,k)} be the k-th iteration of {A(n)}, with {A(n,0)} = {A(n)}. {A(n,1)} = The set of all the subsets of A{(n)}, including {A(n)} and E. {A(n,k)} = The set of all subsets of {A(n,k-1)}, including all of the elements of {A(n,k-1)}. Let A(n,k) be the number of elements in {A(n,k)}. Then A(n,k) = C(n+k,k), with each successive iteration replicating the members of the k-th diagonal of Pascal's Triangle. See examples. - Gregory L. Simay, Aug 06 2018
Binomial(n-1,k) is also the number of permutations avoiding both 213 and 312 with k ascents. - Lara Pudwell, Dec 19 2018
Binomial(n-1,k) is also the number of permutations avoiding both 132 and 213 with k ascents. - Lara Pudwell, Dec 19 2018
Binomial(n,k) is the dimension of the k-th exterior power of a vector space of dimension n. - Stefano Spezia, Dec 22 2018
C(n,k-1) is the number of unoriented colorings of the facets (or vertices) of an n-dimensional simplex using exactly k colors. Each chiral pair is counted as one when enumerating unoriented arrangements. - Robert A. Russell, Oct 20 2020
From Dilcher and Stolarsky: "Two of the most ubiquitous objects in mathematics are the sequence of prime numbers and the binomial coefficients (and thus Pascal's triangle). A connection between the two is given by a well-known characterization of the prime numbers: Consider the entries in the kth row of Pascal's triangle, without the initial and final entries. They are all divisible by k if and only if k is a prime." - Tom Copeland, May 17 2021
Named "Table de M. Pascal pour les combinaisons" by Pierre Remond de Montmort (1708) after the French mathematician, physicist and philosopher Blaise Pascal (1623-1662). - Amiram Eldar, Jun 11 2021
Consider the n-th diagonal of the triangle as a sequence b(n) with n starting at 0. From it form a new sequence by leaving the 0th term as is, and thereafter considering all compositions of n, taking the product of b(i) over the respective numbers i in each composition, adding terms corresponding to compositions with an even number of parts subtracting terms corresponding to compositions with an odd number of parts. Then the n-th row of the triangle is obtained, with every second term multiplied by -1, followed by infinitely many zeros. For sequences starting with 1, this operation is a special case of a self-inverse operation, and therefore the converse is true. - Thomas Anton, Jul 05 2021
C(n,k) is the number of vertices in an n-dimensional unit hypercube, at an L1 distance of k (or: with a shortest path of k 1d-edges) from a given vertex. - Eitan Y. Levine, May 01 2023
C(n+k-1,k-1) is the number of vertices at an L1 distance from a given vertex in an infinite-dimensional box, which has k sides of length 2^m for each m >= 0. Equivalently, given a set of tokens containing k distinguishable tokens with value 2^m for each m >= 0, C(n+k-1,k-1) is the number of subsets of tokens with a total value of n. - Eitan Y. Levine, Jun 11 2023
Numbers in the k-th column, i.e., numbers of the form C(n,k) for n >= k, are known as k-simplex numbers. - Pontus von Brömssen, Jun 26 2023
Let r(k) be the k-th row and c(k) the k-th column. Denote convolution by * and repeated convolution by ^. Then r(k)*r(m)=r(k+m) and c(k)*c(m)=c(k+m+1). This is because r(k) = r(1) ^ k and c(k) = c(0) ^ k+1. - Eitan Y. Levine, Jul 23 2023
Number of permutations of length n avoiding simultaneously the patterns 231 and 312(resp., 213 and 231; 213 and 312) with k descents (equivalently, with k ascents). An ascent (resp., descent) in a permutation a(1)a(2)...a(n) is position i such that a(i)<a(i+1) (resp., a(i)>a(i+1)). - Tian Han, Nov 25 2023
C(n,k) are generalized binomial coefficients of order m=0. Calculated by the formula C(n,k) = Sum_{i=0..n-k} binomial(n+1, n-k-i)*Stirling2(i+ m+ 1, i+1) *(-1)^i, where m = 0 for n>= 0, 0 <= k <= n. - Igor Victorovich Statsenko, Feb 26 2023
The Akiyama-Tanigawa algorithm applied to the diagonals, binomial(n+k,k), yields the powers of n. - Shel Kaphan, May 03 2024
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REFERENCES
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
Amulya Kumar Bag, Binomial theorem in ancient India, Indian Journal of History of Science, vol. 1 (1966), pp. 68-74.
Arthur T. Benjamin and Jennifer Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 63ff.
Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 306.
P. Curtz, Intégration numérique des systèmes différentiels à conditions initiales, Centre de Calcul Scientifique de l'Armement, Arcueil, 1969.
A. W. F. Edwards, Pascal's Arithmetical Triangle, 2002.
William Feller, An Introduction to Probability Theory and Its Application, Vol. 1, 2nd ed. New York: Wiley, p. 36, 1968.
Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, p. 155.
David Hök, Parvisa mönster i permutationer [Swedish], 2007.
Donald E. Knuth, The Art of Computer Programming, Vol. 1, 2nd ed., p. 52.
Sergei K. Lando, Lecture on Generating Functions, Amer. Math. Soc., Providence, R.I., 2003, pp. 60-61.
Blaise Pascal, Traité du triangle arithmétique, avec quelques autres petits traitez sur la mesme matière, Desprez, Paris, 1665.
Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 71.
A. P. Prudnikov, Yu. A. Brychkov and O. I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 6.
John Riordan, Combinatorial Identities, Wiley, 1968, p. 2.
Robert Sedgewick and Philippe Flajolet, An Introduction to the Analysis of Algorithms, Addison-Wesley, Reading, MA, 1996, p. 143.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 115-118.
Douglas B. West, Combinatorial Mathematics, Cambridge, 2021, p. 25.
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LINKS
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
A. Farina, S. Giompapa, A. Graziano, A. Liburdi, M. Ravanelli and F. Zirilli, Tartaglia-Pascal's triangle: a historical perspective with applications, Signal, Image and Video Processing, Vol. 7, No. 1 (January 2013), pp. 173-188.
Subhash Kak, The golden mean and the physics of aesthetics, in: B. Yadav and M. Mohan (eds.), Ancient Indian Leaps into Mathematics, Birkhäuser, Boston, MA, 2009, pp. 111-119; arXiv preprint, arXiv:physics/0411195 [physics.hist-ph], 2004.
Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003. [Cached copy, with permission (pdf only)]
Jack Ramsay, On Arithmetical Triangles, The Pulse of Long Island, June 1965 [Mentions application to design of antenna arrays. Annotated scan.]
Louis W. Shapiro, Seyoum Getu, Wen-Jin Woan and Leon C. Woodson, The Riordan group, Discrete Applied Math., Vol. 34 (1991), pp. 229-239.
Christopher Stover and Eric W. Weisstein, Composition. From MathWorld - A Wolfram Web Resource.
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FORMULA
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a(n, k) = C(n,k) = binomial(n, k).
C(n, k) = C(n-1, k) + C(n-1, k-1).
The triangle is symmetric: C(n,k) = C(n,n-k).
a(n+1, m) = a(n, m) + a(n, m-1), a(n, -1) := 0, a(n, m) := 0, n<m; a(0, 0)=1.
C(n, k) = n!/(k!(n-k)!) if 0<=k<=n, otherwise 0.
G.f.: 1/(1-y-x*y) = Sum_(C(n, k)*x^k*y^n, n, k>=0)
G.f.: 1/(1-x-y) = Sum_(C(n+k, k)*x^k*y^n, n, k>=0).
G.f. for row n: (1+x)^n = Sum_{k=0..n} C(n, k)*x^k.
G.f. for column k: x^k/(1-x)^(k+1); [corrected by Werner Schulte, Jun 15 2022].
E.g.f.: A(x, y) = exp(x+x*y).
E.g.f. for column n: x^n*exp(x)/n!.
In general the m-th power of A007318 is given by: T(0, 0) = 1, T(n, k) = T(n-1, k-1) + m*T(n-1, k), where n is the row-index and k is the column; also T(n, k) = m^(n-k)*C(n, k).
Triangle T(n, k) read by rows; given by A000007 DELTA A000007, where DELTA is Deléham's operator defined in A084938.
Let P(n+1) = the number of integer partitions of (n+1); let p(i) = the number of parts of the i-th partition of (n+1); let d(i) = the number of different parts of the i-th partition of (n+1); let m(i, j) = multiplicity of the j-th part of the i-th partition of (n+1). Define the operator Sum_{i=1..P(n+1), p(i)=k+1} as the sum running from i=1 to i=P(n+1) but taking only partitions with p(i)=(k+1) parts into account. Define the operator Product_{j=1..d(i)} = product running from j=1 to j=d(i). Then C(n, k) = Sum_{p(i)=(k+1), i=1..P(n+1)} p(i)! / [Product_{j=1..d(i)} m(i, j)!]. E.g., C(5, 3) = 10 because n=6 has the following partitions with m=3 parts: (114), (123), (222). For their multiplicities one has: (114): 3!/(2!*1!) = 3; (123): 3!/(1!*1!*1!) = 6; (222): 3!/3! = 1. The sum is 3 + 6 + 1 = 10 = C(5, 3). - Thomas Wieder, Jun 03 2005
G.f.: 1 + x*(1 + x) + x^3*(1 + x)^2 + x^6*(1 + x)^3 + ... . - Michael Somos, Sep 16 2006
Sum_{k=0..floor(n/2)} x^(n-k)*T(n-k,k) = A000007(n), A000045(n+1), A002605(n), A030195(n+1), A057087(n), A057088(n), A057089(n), A057090(n), A057091(n), A057092(n), A057093(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, respectively. Sum_{k=0..floor(n/2)} (-1)^k*x^(n-k)*T(n-k,k) = A000007(n), A010892(n), A009545(n+1), A057083(n), A001787(n+1), A030191(n), A030192(n), A030240(n), A057084(n), A057085(n+1), A057086(n), A084329(n+1) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, respectively. - Philippe Deléham, Sep 16 2006
C(t+p-1, t) = Sum_{i=0..t} C(i+p-2, i) = Sum_{i=1..p} C(i+t-2, t-1). A binomial number is the sum of its left parent and all its right ancestors, which equals the sum of its right parent and all its left ancestors. - Lee Naish (lee(AT)cs.mu.oz.au), Mar 07 2008
Let A(x) = Sum_{n>=0} x^(n*(n+1)/2)*(1+x)^n be the g.f. of the flattened triangle:
A(x) = 1 + (x + x^2) + (x^3 + 2*x^4 + x^5) + (x^6 + 3*x^7 + 3*x^8 + x^9) + ...
then A(x) equals the series Sum_{n>=0} (1+x)^n*x^n*Product_{k=1..n} (1-(1+x)*x^(2*k-1))/(1-(1+x)*x^(2*k));
also, A(x) equals the continued fraction 1/(1- x*(1+x)/(1+ x*(1-x)*(1+x)/(1- x^3*(1+x)/(1+ x^2*(1-x^2)*(1+x)/(1- x^5*(1+x)/(1+ x^3*(1-x^3)*(1+x)/(1- x^7*(1+x)/(1+ x^4*(1-x^4)*(1+x)/(1- ...))))))))).
These formulas are due to (1) a q-series identity and (2) a partial elliptic theta function expression. (End)
Row n of the triangle is the result of applying the ConvOffs transform to the first n terms of the natural numbers (1, 2, 3, ..., n). See A001263 or A214281 for a definition of this transformation. - Gary W. Adamson, Jul 12 2012
Row n (n >= 0) of the triangle is given by the n-th antidiagonal of the infinite matrix P^n, where P = (p_{i,j}), i,j >= 0, is the production matrix
0, 1,
1, 0, 1,
0, 1, 0, 1,
0, 0, 1, 0, 1,
0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 1, 0, 1,
0, 0, 0, 0, 0, 0, 1, 0, 1,
... (End)
Row n of the triangle is also given by the n+1 coefficients of the polynomial P_n(x) defined by the recurrence P_0(x) = 1, P_1(x) = x + 1, P_n(x) = x*P_{n-1}(x) + P_{n-2}(x), n > 1. - L. Edson Jeffery, Aug 12 2013
For a closed-form formula for arbitrary left and right borders of Pascal-like triangles see A228196. - Boris Putievskiy, Aug 18 2013
(1+x)^n = Sum_{k=0..n} (-1)^(n-k)*binomial(n,k)*Sum_{i=0..k} k^(n-i)*binomial(k,i)*x^(n-i)/(n-i)!. - Vladimir Kruchinin, Oct 21 2013
E.g.f.: A(x,y) = exp(x+x*y) = 1 + (x+y*x)/( E(0)-(x+y*x)), where E(k) = 1 + (x+y*x)/(1 + (k+1)/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 08 2013
E.g.f.: E(0) -1, where E(k) = 2 + x*(1+y)/(2*k+1 - x*(1+y)/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
G.f.: 1 + x*(1+x)*(1+x^2*(1+x)/(W(0)-x^2-x^3)), where W(k) = 1 + (1+x)*x^(k+2) - (1+x)*x^(k+3)/W(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
Sum_{n>=0} C(n,k)/n! = e/k!, where e = exp(1), while allowing n < k where C(n,k) = 0. Also Sum_{n>=0} C(n+k-1,k)/n! = e * A000262(k)/k!, and for k>=1 equals e * A067764(k)/A067653(k). - Richard R. Forberg, Jan 01 2014
Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result by A007318(x) = P(x). Then with :xD:^n = x^n*(d/dx)^n and B(n,x), the Bell polynomials (A008277),
A) P(x)= exp(x*dP) = exp[x*(e^M-I)] = exp[M*B(.,x)] = (I+dP)^B(.,x)
B) P(:xD:) = exp(dP:xD:) = exp[(e^M-I):xD:] = exp[M*B(.,:xD:)] = exp[M*xD] = (I+dP)^(xD) with action P(:xD:)g(x) = exp(dP:xD:)g(x) = g[(I+dP)*x] (cf. also A238363).
C) P(x)^y = P(y*x). P(2x) = A038207(x) = exp[M*B(.,2x)], the face vectors of the n-dim hypercubes.
D) P(x) = [St2]*exp(x*M)*[St1] = [St2]*(I+dP)^x*[St1]
E) = [St1]^(-1)*(I+dP)^x*[St1] = [St2]*(I+dP)^x*[St2]^(-1)
where [St1]=padded A008275 just as [St2]=A048993=padded A008277 and exp(x*M) = (I+dP)^x = Sum_{k>=0} C(x,k) dP^k. (End)
Recurrence equation: T(n,k) = T(n-1,k)*(n + k)/(n - k) - T(n-1,k-1) for n >= 2 and 1 <= k < n, with boundary conditions T(n,0) = T(n,n) = 1. Note, changing the minus sign in the recurrence to a plus sign gives a recurrence for the square of the binomial coefficients - see A008459.
There is a relation between the e.g.f.'s of the rows and the diagonals of the triangle, namely, exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(1 + 3*x + 3*x^2/2! + x^3/3!) = 1 + 4*x + 10*x^2/2! + 20*x^3/3! + 35*x^4/4! + .... This property holds more generally for the Riordan arrays of the form ( f(x), x/(1 - x) ), where f(x) is an o.g.f. of the form 1 + f_1*x + f_2*x^2 + .... See, for example, A055248 and A106516.
Let P denote the present triangle. For k = 0,1,2,... define P(k) to be the lower unit triangular block array
/I_k 0\
\ 0 P/ having the k X k identity matrix I_k as the upper left block; in particular, P(0) = P. The infinite product P(0)*P(1)*P(2)*..., which is clearly well-defined, is equal to the triangle of Stirling numbers of the second kind A008277. The infinite product in the reverse order, that is, ...*P(2)*P(1)*P(0), is equal to the triangle of Stirling cycle numbers A130534. (End)
C(a+b,c) = Sum_{k=0..a} C(a,k)*C(b,b-c+k). This is a generalization of equation 1 from section 4.2.5 of the Prudnikov et al. reference, for a=b=c=n: C(2*n,n) = Sum_{k=0..n} C(n,k)^2. See Links section for animation of new formula. - Hermann Stamm-Wilbrandt, Aug 26 2015
The row polynomials of the Pascal matrix P(n,x) = (1+x)^n are related to the Bernoulli polynomials Br(n,x) and their umbral compositional inverses Bv(n,x) by the umbral relation P(n,x) = (-Br(.,-Bv(.,x)))^n = (-1)^n Br(n,-Bv(.,x)), which translates into the matrix relation P = M * Br * M * Bv, where P is the Pascal matrix, M is the diagonal matrix diag(1,-1,1,-1,...), Br is the matrix for the coefficients of the Bernoulli polynomials, and Bv that for the umbral inverse polynomials defined umbrally by Br(n,Bv(.,x)) = x^n = Bv(n,Br(.,x)). Note M = M^(-1). - Tom Copeland, Sep 05 2015
1/(1-x)^k = (r(x) * r(x^2) * r(x^4) * ...) where r(x) = (1+x)^k. - Gary W. Adamson, Oct 17 2016
Boas-Buck type recurrence for column k for Riordan arrays (see the Aug 10 2017 remark in A046521, also for the reference) with the Boas-Buck sequence b(n) = {repeat(1)}. T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} T(j, k), for n >= 1, with T(n, n) = 1. This reduces, with T(n, k) = binomial(n, k), to a known binomial identity (e.g, Graham et al. p. 161). - Wolfdieter Lang, Nov 12 2018
C((p-1)/a, b) == (-1)^b * fact_a(a*b-a+1)/fact_a(a*b) (mod p), where fact_n denotes the n-th multifactorial, a divides p-1, and the denominator of the fraction on the right side of the equation represents the modular inverse. - Isaac Saffold, Jan 07 2019
Binomial sums are Fibonacci numbers A000045:
Sum_{k=0..n} C(n + k, 2*k + 1) = F(2*n).
Sum_{k=0..n} C(n + k, 2*k) = F(2*n + 1). (End)
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EXAMPLE
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Triangle T(n,k) begins:
n\k 0 1 2 3 4 5 6 7 8 9 10 11 ...
0 1
1 1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
5 1 5 10 10 5 1
6 1 6 15 20 15 6 1
7 1 7 21 35 35 21 7 1
8 1 8 28 56 70 56 28 8 1
9 1 9 36 84 126 126 84 36 9 1
10 1 10 45 120 210 252 210 120 45 10 1
11 1 11 55 165 330 462 462 330 165 55 11 1
...
There are C(4,2)=6 ways to distribute 5 balls BBBBB, among 3 different urns, < > ( ) [ ], so that each urn gets at least one ball, namely, <BBB>(B)[B], <B>(BBB)[B], <B>(B)[BBB], <BB>(BB)[B], <BB>(B)[BB], and <B>(BB)[BB].
There are C(4,2)=6 increasing functions from {1,2} to {1,2,3,4}, namely, {(1,1),(2,2)},{(1,1),(2,3)}, {(1,1),(2,4)}, {(1,2),(2,3)}, {(1,2),(2,4)}, and {(1,3),(2,4)}. - Dennis P. Walsh, Apr 07 2011
There are C(4,2)=6 subsets of {1,2,3,4,5} with median element 3, namely, {3}, {1,3,4}, {1,3,5}, {2,3,4}, {2,3,5}, and {1,2,3,4,5}. - Dennis P. Walsh, Dec 15 2011
The successive k-iterations of {A(0)} = E are E;E;E;...; the corresponding number of elements are 1,1,1,... The successive k-iterations of {A(1)} = {a} are (omitting brackets) a;a,E; a,E,E;...; the corresponding number of elements are 1,2,3,... The successive k-iterations of {A(2)} = {a,a} are aa; aa,a,E; aa, a, E and a,E and E;...; the corresponding number of elements are 1,3,6,... - Gregory L. Simay, Aug 06 2018
Boas-Buck type recurrence for column k = 4: T(8, 4) = (5/4)*(1 + 5 + 15 + 35) = 70. See the Boas-Buck comment above. - Wolfdieter Lang, Nov 12 2018
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MAPLE
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MATHEMATICA
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Flatten[Table[Binomial[n, k], {n, 0, 11}, {k, 0, n}]] (* Robert G. Wilson v, Jan 19 2004 *)
Flatten[CoefficientList[CoefficientList[Series[1/(1 - x - x*y), {x, 0, 12}], x], y]] (* Mats Granvik, Jul 08 2014 *)
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PROG
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(Axiom) -- (start)
)set expose add constructor OutputForm
pascal(0, n) == 1
pascal(n, n) == 1
pascal(i, j | 0 < i and i < j) == pascal(i-1, j-1) + pascal(i, j-1)
pascalRow(n) == [pascal(i, n) for i in 0..n]
displayRow(n) == output center blankSeparate pascalRow(n)
for i in 0..20 repeat displayRow i -- (end)
(Python) # See Hobson link. Further programs:
from math import prod, factorial
def C(n, k): return prod(range(n, n-k, -1))//factorial(k) # M. F. Hasler, Dec 13 2019, updated Apr 29 2022, Feb 17 2023
(Haskell)
a007318 n k = a007318_tabl !! n !! k
a007318_row n = a007318_tabl !! n
a007318_list = concat a007318_tabl
a007318_tabl = iterate (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1]
-- Cf. http://www.haskell.org/haskellwiki/Blow_your_mind#Mathematical_sequences
(Maxima) create_list(binomial(n, k), n, 0, 12, k, 0, n); /* Emanuele Munarini, Mar 11 2011 */
(Sage) def C(n, k): return Subsets(range(n), k).cardinality() # Ralf Stephan, Jan 21 2014
(Magma) /* As triangle: */ [[Binomial(n, k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jul 29 2015
(GAP) Flat(List([0..12], n->List([0..n], k->Binomial(n, k)))); # Stefano Spezia, Dec 22 2018
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CROSSREFS
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Equals differences between consecutive terms of A102363. - David G. Williams (davidwilliams(AT)Paxway.com), Jan 23 2006
Row sums give A000079 (powers of 2).
Cf. A083093 (triangle read mod 3), A214292 (first differences of rows).
Partial sums of rows give triangle A008949.
The triangle of the antidiagonals is A011973.
Cf. A008277, A132311, A132312, A052216, A052217, A052218, A052219, A052220, A052221, A052222, A052223, A144225, A202750, A211226, A047999, A026729, A052553, A051920, A193242.
Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A111006, A114197, A162741, A228074, A228196, A228576.
Cf. A115940 (pandigital binomial coefficients C(m,k) with k>1).
Cf. (simplex colorings) A325002 (oriented), [k==n+1] (chiral), A325003 (achiral), A325000 (k or fewer colors), A325009 (orthotope facets, orthoplex vertices), A325017 (orthoplex facets, orthotope vertices).
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 2..12: A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.
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KEYWORD
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AUTHOR
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EXTENSIONS
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Checked all links, deleted 8 that seemed lost forever and were probably not of great importance. - N. J. A. Sloane, May 08 2018
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STATUS
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approved
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A000984
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Central binomial coefficients: binomial(2*n,n) = (2*n)!/(n!)^2.
(Formerly M1645 N0643)
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+10
1033
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1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600, 40116600, 155117520, 601080390, 2333606220, 9075135300, 35345263800, 137846528820, 538257874440, 2104098963720, 8233430727600, 32247603683100, 126410606437752, 495918532948104, 1946939425648112
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,2
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COMMENTS
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Devadoss refers to these numbers as type B Catalan numbers (cf. A000108).
Equal to the binomial coefficient sum Sum_{k=0..n} binomial(n,k)^2.
Number of possible interleavings of a program with n atomic instructions when executed by two processes. - Manuel Carro (mcarro(AT)fi.upm.es), Sep 22 2001
Convolving a(n) with itself yields A000302, the powers of 4. - T. D. Noe, Jun 11 2002
Number of ordered trees with 2n+1 edges, having root of odd degree and nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002
Also number of directed, convex polyominoes having semiperimeter n+2.
Also number of diagonally symmetric, directed, convex polyominoes having semiperimeter 2n+2. - Emeric Deutsch, Aug 03 2002
The second inverse binomial transform of this sequence is this sequence with interpolated zeros. Its g.f. is (1 - 4*x^2)^(-1/2), with n-th term C(n,n/2)(1+(-1)^n)/2. - Paul Barry, Jul 01 2003
Number of possible values of a 2n-bit binary number for which half the bits are on and half are off. - Gavin Scott (gavin(AT)allegro.com), Aug 09 2003
Ordered partitions of n with zeros to n+1, e.g., for n=4 we consider the ordered partitions of 11110 (5), 11200 (30), 13000 (20), 40000 (5) and 22000 (10), total 70 and a(4)=70. See A001700 (esp. Mambetov Bektur's comment). - Jon Perry, Aug 10 2003
Number of nondecreasing sequences of n integers from 0 to n: a(n) = Sum_{i_1=0..n} Sum_{i_2=i_1..n}...Sum_{i_n=i_{n-1}..n}(1). - J. N. Bearden (jnb(AT)eller.arizona.edu), Sep 16 2003
Number of peaks at odd level in all Dyck paths of semilength n+1. Example: a(2)=6 because we have U*DU*DU*D, U*DUUDD, UUDDU*D, UUDUDD, UUU*DDD, where U=(1,1), D=(1,-1) and * indicates a peak at odd level. Number of ascents of length 1 in all Dyck paths of semilength n+1 (an ascent in a Dyck path is a maximal string of up steps). Example: a(2)=6 because we have uDuDuD, uDUUDD, UUDDuD, UUDuDD, UUUDDD, where an ascent of length 1 is indicated by a lower case letter. - Emeric Deutsch, Dec 05 2003
a(n-1) = number of subsets of 2n-1 distinct elements taken n at a time that contain a given element. E.g., n=4 -> a(3)=20 and if we consider the subsets of 7 taken 4 at a time with a 1 we get (1234, 1235, 1236, 1237, 1245, 1246, 1247, 1256, 1257, 1267, 1345, 1346, 1347, 1356, 1357, 1367, 1456, 1457, 1467, 1567) and there are 20 of them. - Jon Perry, Jan 20 2004
The dimension of a particular (necessarily existent) absolutely universal embedding of the unitary dual polar space DSU(2n,q^2) where q>2. - J. Taylor (jt_cpp(AT)yahoo.com), Apr 02 2004.
Number of standard tableaux of shape (n+1, 1^n). - Emeric Deutsch, May 13 2004
Erdős, Graham et al. conjectured that a(n) is never squarefree for sufficiently large n (cf. Graham, Knuth, Patashnik, Concrete Math., 2nd ed., Exercise 112). Sárközy showed that if s(n) is the square part of a(n), then s(n) is asymptotically (sqrt(2)-2)*(sqrt(n))*(Riemann Zeta Function(1/2)). Granville and Ramare proved that the only squarefree values are a(1)=2, a(2)=6 and a(4)=70. - Jonathan Vos Post, Dec 04 2004 [For more about this conjecture, see A261009. - N. J. A. Sloane, Oct 25 2015]
The MathOverflow link contains the following comment (slightly edited): The Erdős squarefree conjecture (that a(n) is never squarefree for n>4) was proved in 1980 by Sárközy, A. (On divisors of binomial coefficients. I. J. Number Theory 20 (1985), no. 1, 70-80.) who showed that the conjecture holds for all sufficiently large values of n, and by A. Granville and O. Ramaré (Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients. Mathematika 43 (1996), no. 1, 73-107) who showed that it holds for all n>4. - Fedor Petrov, Nov 13 2010. [From N. J. A. Sloane, Oct 29 2015]
p divides a((p-1)/2)-1=A030662(n) for prime p=5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, ... = A002144(n) Pythagorean primes: primes of form 4n+1. - Alexander Adamchuk, Jul 04 2006
The number of direct routes from my home to Granny's when Granny lives n blocks south and n blocks east of my home in Grid City. To obtain a direct route, from the 2n blocks, choose n blocks on which one travels south. For example, a(2)=6 because there are 6 direct routes: SSEE, SESE, SEES, EESS, ESES and ESSE. - Dennis P. Walsh, Oct 27 2006
Inverse: With q = -log(log(16)/(pi a(n)^2)), ceiling((q + log(q))/log(16)) = n. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 26 2007
Number of partitions with Ferrers diagrams that fit in an n X n box (including the empty partition of 0). Example: a(2) = 6 because we have: empty, 1, 2, 11, 21 and 22. - Emeric Deutsch, Oct 02 2007
The number of walks of length 2n on an infinite linear lattice that begins and ends at the origin. - Stefan Hollos (stefan(AT)exstrom.com), Dec 10 2007
The number of lattice paths from (0,0) to (n,n) using steps (1,0) and (0,1). - Joerg Arndt, Jul 01 2011
Integral representation: C(2n,n)=1/Pi Integral [(2x)^(2n)/sqrt(1 - x^2),{x,-1, 1}], i.e., C(2n,n)/4^n is the moment of order 2n of the arcsin distribution on the interval (-1,1). - N-E. Fahssi, Jan 02 2008
Straub, Amdeberhan and Moll: "... it is conjectured that there are only finitely many indices n such that C_n is not divisible by any of 3, 5, 7 and 11." - Jonathan Vos Post, Nov 14 2008
Also, in sports, the number of ordered ways for a "Best of 2n-1 Series" to progress. For example, a(2) = 6 means there are six ordered ways for a "best of 3" series to progress. If we write A for a win by "team A" and B for a win by "team B" and if we list the played games chronologically from left to right then the six ways are AA, ABA, BAA, BB, BAB, and ABB. (Proof: To generate the a(n) ordered ways: Write down all a(n) ways to designate n of 2n games as won by team A. Remove the maximal suffix of identical letters from each of these.) - Lee A. Newberg, Jun 02 2009
Number of n X n binary arrays with rows, considered as binary numbers, in nondecreasing order, and columns, considered as binary numbers, in nonincreasing order. - R. H. Hardin, Jun 27 2009
Hankel transform is 2^n. - Paul Barry, Aug 05 2009
It appears that a(n) is also the number of quivers in the mutation class of twisted type BC_n for n>=2.
Number of words on {a,b} of length 2n such that no prefix of the word contains more b's than a's. - Jonathan Nilsson, Apr 18 2012
From Pascal's triangle take row(n) with terms in order a1,a2,..a(n) and row(n+1) with terms b1,b2,..b(n), then 2*(a1*b1 + a2*b2 + ... + a(n)*b(n)) to get the terms in this sequence. - J. M. Bergot, Oct 07 2012. For example using rows 4 and 5: 2*(1*(1) + 4*(5) + 6*(10) + 4*(10) + 1*(5)) = 252, the sixth term in this sequence.
Take from Pascal's triangle row(n) with terms b1, b2, ..., b(n+1) and row(n+2) with terms c1, c2, ..., c(n+3) and find the sum b1*c2 + b2*c3 + ... + b(n+1)*c(n+2) to get A000984(n+1). Example using row(3) and row(5) gives sum 1*(5)+3*(10)+3*(10)+1*(5) = 70 = A000984(4). - J. M. Bergot, Oct 31 2012
a(n) == 2 mod n^3 iff n is a prime > 3. (See Mestrovic link, p. 4.) - Gary Detlefs, Feb 16 2013
Conjecture: For any positive integer n, the polynomial sum_{k=0}^n a(k)x^k is irreducible over the field of rational numbers. In general, for any integer m>1 and n>0, the polynomial f_{m,n}(x) = Sum_{k=0..n} (m*k)!/(k!)^m*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 23 2013
This comment generalizes the comment dated Oct 31 2012 and the second of the sequence's original comments. For j = 1 to n, a(n) = Sum_{k=0..j} C(j,k)* C(2n-j, n-k) = 2*Sum_{k=0..j-1} C(j-1,k)*C(2n-j, n-k). - Charlie Marion, Jun 07 2013
The differences between consecutive terms of the sequence of the quotients between consecutive terms of this sequence form a sequence containing the reciprocals of the triangular numbers. In other words, a(n+1)/a(n)-a(n)/a(n-1) = 2/(n*(n+1)). - Christian Schulz, Jun 08 2013
Number of distinct strings of length 2n using n letters A and n letters B. - Hans Havermann, May 07 2014
Expansion of G.f. A(x) = 1/(1+q*x*c(x)), where parameter q is positive or negative (except q=-1), and c(x) is the g.f. of A000108 for Catalan numbers. The case of q=-1 recovers the g.f. of A000108 as xA^2-A+1=0. The present sequence A000984 refers to q=-2. Recurrence: (1+q)*(n+2)*a(n+2) + ((q*q-4*q-4)*n + 2*(q*q-q-1))*a(n+1) - 2*q*q*(2*n+1)*a(n) = 0, a(0)=1, a(1)=-q. Asymptotics: a(n) ~ ((q+2)/(q+1))*(q^2/(-q-1))^n, q<=-3, a(n) ~ (-1)^n*((q+2)/(q+1))*(q^2/(q+1))^n, q>=5, and a(n) ~ -Kq*2^(2*n)/sqrt(Pi*n^3), where the multiplicative constant Kq is given by K1=1/9 (q=1), K2=1/8 (q=2), K3=3/25 (q=3), K4=1/9 (q=4). These formulas apply to existing sequences A126983 (q=1), A126984 (q=2), A126982 (q=3), A126986 (q=4), A126987 (q=5), A127017 (q=6), A127016 (q=7), A126985 (q=8), A127053 (q=9), and to A007854 (q=-3), A076035 (q=-4), A076036 (q=-5), A127628 (q=-6), A126694 (q=-7), A115970 (q=-8). (End)
a(n)*(2^n)^(j-2) equals S(n), where S(n) is the n-th number in the self-convolved sequence which yields the powers of 2^j for all integers j, n>=0. For example, when n=5 and j=4, a(5)=252; 252*(2^5)^(4-2) = 252*1024 = 258048. The self-convolved sequence which yields powers of 16 is {1, 8, 96, 1280, 17920, 258048, ...}; i.e., S(5) = 258048. Note that the convolved sequences will be composed of numbers decreasing from 1 to 0, when j<2 (exception being j=1, where the first two numbers in the sequence are 1 and all others decreasing). - Bob Selcoe, Jul 16 2014
The variance of the n-th difference of a sequence of pairwise uncorrelated random variables each with variance 1. - Liam Patrick Roche, Jun 04 2015
Number of ordered trees with n edges where vertices at level 1 can be of 2 colors. Indeed, the standard decomposition of ordered trees leading to the equation C = 1 + zC^2 (C is the Catalan function), yields this time G = 1 + 2zCG, from where G = 1/sqrt(1-4z). - Emeric Deutsch, Jun 17 2015
Number of monomials of degree at most n in n variables. - Ran Pan, Sep 26 2015
Let V(n, r) denote the volume of an n-dimensional sphere with radius r, then V(n, 2^n) / Pi = V(n-1, 2^n) * a(n/2) for all even n. - Peter Luschny, Oct 12 2015
a(n) is the number of sets {i1,...,in} of length n such that n >= i1 >= i2 >= ... >= in >= 0. For instance, a(2) = 6 as there are only 6 such sets: (2,2) (2,1) (2,0) (1,1) (1,0) (0,0). - Anton Zakharov, Jul 04 2016
By analytic continuation to the entire complex plane there exist regularized values for divergent sums such as:
Sum_{k>=0} a(k)/(-2)^k = 1/sqrt(3).
Sum_{k>=0} a(k)/(-1)^k = 1/sqrt(5).
Sum_{k>=0} a(k)/(-1/2)^k = 1/3.
Sum_{k>=0} a(k)/(1/2)^k = -1/sqrt(7)i.
Sum_{k>=0} a(k)/(1)^k = -1/sqrt(3)i.
Sum_{k>=0} a(k)/2^k = -i. (End)
Number of sequences (e(1), ..., e(n+1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) > e(j). [Martinez and Savage, 2.18] - Eric M. Schmidt, Jul 17 2017
The o.g.f. for the sequence equals the diagonal of any of the following the rational functions: 1/(1 - (x + y)), 1/(1 - (x + y*z)), 1/(1 - (x + x*y + y*z)) or 1/(1 - (x + y + y*z)). - Peter Bala, Jan 30 2018
Let s denote West's stack-sorting map. a(n) is the number of permutations pi of [n+1] such that s(pi) avoids the patterns 132, 231, and 321. a(n) is also the number of permutations pi of [n+1] such that s(pi) avoids the patterns 132, 312, and 321.
a(n) is the number of permutations of [n+1] that avoid the patterns 1342, 3142, 3412, and 3421. (End)
All binary self-dual codes of length 4n, for n>0, must contain at least a(n) codewords of weight 2n. More to the point, there will always be at least one, perhaps unique, binary self-dual code of length 4n that will contain exactly a(n) codewords that have a hamming weight equal to half the length of the code (2n). This code can be constructed by direct summing the unique binary self-dual code of length 2 (up to permutation equivalence) to itself an even number of times. A permutation equivalent code can be constructed by augmenting two identity matrices of length 2n together. - Nathan J. Russell, Nov 25 2018
Let [b/p] denote the Legendre symbol and 1/b denote the inverse of b mod p. Then, for m and n, where n is not divisible by p,
[(m+n)/p] == [n/p]*Sum_{k=0..(p-1)/2} (-m/(4*n))^k * a(k) (mod p).
Evaluating this identity for m = -1 and n = 1 demonstrates that, for all odd primes p, Sum_{k=0..(p-1)/2} (1/4)^k * a(k) is divisible by p. (End)
Number of vertices of the subgraph of the (2n-1)-dimensional hypercube induced by all bitstrings with n-1 or n many 1s. The middle levels conjecture asserts that this graph has a Hamilton cycle. - Torsten Muetze, Feb 11 2019
a(n) is the number of walks of length 2n from the origin with steps (1,1) and (1,-1) that stay on or above the x-axis. Equivalently, a(n) is the number of walks of length 2n from the origin with steps (1,0) and (0,1) that stay in the first octant. - Alexander Burstein, Dec 24 2019
Number of permutations of length n>0 avoiding the partially ordered pattern (POP) {3>1, 1>2} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is larger than the second element but smaller than the third elements. - Sergey Kitaev, Dec 08 2020
Also the number of integer compositions of 2n+1 with alternating sum 1, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. For example, the a(0) = 1 through a(2) = 6 compositions are:
(1) (2,1) (3,2)
(1,1,1) (1,2,2)
(2,2,1)
(1,1,2,1)
(2,1,1,1)
(1,1,1,1,1)
The following relate to these compositions:
- The unordered version is A000070.
- The alternating sum 0 version is counted by A088218, ranked by A344619.
- Including even indices gives A126869.
- The complement is counted by A202736.
Equivalently, a(n) counts binary numbers with 2n+1 digits and one more 1 than 0's. For example, the a(2) = 6 binary numbers are: 10011, 10101, 10110, 11001, 11010, 11100.
(End)
a(n) is the number of nx2 Young tableaux with a single horizontal wall between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021].
Example for a(2)=6:
3 4 2 4 3 4 3|4 4|3 2|4
1|2, 1|3, 2|1, 1 2, 1 2, 1 3
a(n) is also the number of nx2 Young tableaux with n "walls" between the first and second column.
Example for a(2)=6:
3|4 2|4 4|3 3|4 4|3 4|2
1|2, 1|3, 1|2, 2|1, 2|1, 3|1 (End)
a(n)/4^n is the probability that a fair coin tossed 2n times will come up heads exactly n times and tails exactly n times, or that a random walk with steps of +-1 will return to the starting point after 2n steps (not necessarily for the first time). As n becomes large, this number asymptotically approaches 1/sqrt(n*Pi), using Stirling's approximation for n!.
a(n)/(4^n*(2n-1)) is the probability that a random walk with steps of +-1 will return to the starting point for the first time after 2n steps. The absolute value of the n-th term of A144704 is denominator of this fraction.
Considering all possible random walks of exactly 2n steps with steps of +-1, a(n)/(2n-1) is the number of such walks that return to the starting point for the first time after 2n steps. See the absolute values of A002420 or A284016 for these numbers. For comparison, as mentioned by Stefan Hollos, Dec 10 2007, a(n) is the number of such walks that return to the starting point after 2n steps, but not necessarily for the first time. (End)
Also the size of the shuffle product of two words of length n, such that the union of the two words consist of 2n distinct elements. - Robert C. Lyons, Mar 15 2023
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REFERENCES
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 160.
Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 575, line -3, with a=b=n.
E. Deutsch and L. Shapiro, Seventeen Catalan identities, Bulletin of the Institute of Combinatorics and its Applications, 31, 31-38, 2001.
H. W. Gould, Combinatorial Identities, Morgantown, 1972, (3.66), page 30.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, Second Ed., see Exercise 112.
M. Griffiths, The Backbone of Pascal's Triangle, United Kingdom Mathematics Trust (2008), 3-124.
Leonard Lipshitz and A. van der Poorten. "Rational functions, diagonals, automata and arithmetic." In Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990): 339-358.
J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Dennis E. Davenport, Lara K. Pudwell, Louis W. Shapiro, and Leon C. Woodson, The Boundary of Ordered Trees, Journal of Integer Sequences, Vol. 18 (2015), Article 15.5.8.
L. W. Shapiro, S. Getu, Wen-Jin Woan and L. C. Woodson, The Riordan Group, Discrete Appl. Maths. 34 (1991) 229-239.
M. Wallner, Lattice Path Combinatorics, Diplomarbeit, Institut für Diskrete Mathematik und Geometrie der Technischen Universität Wien, 2013.
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FORMULA
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a(n)/(n+1) = A000108(n), the Catalan numbers.
G.f.: A(x) = (1 - 4*x)^(-1/2) = 1F0(1/2;;4x).
D-finite with recurrence: n*a(n) + 2*(1-2*n)*a(n-1)=0.
a(n) = 2^n/n! * Product_{k=0..n-1} (2*k+1).
Using Stirling's formula in A000142 it is easy to get the asymptotic expression a(n) ~ 4^n / sqrt(Pi * n). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
Integral representation as n-th moment of a positive function on the interval [0, 4]: a(n) = Integral_{x=0..4}(x^n*((x*(4-x))^(-1/2))/Pi), n=0, 1, ... This representation is unique. - Karol A. Penson, Sep 17 2001
a(n) = Max_{ (i+j)!/(i!j!) | 0<=i,j<=n }. - Benoit Cloitre, May 30 2002
E.g.f.: exp(2*x)*I_0(2x), where I_0 is Bessel function. - Michael Somos, Sep 08 2002
E.g.f.: I_0(2*x) = Sum a(n)*x^(2*n)/(2*n)!, where I_0 is Bessel function. - Michael Somos, Sep 09 2002
Determinant of n X n matrix M(i, j) = binomial(n+i, j). - Benoit Cloitre, Aug 28 2003
Given m = C(2*n, n), let f be the inverse function, so that f(m) = n. Letting q denote -log(log(16)/(m^2*Pi)), we have f(m) = ceiling( (q + log(q)) / log(16) ). - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Oct 30 2003
a(n+1) = Sum_{j=n..n*2+1} binomial(j, n). E.g., a(4) = C(7,3) + C(6,3) + C(5,3) + C(4,3) + C(3,3) = 35 + 20 + 10 + 4 + 1 = 70. - Jon Perry, Jan 20 2004
a(n) = (-1)^(n)*Sum_{j=0..(2*n)} (-1)^j*binomial(2*n, j)^2. - Helena Verrill (verrill(AT)math.lsu.edu), Jul 12 2004
a(n) = Sum_{k=0..n} binomial(2n+1, k)*sin((2n-2k+1)*Pi/2). - Paul Barry, Nov 02 2004
a(n-1) = (1/2)*(-1)^n*Sum_{0<=i, j<=n}(-1)^(i+j)*binomial(2n, i+j). - Benoit Cloitre, Jun 18 2005
G.f.: c(x)^2/(2*c(x)-c(x)^2) where c(x) is the g.f. of A000108. - Paul Barry, Feb 03 2006
G.f.: 1/(1-2x-2x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-... (continued fraction);
G.f.: 1/(1-2x/(1-x/(1-x/(1-x/(1-... (continued fraction). (End)
Let A(x) be the g.f. and B(x) = A(-x), then B(x) = sqrt(1-4*x*B(x)^2). - Vladimir Kruchinin, Jan 16 2011
a(n) = (-4)^n*sqrt(Pi)/(gamma((1/2-n))*gamma(1+n)). - Gerry Martens, May 03 2011
a(n) = upper left term in M^n, M = the infinite square production matrix:
2, 2, 0, 0, 0, 0, ...
1, 1, 1, 0, 0, 0, ...
1, 1, 1, 1, 0, 0, ...
1, 1, 1, 1, 1, 0, ...
1, 1, 1, 1, 1, 1, ....
a(n) = Hypergeometric([-n,-n],[1],1). - Peter Luschny, Nov 01 2011
G.f.: 1 + 2*x/(U(0)-2*x) where U(k) = 2*(2*k+1)*x + (k+1) - 2*(k+1)*(2*k+3)*x/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Jun 28 2012
a(n) = Sum_{k=0..n} binomial(n,k)^2*H(k)/(2*H(n)-H(2*n)), n>0, where H(n) is the n-th harmonic number. - Gary Detlefs, Mar 19 2013
G.f.: Q(0)*(1-4*x), where Q(k) = 1 + 4*(2*k+1)*x/( 1 - 1/(1 + 2*(k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 11 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 2*x*(2*k+1)/(2*x*(2*k+1) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
E.g.f.: E(0)/2, where E(k) = 1 + 1/(1 - 2*x/(2*x + (k+1)^2/(2*k+1)/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
Special values of Jacobi polynomials, in Maple notation: a(n) = 4^n*JacobiP(n,0,-1/2-n,-1). - Karol A. Penson, Jul 27 2013
a(n) = 2^(4*n)/((2*n+1)*Sum_{k=0..n} (-1)^k*C(2*n+1,n-k)/(2*k+1)). - Mircea Merca, Nov 12 2013
a(n) = C(2*n-1,n-1)*C(4*n^2,2)/(3*n*C(2*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014
0 = a(n)*(16*a(n+1) - 6*a(n+2)) + a(n+1)*(-2*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 17 2014
G.f.: Sum_{n>=0} x^n/(1-x)^(2*n+1) * Sum_{k=0..n} C(n,k)^2 * x^k. - Paul D. Hanna, Nov 08 2014
a(n) = 4^n*hypergeom([-n,1/2],[1],1). - Peter Luschny, May 19 2015
a(n) = Sum_{k=0..floor(n/2)} C(n,k)*C(n-k,k)*2^(n-2*k). - Robert FERREOL, Aug 29 2015
a(n) ~ 4^n*(2-2/(8*n+2)^2+21/(8*n+2)^4-671/(8*n+2)^6+45081/(8*n+2)^8)/sqrt((4*n+1) *Pi). - Peter Luschny, Oct 14 2015
A(-x) = 1/x * series reversion( x*(2*x + sqrt(1 + 4*x^2)) ). Compare with the o.g.f. B(x) of A098616, which satisfies B(-x) = 1/x * series reversion( x*(2*x + sqrt(1 - 4*x^2)) ). See also A214377. - Peter Bala, Oct 19 2015
Sum_{n>=0} (-1)^n/a(n) = 4*(5 - sqrt(5)*log(phi))/25 = 0.6278364236143983844442267..., where phi is the golden ratio. - Ilya Gutkovskiy, Jul 04 2016
This sequence occurs as the closed-form expression for several binomial sums:
a(n) = Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n,k)*binomial(2*n + 1,k).
a(n) = 2*Sum_{k = 0..2*n-1} (-1)^(n+k)*binomial(2*n - 1,k)*binomial(2*n,k) for n >= 1.
a(n) = 2*Sum_{k = 0..n-1} binomial(n - 1,k)*binomial(n,k) for n >= 1.
a(n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x + k,n)*binomial(y + k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x - k,n)*binomial(y - k,n) for arbitrary x and y.
For m = 3,4,5,... both Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x + k,n)*binomial(y + k,n) and Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x - k,n)*binomial(y - k,n) appear to equal Kronecker's delta(n,0).
a(n) = (-1)^n*Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x + k,n)*binomial(y - k,n) for arbitrary x and y.
For m = 3,4,5,... Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x + k,n)*binomial(y - k,n) appears to equal Kronecker's delta(n,0).
a(n) = Sum_{k = 0..2n} (-1)^k*binomial(2*n,k)*binomial(3*n - k,n)^2 = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)* binomial(n + k,n)^2. (Gould, Vol. 7, 5.23).
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n,n + k)*binomial(n + k,n)^2. (End)
Sum_{k>=0} a(k)/(p/q)^k = sqrt(p/(p-4q)) for q in N, p in Z/{-4q< (some p) <-2}.
...
Sum_{k>=0} a(k)/(-4)^k = 1/sqrt(2).
Sum_{k>=0} a(k)/(17/4)^k = sqrt(17).
Sum_{k>=0} a(k)/(18/4)^k = 3.
Sum_{k>=0} a(k)/5^k = sqrt(5).
Sum_{k>=0} a(k)/6^k = sqrt(3).
Sum_{k>=0} a(k)/8^k = sqrt(2).
...
Sum_{k>=0} a(k)/(p/q)^k = sqrt(p/(p-4q)) for p>4q.(End)
Boas-Buck recurrence: a(n) = (2/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n, 0). See a comment there. - Wolfdieter Lang, Aug 10 2017
a(n) = Sum_{k = 0..n} (-1)^(n-k) * binomial(2*n+1, k) for n in N. - Rene Adad, Sep 30 2017
G.f. for {1/a(n)}: 4*(sqrt(4-x) + sqrt(x)*arcsin(sqrt(x)/2)) / (4-x)^(3/2).
E.g.f. for {1/a(n)}: 1 + exp(x/4)*sqrt(Pi*x)*erf(sqrt(x)/2)/2.
Sum_{n>=0} (-1)^n/a(n) = 4*(1/5 - arcsinh(1/2)/(5*sqrt(5))). (End)
a(n) = 2^(2*n)*Product_{k=1..2*n} k^((-1)^(k+1)) = A056040(2*n).
a(n) = 4^n*binomial(n-1/2,-1/2) = 4^n*GegenbauerC(n,1/4,1). - Gerry Martens, Oct 19 2022
Occurs on the right-hand side of the binomial sum identities Sum_{k = -n..n} (-1)^k * (n + x - k) * binomial(2*n, n+k)^2 = (x + n)*a(n) and Sum_{k = -n..n} (-1)^k * (n + x - k)^2 * binomial(2*n, n+k)^3 = x*(x + 2*n)*a(n) (x arbitrary). Compare with the identity: Sum_{k = -n..n} (-1)^k * binomial(2*n, n+k)^2 = a(n). - Peter Bala, Jul 31 2023
4^n*a(n) = Sum_{k = 0..2*n} (-1)^k*a(k)*a(2*n-k).
16^n = Sum_{k = 0..2*n} a(k)*a(2*n-k). (End)
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EXAMPLE
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G.f.: 1 + 2*x + 6*x^2 + 20*x^3 + 70*x^4 + 252*x^5 + 924*x^6 + ...
For n=2, a(2) = 4!/(2!)^2 = 24/4 = 6, and this is the middle coefficient of the binomial expansion (a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4. - Michael B. Porter, Jul 06 2016
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MAPLE
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with(combstruct); [seq(count([S, {S=Prod(Set(Z, card=i), Set(Z, card=i))}, labeled], size=(2*i)), i=0..20)];
with(combstruct); [seq(count([S, {S=Sequence(Union(Arch, Arch)), Arch=Prod(Epsilon, Sequence(Arch), Z)}, unlabeled], size=i), i=0..25)];
with(combstruct):bin := {B=Union(Z, Prod(B, B))}: seq (count([B, bin, unlabeled], size=n)*n, n=1..25); # Zerinvary Lajos, Dec 05 2007
A000984List := proc(m) local A, P, n; A := [1, 2]; P := [1];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), 2*P[-1]]);
A := [op(A), 2*P[-1]] od; A end: A000984List(28); # Peter Luschny, Mar 24 2022
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MATHEMATICA
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CoefficientList[Series[1/Sqrt[1-4x], {x, 0, 25}], x] (* Harvey P. Dale, Mar 14 2011 *)
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PROG
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(Magma) a:= func< n | Binomial(2*n, n) >; [ a(n) : n in [0..10]];
(PARI) A000984(n)=binomial(2*n, n) \\ much more efficient than (2n)!/n!^2. \\ M. F. Hasler, Feb 26 2014
(PARI) fv(n, p)=my(s); while(n\=p, s+=n); s
(PARI) fv(n, p)=my(s); while(n\=p, s+=n); s
(Haskell)
(Python)
from __future__ import division
for n in range(10**3):
b = b*(4*n+2)//(n+1)
(GAP) List([1..1000], n -> Binomial(2*n, n)); # Muniru A Asiru, Jan 30 2018
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CROSSREFS
|
Bisection of A001405 and of A226302. See also A025565, the same ordered partitions but without all in which are two successive zeros: 11110 (5), 11200 (18), 13000 (2), 40000 (0) and 22000 (1), total 26 and A025565(4)=26.
See A261009 for a conjecture about this sequence.
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.
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KEYWORD
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nonn,easy,core,nice,walk,frac
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AUTHOR
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STATUS
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approved
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A001499
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Number of n X n matrices with exactly 2 1's in each row and column, other entries 0.
(Formerly M4286 N1792)
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+10
22
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1, 0, 1, 6, 90, 2040, 67950, 3110940, 187530840, 14398171200, 1371785398200, 158815387962000, 21959547410077200, 3574340599104475200, 676508133623135814000, 147320988741542099484000, 36574751938491748341360000, 10268902998771351157327104000
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OFFSET
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0,4
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COMMENTS
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Or, number of labeled 2-regular relations of order n.
Also number of ways to arrange 2n rooks on an n X n chessboard, with no more than 2 rooks in each row and column (no 3 in a line). - Vaclav Kotesovec, Aug 03 2013
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REFERENCES
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R. Bricard, L'Intermédiaire des Mathématiciens, 8 (1901), 312-313.
L. Comtet, Advanced Combinatorics, Reidel, 1974, Sect. 6.3 Multipermutations, pp. 235-236, P(n,2), bipermutations.
L. Erlebach and O. Ruehr, Problem 79-5, SIAM Review. Solution by D. E. Knuth. Reprinted in Problems in Applied Mathematics, ed. M. Klamkin, SIAM, 1990, p. 350.
Shanzhen Gao and Kenneth Matheis, Closed formulas and integer sequences arising from the enumeration of (0,1)-matrices with row sum two and some constant column sums. In Proceedings of the Forty-First Southeastern International Conference on Combinatorics, Graph Theory and Computing. Congr. Numer. 202 (2010), 45-53.
J. T. Lewis, Maximal L-free subsets of a squarefree array, Congressus Numerantium, 141 (1999), 151-155.
R. W. Robinson, Numerical implementation of graph counting algorithms, AGRC Grant, Math. Dept., Univ. Newcastle, Australia, 1976.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Cor. 5.5.11 (b).
M. L. Stein and P. R. Stein, Enumeration of Stochastic Matrices with Integer Elements. Report LA-4434, Los Alamos Scientific Laboratory of the University of California, Los Alamos, NM, Jun 1970.
J. H. van Lint and R. M. Wilson, A Course in Combinatorics (Cambridge University Press, Cambridge, 1992), pp. 152-153. [The second edition is said to be a better reference.]
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LINKS
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Sally Cockburn and Joshua Lesperance, Deranged Socks, Mathematics Magazine, Vol. 86, no. 2, April 2013, pp. 97-109.
L. Erlebach and O. Ruehr, Problem 79-5, SIAM Review, Vol. 21, No. 1 (Jan., 1979), p. 140. Solution by D. E. Knuth, SIAM Review, Vol. 22, No. 1 (Jan., 1980), pp. 101-102.
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FORMULA
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a(n) = (n! (n-1) Gamma(n-1/2) / Gamma(1/2) ) * 1F1[2-n; 3/2-n; -1/2] [Erlebach and Ruehr]. This representation is exact, asymptotic and convergent.
D-finite with recurrence 2*a(n) -2*n*(n-1)*a(n-1) -n*(n-1)^2*a(n-2)=0.
a(n) ~ 2 sqrt(Pi) n^(2n + 1/2) e^(-2n - 1/2) [Knuth]
a(n) = (1/2)*n*(n-1)^2 * ( (2*n-3)*a(n-2) + (n-2)^2*a(n-3) ) (from Anand et al.)
Sum_{n >= 0} a(n)*x^n/(n!)^2 = exp(-x/2)/sqrt(1-x); a(n) = n(n-1)/2 [ 2 a(n-1) + (n-1) a(n-2) ] (Bricard)
b_n = a_n/n! satisfies b_n = (n-1)(b_{n-1} + b_{n-2}/2); e.g.f. for {b_n} and for derangements (A000166) are related by D(x) = B(x)^2.
a(n) = 4^(-n) * n!^2 * Sum_{i=0..n} (-2)^i * (2*n - 2*i)! / (i!*(n-i)!^2). - Shanzhen Gao, Feb 15 2010
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MATHEMATICA
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a[n_] := (n-1)*n!*Gamma[n-1/2]*Hypergeometric1F1[2-n, 3/2-n, -1/2]/Sqrt[Pi]; Table[a[n], {n, 0, 17}] (* Jean-François Alcover, Oct 06 2011, after first formula *)
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PROG
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(PARI) a(n)=if(n<2, n==0, (n^2-n)*(a(n-1)+(n-1)/2*a(n-2)))
(PARI) seq(n)={Vec(serlaplace(serlaplace(exp(-x/2 + O(x*x^n))/sqrt(1-x + O(x*x^n)))))}; \\ Andrew Howroyd, Sep 09 2018
(Haskell)
a001499 n = a001499_list !! n
a001499_list = 1 : 0 : 1 : zipWith (*) (drop 2 a002411_list)
(zipWith (+) (zipWith (*) [3, 5 ..] $ tail a001499_list)
(zipWith (*) (tail a000290_list) a001499_list))
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CROSSREFS
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KEYWORD
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nonn,nice,easy
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AUTHOR
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STATUS
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approved
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A094527
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Triangle T(n,k), read by rows, defined by T(n,k) = binomial(2*n,n-k).
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+10
15
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1, 2, 1, 6, 4, 1, 20, 15, 6, 1, 70, 56, 28, 8, 1, 252, 210, 120, 45, 10, 1, 924, 792, 495, 220, 66, 12, 1, 3432, 3003, 2002, 1001, 364, 91, 14, 1, 12870, 11440, 8008, 4368, 1820, 560, 120, 16, 1, 48620, 43758, 31824, 18564, 8568, 3060, 816, 153, 18, 1, 184756, 167960
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OFFSET
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0,2
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COMMENTS
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Right-hand side of even-numbered rows of Pascal's triangle.
Triangle T(n,k), 0 <= k <= n, read by rows defined by: T(0,0)=1, T(n,k)=0 if k < 0 or if k > n, T(n,0) = 2*T(n-1,0) + 2*T(n-1,1), T(n,k) = T(n-1,k-1) + 2*T(n-1,k) + T(n-1,k+1) for k >= 1. - Philippe Deléham, Mar 14 2007
Central coefficients T(2n,n) are binomial(4n,n) (A005810).
The A- and Z-sequence for this Riordan triangle is [1,2,1] and [2,2], respectively. For the notion of Z- and A-sequences for Riordan arrays see the W. Lang link under A006232 with details and references. See also the Philippe Deléham comment above. - Wolfdieter Lang, Nov 22 2012
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LINKS
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FORMULA
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Riordan array (1/sqrt(1-4x)), (1-2x-sqrt(1-4x))/(2x)). Column k has e.g.f. exp(2x)Bessel_I(k, 2x). - Paul Barry, Jul 14 2005
Product of Riordan arrays (1/(1-x), x/(1-x)) (Pascal's triangle, A007318) and (1/sqrt(1-2x-3x^2), (1-x-sqrt(1-2x-3x^2))/(2x)) (A094531). Inverse is A110162. - Paul Barry, Jul 14 2005
T(n,k) = Sum_{j=0..n} C(n,j)*C(n,j-k). - Paul Barry, Mar 07 2006
The o.g.f. for the row polynomials P(n,x) := Sum_{k=0..n} T(n,k)*x^k is G(z,x) = (-x + (1+x)*z + x*z*c(z))/(sqrt(1-4*z)*((1+x)^2*z -x)) with c the o.g.f. of A000108 (Catalan). This follows from the Riordan property.
The o.g.f. for column no. k is (c(x)-1)^k/sqrt(1-4*x) (from the Riordan property). (End)
Riordan array has the form ( x*h'(x)/h(x), h(x) ) with h(x) = ( 1 - 2*x - sqrt(1 - 4*x) )/(2*x) and so belongs to the hitting time subgroup of the Riordan group (see Peart and Woan, Example 5.1).
T(n,k) = [x^(n-k)] f(x)^n with f(x) = (1 + x)^2. In general the (n,k)th entry of the hitting time array ( x*h'(x)/h(x), h(x) ) has the form [x^(n-k)] f(x)^n, where f(x) = x/( series reversion of h(x) ). (End)
n-th row polynomial R(n,t) = [x^n] ( (1 + (1 + t)*x)^2/(1 + t*x) )^n.
exp ( Sum_{n >= 1} R(n,t)*x^n/n ) = 1 + (2 + t)*x + (5 + 4*t + t^2)*x^2 + ... is the o.g.f. for A039598. (End)
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EXAMPLE
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The triangle T(n,k) begins:
n\k 0 1 2 3 4 5 6 7 8 9 10
0: 1
1: 2 1
2: 6 4 1
3: 20 15 6 1
4: 70 56 28 8 1
5: 252 210 120 45 10 1
6: 924 792 495 220 66 12 1
7: 3432 3003 2002 1001 364 91 14 1
8: 12870 11440 8008 4368 1820 560 120 16 1
9: 48620 43758 31824 18564 8568 3060 816 153 18 1
10: 184756 167960 125970 77520 38760 15504 4845 1140 190 20 1
Production array is
2, 1,
2, 2, 1,
0, 1, 2, 1,
0, 0, 1, 2, 1,
0, 0, 0, 1, 2, 1,
0, 0, 0, 0, 1, 2, 1,
0, 0, 0, 0, 0, 1, 2, 1 (End)
Recurrence from the Riordan A-sequence [1,2,1]: T(4,1) = 56 = 1*T(3,0) + 2*T(3,1) + 1*T(3,2) = 1*20 + 2*15 + 1*6.
Recurrence from the Riordan Z-sequence [2,2]: T(7,0) = 3432 = 2*T(6,0) + 2*T(6,1) = 2*924 + 2*792. See the Philippe Deléham comment above. (End)
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MAPLE
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binomial(2*n, n-k) ;
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MATHEMATICA
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T[n_, k_] := Binomial[2*n, n - k];
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A138034
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Expansion of (1+3*x^2)/(1-x+x^2).
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+10
11
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1, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3
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OFFSET
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0,3
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COMMENTS
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LINKS
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FORMULA
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G.f.: (1+3*x^2)/(1-x+x^2). a(n)=a(n-1)-a(n-2), n>2.
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MATHEMATICA
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CoefficientList[Series[(1 + 3*x^2)/(1 - x + x^2), {x, 0, 100}], x] (* Wesley Ivan Hurt, Jan 15 2017 *)
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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Karem Boubaker (mmbb11112000(AT)yahoo.fr), Mar 01 2008; corrected Mar 03 2008
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STATUS
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approved
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A162547
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Somos-4 variant: if n!=4k+1, then a(n) = (4*a(n-1)*a(n-3) - 4*a(n-2)^2) / a(n-4), otherwise a(n) = 0, with a(-2) = a(-1) = a(0) = 1.
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+10
5
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1, 0, -4, -16, -64, 0, 4096, 65536, 1048576, 0, -1073741824, -68719476736, -4398046511104, 0, 72057594037927936, 18446744073709551616, 4722366482869645213696, 0, -1237940039285380274899124224, -1267650600228229401496703205376, -1298074214633706907132624082305024, 0
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OFFSET
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0,3
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COMMENTS
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LINKS
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FORMULA
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a(n) = a(-2 - n), a(n) * a(n+5) = 4 * a(n+1) * a(n+4) for all n in Z. - Michael Somos, Jul 20 2014
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MATHEMATICA
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a[ n_] := With[{m = n + 1, k = Quotient[n + 2, 4]}, Boole[Mod[m, 4] != 2] (-1)^k 4^(k (m - 2 k))]; (* Michael Somos, Jun 26 2017 *)
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PROG
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(PARI) {a(n) = my(m=n+1, k=(n+2)\4); (m%4!=2) * (-1)^k * 4^(k*(m - 2*k))}; /* Michael Somos, Jul 20 2014 */
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CROSSREFS
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KEYWORD
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easy,sign
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AUTHOR
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STATUS
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approved
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A300192
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Triangle read by rows: row n consists of the coefficients of the expansion of the polynomial (x^2 + 2*x + 1)^n + (x^2 - 1)*(x + 1)^n.
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+10
5
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0, 0, 1, 0, 1, 2, 1, 0, 2, 6, 6, 2, 0, 3, 13, 22, 18, 7, 1, 0, 4, 23, 56, 75, 60, 29, 8, 1, 0, 5, 36, 115, 215, 261, 215, 121, 45, 10, 1, 0, 6, 52, 206, 495, 806, 938, 798, 496, 220, 66, 12, 1, 0, 7, 71, 336, 987, 2016, 3031, 3452, 3010, 2003, 1001, 364, 91
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OFFSET
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0,6
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REFERENCES
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M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972.
J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996.
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LINKS
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Milan Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013.
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FORMULA
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T(n,k) = binomial(2*n,k) + binomial(n,k-2) - binomial(n,k).
T(n,k) = T(n-1,k-1)+ T(n-1,k) + A034871(n-1,k-1), with T(n,0) = T(0,1) = 0 and T(0,2) = 1
T(n,n+k) = binomial(2*n, n-k) = A094527(n,k), for k >= 3 and n>=k.
G.f.: 1/(1 - y*(x^2 + 2*x + 1)) + (x^2 - 1)/(1 - y*(x + 1)).
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EXAMPLE
|
The triangle T(n, k) begins:
n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
0: 0 0 1
1: 0 1 2 1
2: 0 2 6 6 2
3: 0 3 13 22 18 7 1
4: 0 4 23 56 75 60 29 8 1
5: 0 5 36 115 215 261 215 121 45 10 1
6: 0 6 52 206 495 806 938 798 496 220 66 12 1
7: 0 7 71 336 987 2016 3031 3452 3010 2003 1001 364 91 14 1
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MAPLE
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T := (n, k) -> binomial(2*n, k) + binomial(n, k - 2) - binomial(n, k);
for n from 0 to 10 do seq(T(n, k), k = 0 .. max(2*n, n + 2)) od;
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PROG
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(Maxima)
T(n, k) := binomial(2*n, k) + binomial(n, k - 2) - binomial(n, k)$
a : []$
for n:0 thru 10 do
a : append(a, makelist(T(n, k), k, 0, max(2*n, n + 2)))$
a;
(PARI) row(n) = Vecrev((x^2 + 2*x + 1)^n + (x^2 - 1)*(x + 1)^n); \\ Michel Marcus, Nov 12 2022
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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A101500
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A Chebyshev transform of the central binomial numbers.
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+10
3
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1, 2, 5, 16, 53, 178, 609, 2112, 7393, 26066, 92437, 329360, 1178149, 4228322, 15218305, 54907136, 198527617, 719170850, 2609577701, 9483269008, 34508808789, 125727351186, 458573578977, 1674270763584, 6118472289889, 22378379004146, 81913223571701
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OFFSET
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0,2
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COMMENTS
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A Chebyshev transform of A000984. Under the Chebyshev transform, we map a g.f. g(x) to (1/(1+x^2))g(x/(1+x^2).
Also equal to the Riordan array (1/(1-x)^2,x/(1-x)^2) applied to aerated central binomial coefficients (with g.f. 1/sqrt(1-4x^2)). - Paul Barry, Jul 06 2009
Directed 2-D walks with n steps starting at (0,0) and ending on the X-axis using steps N,S,E,W and avoiding N followed by S. - David Scambler, Jun 24 2013
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LINKS
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FORMULA
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G.f.: 1/(sqrt(1+x^2)*sqrt(1-4*x+x^2)).
a(n) = Sum_{k=0..floor(n/2)} C(n-k, k)*(-1)^k*C(2(n-2k), n-2k).
From Paul Barry, Jul 06 2009: (Start)
G.f.: 1/((1-x)^2-2*x^2/((1-x)^2-x^2/((1-x)^2-x^2/((1-x)^2-... (continued fraction);
a(n) = Sum_{k=0..n} C(n+k+1,n-k)*C(k,k/2)*(1+(-1)^k)/2. (End)
Conjecture: n*a(n) +2*(-2*n+1)*a(n-1) +2*(n-1)*a(n-2) +2*(-2*n+3)*a(n-3) +(n-2)*a(n-4)=0. - R. J. Mathar, Nov 16 2012
a(n) ~ sqrt(1/2 + 7/(8*sqrt(3))) * (2+sqrt(3))^n / sqrt(Pi*n). - Vaclav Kotesovec, Feb 08 2014
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MATHEMATICA
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CoefficientList[Series[1/Sqrt[(1+x^2)*(1-4*x+x^2)], {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 08 2014 *)
Table[1/2^n* Sum[(-1)^k*Binomial[2 k, k]* Sum[Binomial[n - 2 k, j]^2*3^j, {j, 0, n - 2 k}], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Jun 30 2018 *)
Table[Sum[Binomial[n - k, k]*(-1)^k*Binomial[2 (n - 2 k), n - 2 k], {k, 0, Floor[n/2]}], {n, 0, 20}] (* Vaclav Kotesovec, Jun 30 2018 *)
a[ n_] := Sum[Binomial[n + k + 1, 2k + 1] Binomial[k, Quotient[k, 2]], {k, 0, n, 2}]; (* Michael Somos, Jun 30 2018 *)
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PROG
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(PARI) A101500(maxx)={n=0; while(n<=maxx, z=sum(k=0, floor(n/2), binomial(n-k, k)*binomial(2*(n-2*k), n-2*k)*(-1)^k ) ; print1(z, ", "); n+=1); } \\ Bill McEachen, Jan 02 2016
(PARI) x='x+O('x^40); Vec(1/(sqrt(1+x^2)*sqrt(1-4*x+x^2))); \\ Michel Marcus, Jan 03 2016
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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