A046521 -id:A046521 - cp's OEIS frontend

A007318 Pascal's triangle read by rows: C(n,k) = binomial(n,k) = n!/(k!*(n-k)!), 0 <= k <= n.

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

Offset: 0

N. J. A. Sloane and Mira Bernstein, Apr 28 1994

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
Inverse of A130595 (as an infinite lower triangular matrix). - Philippe Deléham, Aug 21 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)
From Milan Janjic, May 07 2008: (Start)
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
Central binomial coefficients: T(2*n,n) = A000984(n), T(n, floor(n/2)) = A001405(n). - Reinhard Zumkeller, Nov 09 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
See A193242. - Alexander R. Povolotsky, Feb 05 2013
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
From Daniel Forgues, Mar 12 2015: (Start)
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)
A258993(n,k) = T(n+k,n-k), n > 0. - Reinhard Zumkeller, Jun 22 2015
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.
Satisfies Benford's law [Diaconis, 1977] - N. J. A. Sloane, Feb 09 2017
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 k-th 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)). - 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

			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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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.
Infinite matrix squared: A038207, cubed: A027465.
Cf. A101164. If rows are sorted we get A061554 or A107430.
Another version: A108044.
Cf. A008277, A132311, A132312, A052216, A052217, A052218, A052219, A052220, A052221, A052222, A052223, A144225, A202750, A211226, A047999, A026729, A052553, A051920, A193242.
Triangle sums (see the comments): A000079 (Row1); A000007 (Row2); A000045 (Kn11 & Kn21); A000071 (Kn12 & Kn22); A001924 (Kn13 & Kn23); A014162 (Kn14 & Kn24); A014166 (Kn15 & Kn25); A053739 (Kn16 & Kn26); A053295 (Kn17 & Kn27); A053296 (Kn18 & Kn28); A053308 (Kn19 & Kn29); A053309 (Kn110 & Kn210); A001519 (Kn3 & Kn4); A011782 (Fi1 & Fi2); A000930 (Ca1 & Ca2); A052544 (Ca3 & Ca4); A003269 (Gi1 & Gi2); A055988 (Gi3 & Gi4); A034943 (Ze1 & Ze2); A005251 (Ze3 & Ze4). - Johannes W. Meijer, Sep 22 2010
Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A111006, A114197, A162741, A228074, A228196, A228576.
Cf. A137948, A245334.
Cf. A085478, A258993.
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.

-- (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)

Flat(List([0..12],n->List([0..n],k->Binomial(n,k)))); # Stefano Spezia, Dec 22 2018

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
-- Reinhard Zumkeller, Nov 09 2011, Oct 22 2010

/* As triangle: */ [[Binomial(n, k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jul 29 2015

Maple
```
A007318 := (n,k)->binomial(n,k);
```

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 *)

create_list(binomial(n,k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */

C(n,k)=binomial(n,k) \\ Charles R Greathouse IV, Jun 08 2011

# 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

from math import comb, isqrt
def A007318(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),n-comb(r+1,2)) # Chai Wah Wu, Nov 11 2024

def C(n,k): return Subsets(range(n), k).cardinality() # Ralf Stephan, Jan 21 2014

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, nC(n, k) = n!/(k!(n-k)!) if 0<=k<=n, otherwise 0.
C(n, k) = ((n-k+1)/k) * C(n, k-1) with C(n, 0) = 1. - Michael B. Porter, Mar 23 2025
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
C(n, k) = Sum_{j=0..k} (-1)^j*C(n+1+j, k-j)*A000108(j). - Philippe Deléham, Oct 10 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(n,k) <= A062758(n) for n > 1. - Reinhard Zumkeller, Mar 04 2008
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
From Paul D. Hanna, Mar 24 2011: (Start)
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)
For n > 0: T(n,k) = A029600(n,k) - A029635(n,k), 0 <= k <= n. - Reinhard Zumkeller, Apr 16 2012
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
From L. Edson Jeffery, Aug 02 2012: (Start)
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
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 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
Sum_{n>=k} 1/C(n,k) = k/(k-1) for k>=1. - Richard R. Forberg, Feb 10 2014
From Tom Copeland, Apr 26 2014: (Start)
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)
   with dP = A132440, M = A238385-I, and I = identity matrix, and
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)
T(n,k) = A245334(n,k) / A137948(n,k), 0 <= k <= n. - Reinhard Zumkeller, Aug 31 2014
From Peter Bala, Dec 21 2014: (Start)
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
C(n,k-1) = A325002(n,k) - [k==n+1] = (A325002(n,k) + A325003(n,k)) / 2 = [k==n+1] + A325003(n,k). - Robert A. Russell, Oct 20 2020
From Hermann Stamm-Wilbrandt, May 13 2021: (Start)
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)
C(n,k) = Sum_{i=0..k} A000108(i) * C(n-2i-1, k-i), for 0 <= k <= floor(n/2)-1. - Tushar Bansal, May 17 2025

Checked all links, deleted 8 that seemed lost forever and were probably not of great importance. - N. J. A. Sloane, May 08 2018

A000984 Central binomial coefficients: binomial(2n,n) = (2n)!/(n!)^2.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

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)) * zeta(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
So this is the 2-dimensional analog of A008793. - William Entriken, Aug 06 2013
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
Also the Catalan transform of A000079. - R. J. Mathar, Nov 06 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
Equals INVERT transform of A081696: (1, 1, 3, 9, 29, 97, 333, ...). - Gary W. Adamson, May 15 2009
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.
Central terms of Pascal's triangle: a(n) = A007318(2*n,n). - Reinhard Zumkeller, Nov 09 2011
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
From Fung Lam, May 19 2014: (Start)
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
From Ralf Steiner, Apr 07 2017: (Start)
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
From Colin Defant, Sep 16 2018: (Start)
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
From Isaac Saffold, Dec 28 2018: (Start)
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
From Gus Wiseman, Jul 21 2021: (Start)
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 -1 version is counted by A001791, ranked by A345910/A345912.
- The alternating sum 0 version is counted by A088218, ranked by A344619.
- Including even indices gives A126869.
- The complement is counted by A202736.
- Ranked by A345909 (reverse: A345911).
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)
From Michael Wallner, Jan 25 2022: (Start)
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)
From Shel Kaphan, Jan 12 2023: (Start)
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)
p divides a((p-1)/2) + 1 for primes p of the form 4*k+3 (A002145). - Jules Beauchamp, Feb 11 2023
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
a(n) is the number of vertices of the n-dimensional cyclohedron W_{n+1}. - Jose Bastidas, Mar 25 2025
Consider a stack of pancakes of height n, where the only allowed operation is reversing the top portion of the stack. First, perform a series of reversals of increasing sizes, followed by a series of reversals of decreasing sizes. The number of distinct permutations of the initial stack that can be reached through these operations is a(n). - Thomas Baruchel, May 12 2025

			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

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.
Arthur T. Benjamin and Jennifer 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.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 101.
Emeric Deutsch and Louis W. Shapiro, Seventeen Catalan identities, Bulletin of the Institute of Combinatorics and its Applications, 31 (2001), 31-38.
Henry W. Gould, Combinatorial Identities, Morgantown, 1972, (3.66), page 30.
Ronald. L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, Second Ed., see Exercise 112.
Martin Griffiths, The Backbone of Pascal's Triangle, United Kingdom Mathematics Trust (2008), 3-124.
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Index entries for "core" sequences.

Cf. A000108, A002420, A002457, A030662, A002144, A135091, A081696, A182400. Differs from A071976 at 10th term.
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.
Cf. A226078, A051924 (first differences).
Row sums of A059481, A008459, A152229, A158815, A205946.
Cf. A258290 (arithmetic derivative). Cf. A098616, A214377.
See A261009 for a conjecture about this sequence.
Cf. A046521 (first column).
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.
Cf. A000346, A001700, A001791, A008549, A032443, A073016, A097805, A126869.
Cf. A003418, A056040, A001316, A261130, A356637.

List([1..1000], n -> Binomial(2*n,n)); # Muniru A Asiru, Jan 30 2018

a000984 n = a007318_row (2*n) !! n  -- Reinhard Zumkeller, Nov 09 2011

a:= func< n | Binomial(2*n,n) >; [ a(n) : n in [0..10]];

A000984 := n-> binomial(2*n,n); seq(A000984(n), n=0..30);
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

Table[Binomial[2n, n], {n, 0, 24}] (* Alonso del Arte, Nov 10 2005 *)
CoefficientList[Series[1/Sqrt[1-4x],{x,0,25}],x]  (* Harvey P. Dale, Mar 14 2011 *)

A000984(n):=(2*n)!/(n!)^2$ makelist(A000984(n),n,0,30); /* Martin Ettl, Oct 22 2012 */

A000984(n)=binomial(2*n,n) \\ much more efficient than (2n)!/n!^2. \\ M. F. Hasler, Feb 26 2014

fv(n,p)=my(s);while(n\=p,s+=n);s
a(n)=prodeuler(p=2,2*n,p^(fv(2*n,p)-2*fv(n,p))) \\ Charles R Greathouse IV, Aug 21 2013

fv(n,p)=my(s);while(n\=p,s+=n);s
a(n)=my(s=1);forprime(p=2,2*n,s*=p^(fv(2*n,p)-2*fv(n,p)));s \\ Charles R Greathouse IV, Aug 21 2013

from _future_ import division
A000984_list, b = [1], 1
for n in range(10**3):
    b = b*(4*n+2)//(n+1)
    A000984_list.append(b) # Chai Wah Wu, Mar 04 2016

a(n)/(n+1) = A000108(n), the Catalan numbers.
G.f.: A(x) = (1 - 4*x)^(-1/2) = 1F0(1/2;;4x).
a(n+1) = 2*A001700(n) = A030662(n) + 1. a(2*n) = A001448(n), a(2*n+1) = 2*A002458(n) =A099976.
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).
a(n) = a(n-1)*(4-2/n) = Product_{k=1..n} (4-2/k) = 4*a(n-1) + A002420(n) = A000142(2*n)/(A000142(n)^2) = A001813(n)/A000142(n) = sqrt(A002894(n)) = A010050(n)/A001044(n) = (n+1)*A000108(n) = -A005408(n-1)*A002420(n). - Henry Bottomley, Nov 10 2000
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
Sum_{n>=1} 1/a(n) = (2*Pi*sqrt(3) + 9)/27. [Lehmer 1985, eq. (15)] - Benoit Cloitre, May 01 2002 (= A073016. - Bernard Schott, Jul 20 2022)
a(n) = Max_{ (i+j)!/(i!j!) | 0<=i,j<=n }. - Benoit Cloitre, May 30 2002
a(n) = Sum_{k=0..n} binomial(n+k-1,k), row sums of A059481. - Vladeta Jovovic, Aug 28 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
a(n) = Sum_{k=0..n} binomial(n, k)^2. - Benoit Cloitre, Jan 31 2003
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) = 2*Sum_{k=0..(n-1)} a(k)*a(n-k+1)/(k+1). - Philippe Deléham, Jan 01 2004
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
a(n) = C(2n, n-1) + C(n) = A001791(n) + A000108(n). - Lekraj Beedassy, Aug 02 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
a(n) = A006480(n) / A005809(n). - Zerinvary Lajos, Jun 28 2007
a(n) = Sum_{k=0..n} A106566(n,k)*2^k. - Philippe Deléham, Aug 25 2007
a(n) = Sum_{k>=0} A039599(n, k). a(n) = Sum_{k>=0} A050165(n, k). a(n) = Sum_{k>=0} A059365(n, k)*2^k, n>0. a(n+1) = Sum_{k>=0} A009766(n, k)*2^(n-k+1). - Philippe Deléham, Jan 01 2004
a(n) = 4^n*Sum_{k=0..n} C(n,k)(-4)^(-k)*A000108(n+k). - Paul Barry, Oct 18 2007
a(n) = Sum_{k=0..n} A039598(n,k)*A059841(k). - Philippe Deléham, Nov 12 2008
A007814(a(n)) = A000120(n). - Vladimir Shevelev, Jul 20 2009
From Paul Barry, Aug 05 2009: (Start)
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)
If n>=3 is prime, then a(n) == 2 (mod 2*n). - Vladimir Shevelev, Sep 05 2010
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, .... - Gary W. Adamson, Jul 14 2011
a(n) = Hypergeometric([-n,-n],[1],1). - Peter Luschny, Nov 01 2011
E.g.f.: hypergeometric([1/2],[1],4*x). - Wolfdieter Lang, Jan 13 2012
a(n) = 2*Sum_{k=0..n-1} a(k)*A000108(n-k-1). - Alzhekeyev Ascar M, Mar 09 2012
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
Sum_{n>=0} a(n)/n! = A234846. - Richard R. Forberg, Feb 10 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
a(n+1) = 4*a(n) - 2*A000108(n). Also a(n) = 4^n*Product_{k=1..n}(1-1/(2*k)). - Stanislav Sykora, Aug 09 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*binomial(-1/2,n). - Jean-François Alcover, Feb 10 2015
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
a(n) = GegenbauerC(n,-n,-1). - Peter Luschny, May 07 2016
a(n) = gamma(1+2*n)/gamma(1+n)^2. - Andres Cicuttin, May 30 2016
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
From Peter Bala, Jul 22 2016: (Start)
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)
From Ralf Steiner, Apr 07 2017: (Start)
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
a(n) = A034870(n,n). - Franck Maminirina Ramaharo, Nov 26 2018
From Jianing Song, Apr 10 2022: (Start)
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)
From Peter Luschny, Sep 08 2022: (Start)
a(n) = 2^(2*n)*Product_{k=1..2*n} k^((-1)^(k+1)) = A056040(2*n).
a(n) = A001316(n) * A356637(n) * A261130(n) for n >= 2. (End)
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
From Peter Bala, Mar 31 2024: (Start)
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)
From Gary Detlefs, May 28 2024: (Start)
a(n) = Sum_{k=0..floor(n/2)} binomial(n,2k)*binomial(2*k,k)*2^(n-2*k). (H. W. Gould) - Gary Detlefs, May 28 2024
a(n) = Sum_{k=0..2*n} (-1)^k*binomial(2n,k)*binomial(2*n+2*k,n+k)*3^(2*n-k). (H. W. Gould) (End)
a(n) = Product_{k>=n+1} k^2/(k^2 - n^2). - Antonio Graciá Llorente, Sep 08 2024
a(n) = Product_{k=1..n} A003418(floor(2*n/k))^((-1)^(k+1)) (Golomb, 2003). - Amiram Eldar, Aug 08 2025

A049310 Triangle of coefficients of Chebyshev's S(n,x) := U(n,x/2) polynomials (exponents in increasing order).

Original entry on oeis.org

1, 0, 1, -1, 0, 1, 0, -2, 0, 1, 1, 0, -3, 0, 1, 0, 3, 0, -4, 0, 1, -1, 0, 6, 0, -5, 0, 1, 0, -4, 0, 10, 0, -6, 0, 1, 1, 0, -10, 0, 15, 0, -7, 0, 1, 0, 5, 0, -20, 0, 21, 0, -8, 0, 1, -1, 0, 15, 0, -35, 0, 28, 0, -9, 0, 1, 0, -6, 0, 35, 0, -56, 0, 36, 0, -10, 0, 1, 1, 0, -21, 0, 70, 0, -84, 0

Offset: 0

Wolfdieter Lang

G.f. for row polynomials S(n,x) (signed triangle): 1/(1-x*z+z^2). Unsigned triangle |a(n,m)| has Fibonacci polynomials F(n+1,x) as row polynomials with g.f. 1/(1-x*z-z^2). |a(n,m)| triangle has rows of Pascal's triangle A007318 in the even-numbered diagonals (odd-numbered ones have only 0's).
Row sums (unsigned triangle) A000045(n+1) (Fibonacci). Row sums (signed triangle) S(n,1) sequence = periodic(1,1,0,-1,-1,0) = A010892.
Alternating row sums A049347(n) = S(n,-1) = periodic(1,-1,0). - Wolfdieter Lang, Nov 04 2011
S(n,x) is the characteristic polynomial of the adjacency matrix of the n-path. - Michael Somos, Jun 24 2002
S(n,x) is also the matching polynomial of the n-path. - Eric W. Weisstein, Apr 10 2017
|T(n,k)| = number of compositions of n+1 into k+1 odd parts. Example: |T(7,3)| = 10 because we have (1,1,3,3), (1,3,1,3), (1,3,3,1), (3,1,1,3), (3,1,3,1), (3,3,1,1), (1,1,1,5), (1,1,5,1), (1,5,1,1) and (5,1,1,1). - Emeric Deutsch, Apr 09 2005
S(n,x)= R(n,x) + S(n-2,x), n >= 2, S(-1,x)=0, S(0,x)=1, R(n,x):=2*T(n,x/2) = Sum_{m=0..n} A127672(n,m)*x^m (monic integer Chebyshev T-Polynomials). This is the rewritten so-called trace of the transfer matrix formula for the T-polynomials. - Wolfdieter Lang, Dec 02 2010
In a regular N-gon inscribed in a unit circle, the side length is d(N,1) = 2*sin(Pi/N). The length ratio R(N,k):=d(N,k)/d(N,1) for the (k-1)-th diagonal, with k from {2,3,...,floor(N/2)}, N >= 4, equals S(k-1,x) = sin(k*Pi/N)/sin(Pi/N) with x=rho(N):=R(N,2) = 2*cos(Pi/N). Example: N=7 (heptagon), rho=R(7,2), sigma:=R(N,3) = S(2,rho) = rho^2 - 1. Motivated by the quoted paper by P. Steinbach. - Wolfdieter Lang, Dec 02 2010
From Wolfdieter Lang, Jul 12 2011: (Start)
In q- or basic analysis, q-numbers are [n]_q := S(n-1,q+1/q) = (q^n-(1/q)^n)/(q-1/q), with the row polynomials S(n,x), n >= 0.
The zeros of the row polynomials S(n-1,x) are (from those of Chebyshev U-polynomials):
  x(n-1;k) = +- t(k,rho(n)), k = 1..ceiling((n-1)/2), n >= 2, with t(n,x) the row polynomials of A127672 and rho(n):= 2*cos(Pi/n). The simple vanishing zero for even n appears here as +0 and -0.
Factorization of the row polynomials S(n-1,x), x >= 1, in terms of the minimal polynomials of cos(2 Pi/2), called Psi(n,x), with coefficients given by A181875/A181876:
  S(n-1,x) = (2^(n-1))*Product_{n>=1}(Psi(d,x/2), 2 < d | 2n).
(From the rewritten eq. (3) of the Watkins and Zeitlin reference, given under A181872.) [See the W. Lang ArXiv link, Proposition 9, eq. (62). - Wolfdieter Lang, Apr 14 2018]
  (End)
The discriminants of the S(n,x) polynomials are found in A127670. - Wolfdieter Lang, Aug 03 2011
This is an example for a subclass of Riordan convolution arrays (lower triangular matrices) called Bell arrays. See the L. W. Shapiro et al. reference under A007318. If a Riordan array is named (G(z),F(z)) with F(z)=z*Fhat(z), the o.g.f. for the row polynomials is G(z)/(1-x*z*Fhat(z)), and it becomes a Bell array if G(z)=Fhat(z). For the present Bell type triangle G(z)=1/(1+z^2) (see the o.g.f. comment above). This leads to the o.g.f. for the column no. k, k >= 0, x^k/(1+x^2)^(k+1) (see the formula section), the one for the row sums and for the alternating row sums (see comments above). The Riordan (Bell) A- and Z-sequences (defined in a W. Lang link under A006232, with references) have o.g.f.s 1-x*c(x^2) and -x*c(x^2), with the o.g.f. of the Catalan numbers A000108. Together they lead to a recurrence given in the formula section. - Wolfdieter Lang, Nov 04 2011
The determinant of the N x N matrix S(N,[x[1], ..., x[N]]) with elements S(m-1,x[n]), for n, m = 1, 2, ..., N, and for any x[n], is identical with the determinant of V(N,[x[1], ..., x[N]]) with elements x[n]^(m-1) (a Vandermondian, which equals Product_{1 <= i < j<= N} (x[j] - x[i])). This is a special instance of a theorem valid for any N >= 1 and any monic polynomial system p(m,x), m>=0, with p(0,x) = 1. For this theorem see the Vein-Dale reference, p. 59. Thanks to L. Edson Jeffery for an email asking for a proof of the non-singularity of the matrix S(N,[x[1], ...., x[N]]) if and only if the x[j], j = 1..N, are pairwise distinct. - Wolfdieter Lang, Aug 26 2013
These S polynomials also appear in the context of modular forms. The rescaled Hecke operator T*n = n^((1-k)/2)*T_n acting on modular forms of weight k satisfies T*(p^n) = S(n, T*p), for each prime p and positive integer n. See the Koecher-Krieg reference, p. 223. - _Wolfdieter Lang, Jan 22 2016
For a shifted o.g.f. (mod signs), its compositional inverse, and connections to Motzkin and Fibonacci polynomials, non-crossing partitions and other combinatorial structures, see A097610. - Tom Copeland, Jan 23 2016
From M. Sinan Kul, Jan 30 2016; edited by Wolfdieter Lang, Jan 31 2016 and Feb 01 2016: (Start)
Solutions of the Diophantine equation u^2 + v^2 - k*u*v = 1 for integer k given by (u(k,n), v(k,n)) = (S(n,k), S(n-1,k)) because of the Cassini-Simson identity: S(n,x)^2 - S(n+1,x)*S(n-1, x) = 1, after use of the S-recurrence. Note that S(-n, x) = -S(-n-2, x), n >= 1, and the periodicity of some S(n, k) sequences.
Hence another way to obtain the row polynomials would be to take powers of the matrix [x, -1; 1,0]: S(n, x) = (([x, -1; 1, 0])^n)[1,1], n >= 0.
See also a Feb 01 2016 comment on A115139 for a well-known S(n, x) sum formula.
Then we have with the present T triangle
  A039834(n) = -i^(n+1)*T(n-1, k) where i is the imaginary unit and n >= 0.
  A051286(n) = Sum_{i=0..n} T(n,i)^2 (see the Philippe Deléham, Nov 21 2005 formula),
  A181545(n) = Sum_{i=0..n+1} abs(T(n,i)^3),
  A181546(n) = Sum_{i=0..n+1} T(n,i)^4,
  A181547(n) = Sum_{i=0..n+1} abs(T(n,i)^5).
S(n, 0) = A056594(n), and for k = 1..10 the sequences S(n-1, k) with offset n = 0 are A128834, A001477, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189.
(End)
For more on the Diophantine equation presented by Kul, see the Ismail paper. - Tom Copeland, Jan 31 2016
The o.g.f. for the Legendre polynomials L(n,x) is 1 / sqrt(1- 2x*z + z^2), and squaring it gives the o.g.f. of U(n,x), A053117, so Sum_{k=0..n} L(k,x/2) L(n-k,x/2) = S(n,x). This gives S(n,x) = L(n/2,x/2)^2 + 2*Sum_{k=0..n/2-1} L(k,x/2) L(n-k,x/2) for n even and S(n,x) = 2*Sum_{k=0..(n-1)/2} L(k,x/2) L(n-k,x/2) for odd n. For a connection to elliptic curves and modular forms, see A053117. For the normalized Legendre polynomials, see A100258. For other properties and relations to other polynomials, see Allouche et al. - Tom Copeland, Feb 04 2016
LG(x,h1,h2) = -log(1 - h1*x + h2*x^2) = Sum_{n>0} F(n,-h1,h2,0,..,0) x^n/n is a log series generator of the bivariate row polynomials of A127672 with A127672(0,0) = 0 and where F(n,b1,b2,..,bn) are the Faber polynomials of A263916. Exp(LG(x,h1,h2)) = 1 / (1 - h1*x + h2*x^2 ) is the o.g.f. of the bivariate row polynomials of this entry. - Tom Copeland, Feb 15 2016 (Instances of the bivariate o.g.f. for this entry are on pp. 5 and 18 of Sunada. - Tom Copeland, Jan 18 2021)
For distinct odd primes p and q the Legendre symbol can be written as Legendre(q,p) = Product_{k=1..P} S(q-1, 2*cos(2*Pi*k/p)), with P = (p-1)/2. See the Lemmermeyer reference, eq. (8.1) on p. 236. Using the zeros of S(q-1, x) (see above) one has S(q-1, x) = Product_{l=1..Q} (x^2 - (2*cos(Pi*l/q))^2), with Q = (q-1)/2. Thus S(q-1, 2*cos(2*Pi*k/p)) = ((-4)^Q)*Product_{l=1..Q} (sin^2(2*Pi*k/p) - sin^2(Pi*l/q)) = ((-4)^Q)*Product_{m=1..Q} (sin^2(2*Pi*k/p) - sin^2(2*Pi*m/q)). For the proof of the last equality see a W. Lang comment on the triangle A057059 for n = Q and an obvious function f. This leads to Eisenstein's proof of the quadratic reciprocity law Legendre(q,p) = ((-1)^(P*Q)) * Legendre(p,q), See the Lemmermeyer reference, pp. 236-237. - Wolfdieter Lang, Aug 28 2016
For connections to generalized Fibonacci polynomials, compare their generating function on p. 5 of the Amdeberhan et al. link with the o.g.f. given above for the bivariate row polynomials of this entry. - Tom Copeland, Jan 08 2017
The formula for Ramanujan's tau function (see A000594) for prime powers is tau(p^k) = p^(11*k/2)*S(k, p^(-11/2)*tau(p)) for k >= 1, and p = A000040(n), n >= 1. See the Hardy reference, p. 164, eqs. (10.3.4) and (10.3.6) rewritten in terms of S. - Wolfdieter Lang, Jan 27 2017
From Wolfdieter Lang, May 08 2017: (Start)
The number of zeros Z(n) of the S(n, x) polynomials in the open interval (-1,+1) is 2*b(n) for even n >= 0 and 1 + 2*b(n) for odd n >= 1, where b(n) = floor(n/2) - floor((n+1)/3). This b(n) is the number of integers k in the interval (n+1)/3 < k <= floor(n/2). See a comment on the zeros of S(n, x) above, and b(n) = A008615(n-2), n >= 0. The numbers Z(n) have been proposed (with a conjecture related to A008611) by Michel Lagneau, as the number of zeros of Fibonacci polynomials on the imaginary axis (-I,+I), with I=sqrt(-1). They are Z(n) = A008611(n-1), n >= 0, with A008611(-1) = 0. Also Z(n) = A194960(n-4), n >= 0. Proof using the A008611 version. A194960 follows from this.
In general the number of zeros Z(a;n) of S(n, x) for n >= 0 in the open interval (-a,+a) for a from the interval (0,2) (x >= 2 never has zeros, and a=0 is trivial: Z(0;n) = 0) is with b(a;n) = floor(n//2) - floor((n+1)*arccos(a/2)/Pi), as above Z(a;n) = 2*b(a;n) for even n >= 0 and 1 + 2*b(a;n) for odd n >= 1. For the closed interval [-a,+a] Z(0;n) = 1 and for a from (0,1) one uses for Z(a;n) the values b(a;n) = floor(n/2) - ceiling((n+1)*arccos(a/2)/Pi) + 1. (End)
The Riordan row polynomials S(n, x) (Chebyshev S) belong to the Boas-Buck class (see a comment and references in A046521), hence they satisfy the Boas-Buck identity: (E_x - n*1)*S(n, x) = (E_x + 1)*Sum_{p=0..n-1} (1 - (-1)^p)*(-1)^((p+1)/2)*S(n-1-p, x), for n >= 0, where E_x = x*d/dx (Euler operator). For the triangle T(n, k) this entails a recurrence for the sequence of column k, given in the formula section. - Wolfdieter Lang, Aug 11 2017
The e.g.f. E(x,t) := Sum_{n>=0} (t^n/n!)*S(n,x) for the row polynomials is obtained via inverse Laplace transformation from the above given o.g.f. as E(x,t) = ((1/xm)*exp(t/xm) - (1/xp)*exp(t/xp) )/(xp - xm) with xp = (x + sqrt(x^2-4))/2 and xm = (x - sqrt(x^2-4))/2. - Wolfdieter Lang, Nov 08 2017
From Wolfdieter Lang, Apr 12 2018: (Start)
Factorization of row polynomials S(n, x), for n >= 1, in terms of C polynomials (not Chebyshev C) with coefficients given in A187360. This is obtained from the factorization into Psi polynomials (see the Jul 12 2011 comment above) but written in terms of minimal polynomials of 2*cos(2*Pi/n) with coefficients in A232624:
S(2*k, x) = Product_{2 <= d | (2*k+1)} C(d, x)*(-1)^deg(d)*C(d, -x), with deg(d) = A055034(d) the degree of C(d, x).
S(2*k+1, x) = Product_{2 <= d | 2*(k+1)} C(d, x) * Product_{3 <= 2*d + 1 | (k+1)} (-1)^(deg(2*d+1))*C(2*d+1, -x).
Note that (-1)^(deg(2*d+1))*C(2*d+1, -x)*C(2*d+1, x) pairs always appear.
The number of C factors of S(2*k, x), for k >= 0, is 2*(tau(2*k+1) - 1) = 2*(A099774(k+1) - 1) = 2*A095374(k), and for S(2*k+1, x), for k >= 0, it is tau(2*(k+1)) + tau_{odd}(k+1) - 2 = A302707(k), with tau(2*k+1) = A099774(k+1), tau(n) = A000005 and tau(2*(k+1)) = A099777(k+1).
For the reverse problem, the factorization of C polynomials into S polynomials, see A255237. (End)
The S polynomials with general initial conditions S(a,b;n,x) = x*S(a,b;n-1,x) - S(a,b;n-2,x), for n >= 1, with S(a,b;-1,x) = a and S(a,b;0,x) = b are S(a,b;n,x) = b*S(n, x) - a*S(n-1, x), for n >= -1. Recall that S(-2, x) = -1 and S(-1, x) = 0. The o.g.f. is G(a,b;z,x) = (b - a*z)/(1 - x*z + z^2). - Wolfdieter Lang, Oct 18 2019
Also the convolution triangle of A101455. - Peter Luschny, Oct 06 2022
From Wolfdieter Lang, Apr 26 2023: (Start)
Multi-section of S-polynomials: S(m*n+k, x) = S(m+k, x)*S(n-1, R(m, x)) - S(k, x)*S(n-2, R(m, x)), with R(n, x) = S(n, x) - S(n-2, x) (see A127672), S(-2, x) = -1, and S(-1, x) = 0, for n >= 0, m >= 1, and k = 0, 1, ..., m-1.
O.g.f. of {S(m*n+k, y)}_{n>=0}: G(m,k,y,x) = (S(k, y) - (S(k, y)*R(m, y) - S(m+k, y))*x)/(1 - R(m,y)*x + x^2).
See eqs. (40) and (49), with r = x or y and s =-1, of the G. Detlefs and W. Lang link at A034807. (End)
S(n, x) for complex n and complex x: S(n, x) = ((-i/2)/sqrt(1 - (x/2)^2))*(q(x/2)*exp(+n*log(q(x/2))) - (1/q(x/2))*exp(-n*log(q(x/2)))), with q(x) = x + sqrt(1 - x^2)*i. Here log(z) = |z| + Arg(z)*i, with Arg(z) from [-Pi,+Pi) (principal branch). This satisfies the recurrence relation for S because it is derived from the Binet - de Moivre formula for S. Examples: S(n/m, 0) = cos((n/m)*Pi/4), for n >= 0 and m >= 1. S(n*i, 0) = (1/2)*(1 + exp(n*Pi))*exp(-(n/2)*Pi), for n >= 0. S(1+i, 2+i) = 0.6397424847... + 1.0355669490...*i. Thanks to Roberto Alfano for asking a question leading to this formula. - Wolfdieter Lang, Jun 05 2023
Lim_{n->oo} S(n, x)/S(n-1, x) = r(x) = (x - sqrt(x^2 -4))/2, for |x| >= 2. For x = +-2, this limit is +-1. - Wolfdieter Lang, Nov 15 2023

			The triangle T(n, k) begins:
  n\k  0  1   2   3   4   5   6    7   8   9  10  11
  0:   1
  1:   0  1
  2:  -1  0   1
  3:   0 -2   0   1
  4:   1  0  -3   0   1
  5:   0  3   0  -4   0   1
  6:  -1  0   6   0  -5   0   1
  7:   0 -4   0  10   0  -6   0    1
  8:   1  0 -10   0  15   0  -7    0   1
  9:   0  5   0 -20   0  21   0   -8   0   1
  10: -1  0  15   0 -35   0  28    0  -9   0   1
  11:  0 -6   0  35   0 -56   0   36   0 -10   0   1
  ... Reformatted and extended by _Wolfdieter Lang_, Oct 24 2012
For more rows see the link.
E.g., fourth row {0,-2,0,1} corresponds to polynomial S(3,x)= -2*x + x^3.
From _Wolfdieter Lang_, Jul 12 2011: (Start)
Zeros of S(3,x) with rho(4)= 2*cos(Pi/4) = sqrt(2):
  +- t(1,sqrt(2)) = +- sqrt(2) and
  +- t(2,sqrt(2)) = +- 0.
Factorization of S(3,x) in terms of Psi polynomials:
S(3,x) = (2^3)*Psi(4,x/2)*Psi(8,x/2) = x*(x^2-2).
(End)
From _Wolfdieter Lang_, Nov 04 2011: (Start)
A- and Z- sequence recurrence:
T(4,0) = - (C(0)*T(3,1) + C(1)*T(3,3)) = -(-2 + 1) = +1,
T(5,3) = -3 - 1*1 = -4.
(End)
Boas-Buck recurrence for column k = 2, n = 6: S(6, 2) = (3/4)*(0 - 2* S(4 ,2) + 0 + 2*S(2, 2)) = (3/4)*(-2*(-3) + 2) = 6. - _Wolfdieter Lang_, Aug 11 2017
From _Wolfdieter Lang_, Apr 12 2018: (Start)
Factorization into C polynomials (see the Apr 12 2018 comment):
S(4, x) = 1 - 3*x^2 + x^4 = (-1 + x + x^2)*(-1 - x + x^2) = (-C(5, -x)) * C(5, x); the number of factors is 2 = 2*A095374(2).
S(5, x) = 3*x - 4*x^3 + x^5 = x*(-1 + x)*(1 + x)*(-3 + x^2) = C(2, x)*C(3, x)*(-C(3, -x))*C(6, x); the number of factors is 4 = A302707(2). (End)

G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, AMS Chelsea Publishing, Providence, Rhode Island, 2002, p. 164.
Max Koecher and Aloys Krieg, Elliptische Funktionen und Modulformen, 2. Auflage, Springer, 2007, p. 223.
Franz Lemmermeyer, Reciprocity Laws. From Euler to Eisenstein, Springer, 2000.
D. S. Mitrinovic, Analytic Inequalities, Springer-Verlag, 1970; p. 232, Sect. 3.3.38.
Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990, pp. 60 - 61.
R. Vein and P. Dale, Determinants and Their Applications in Mathematical Physics, Springer, 1999.

T. D. Noe, Rows 0 to 100 of the triangle, flattened.
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], Table 22.8, p.797.
J. Allouche and G. Skordev, Schur congruences, Carlitz sequences of polynomials and automaticity, Discrete Mathematics, Vol. 214, Issue 1-3, 21 March 2000, p.21-49.
T. Amdeberhan, X. Chen, V. Moll, and B. Sagan, Generalized Fibonacci polynomials and Fibonomial coefficients, arXiv preprint arXiv:1306.6511 [math.CO], 2013.
Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
Paul Barry, On the inversion of Riordan arrays, arXiv:2101.06713 [math.CO], 2021.
Paul Barry and A. Hennessy, Meixner-Type Results for Riordan Arrays and Associated Integer Sequences, J. Int. Seq. 13 (2010) # 10.9.4, section 5.
C. Beck, Spatio-temporal Chaos and Vacuum Fluctuations of Quantized Fields , arXiv preprint arXiv:0207081 [hep-th], 2002.
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group, arXiv:2112.11595 [math.CO], 2021.
Tom Copeland, Addendum to Elliptic Lie Triad
Jerry Ray Dias, Properties and relationships of conjugated polyenes having a reciprocal eigenvalue spectrum - dendralene and radialene hydrocarbons, Croatica Chem. Acta, 77 (2004), 325-330. [p. 328].
S. R. Finch, P. Sebah and Z.-Q. Bai, Odd Entries in Pascal's Trinomial Triangle, arXiv:0802.2654 [math.NT], 2008.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
M. Ismail, One parameter generalizations of the Fibonacci and Lucas numbers, arXiv preprint arXiv:0606743v1 [math.CA], 2006.
Wolfdieter Lang, First rows of the triangle.
Wolfdieter Lang, Chebyshev S-polynomials: ten applications.
Wolfdieter Lang, The field Q(2cos(pi/n)), its Galois group and length ratios in the regular n-gon, arXiv:1210.1018 [math.GR], 2012-2017.
R. Sazdanovic, A categorification of the polynomial ring, slide presentation, 2011. [From _Tom Copeland_, Dec 27 2015]
P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 22-31.
T. Sunada, Discrete Geometric Analysis, 2008.
Eric Weisstein's World of Mathematics, Adjacency Matrix, Characteristic Polynomial, Chebyshev Polynomial of the Second Kind, Fibonacci Polynomial, Matching Polynomial, and Path Graph
Index entries for sequences related to Chebyshev polynomials.
Index entries for "core" sequences

Cf. A000005, A000217, A000292, A000332, A000389, A001227, A007318, A008611, A008615, A101455, A010892, A011973, A053112 (without zeros), A053117, A053119 (reflection), A053121 (inverse triangle), A055034, A097610, A099774, A099777, A100258, A112552 (first column clipped), A127672, A168561 (absolute values), A187360. A194960, A232624, A255237.

Triangles of coefficients of Chebyshev's S(n,x+k) for k = 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5: A207824, A207823, A125662, A078812, A101950, A049310, A104562, A053122, A207815, A159764, A123967.

A049310:= func< n,k | ((n+k) mod 2) eq 0 select (-1)^(Floor((n+k)/2)+k)*Binomial(Floor((n+k)/2), k) else 0 >;
[A049310(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 25 2022

A049310 := proc(n,k): binomial((n+k)/2,(n-k)/2)*cos(Pi*(n-k)/2)*(1+(-1)^(n-k))/2 end: seq(seq(A049310(n,k), k=0..n),n=0..11); # Johannes W. Meijer, Aug 08 2011
# Uses function PMatrix from A357368. Adds a row above and a column to the left.
PMatrix(10, n -> ifelse(irem(n, 2) = 0, 0, (-1)^iquo(n-1, 2))); # Peter Luschny, Oct 06 2022

t[n_, k_] /; EvenQ[n+k] = ((-1)^((n+k)/2+k))*Binomial[(n+k)/2, k]; t[n_, k_] /; OddQ[n+k] = 0; Flatten[Table[t[n, k], {n, 0, 12}, {k, 0, n}]][[;; 86]] (* Jean-François Alcover, Jul 05 2011 *)
Table[Coefficient[(-I)^n Fibonacci[n + 1, - I x], x, k], {n, 0, 10}, {k, 0, n}] //Flatten (* Clark Kimberling, Aug 02 2011; corrected by Eric W. Weisstein, Apr 06 2017 *)
CoefficientList[ChebyshevU[Range[0, 10], -x/2], x] // Flatten (* Eric W. Weisstein, Apr 06 2017 *)
CoefficientList[Table[(-I)^n Fibonacci[n + 1, -I x], {n, 0, 10}], x] // Flatten (* Eric W. Weisstein, Apr 06 2017 *)

{T(n, k) = if( k<0 || k>n || (n + k)%2, 0, (-1)^((n + k)/2 + k) * binomial((n + k)/2, k))} /* Michael Somos, Jun 24 2002 */

@CachedFunction
def A049310(n,k):
    if n< 0: return 0
    if n==0: return 1 if k == 0 else 0
    return A049310(n-1,k-1) - A049310(n-2,k)
for n in (0..9): [A049310(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

T(n,k) := 0 if n < k or n+k odd, otherwise ((-1)^((n+k)/2+k))*binomial((n+k)/2, k); T(n, k) = -T(n-2, k)+T(n-1, k-1), T(n, -1) := 0 =: T(-1, k), T(0, 0)=1, T(n, k)= 0 if n < k or n+k odd; g.f. k-th column: (1 / (1 + x^2)^(k + 1)) * x^k. - Michael Somos, Jun 24 2002
T(n,k) = binomial((n+k)/2, (n-k)/2)*cos(Pi*(n-k)/2)*(1+(-1)^(n-k))/2. - Paul Barry, Aug 28 2005
Sum_{k=0..n} T(n,k)^2 = A051286(n). - Philippe Deléham, Nov 21 2005
Recurrence for the (unsigned) Fibonacci polynomials: F(1)=1, F(2)=x; for n > 2, F(n) = x*F(n-1) + F(n-2).
From Wolfdieter Lang, Nov 04 2011: (Start)
The Riordan A- and Z-sequences, given in a comment above, lead together to the recurrence:
T(n,k) = 0 if n < k, if k=0 then T(0,0)=1 and
  T(n,0)= -Sum_{i=0..floor((n-1)/2)} C(i)*T(n-1,2*i+1), otherwise T(n,k) = T(n-1,k-1) - Sum_{i=1..floor((n-k)/2)} C(i)*T(n-1,k-1+2*i), with the Catalan numbers C(n)=A000108(n).
(End)
The row polynomials satisfy also S(n,x) = 2*(T(n+2, x/2) - T(n, x/2))/(x^2-4) with the Chebyshev T-polynomials. Proof: Use the trace formula 2*T(n, x/2) = S(n, x) - S(n-2, x) (see the Dec 02 2010 comment above) and the S-recurrence several times. This is a formula which expresses the S- in terms of the T-polynomials. - Wolfdieter Lang, Aug 07 2014
From Tom Copeland, Dec 06 2015: (Start)
The non-vanishing, unsigned subdiagonals Diag_(2n) contain the elements D(n,k) = Sum_{j=0..k} D(n-1,j) = (k+1) (k+2) ... (k+n) / n! = binomial(n+k,n), so the o.g.f. for the subdiagonal is (1-x)^(-(n+1)). E.g., Diag_4 contains D(2,3) = D(1,0) + D(1,1) + D(1,2) + D(1,3) = 1 + 2 + 3 + 4 = 10 = binomial(5,2). Diag_4 is shifted A000217; Diag_6, shifted A000292: Diag_8, shifted A000332; and Diag_10, A000389.
The non-vanishing antidiagonals are signed rows of the Pascal triangle A007318.
For a reversed, unsigned version with the zeros removed, see A011973. (End)
The Boas-Buck recurrence (see a comment above) for the sequence of column k is: S(n, k) = ((k+1)/(n-k))*Sum_{p=0..n-1-k} (1 - (-1)^p)*(-1)^((p+1)/2) * S(n-1-p, k), for n > k >= 0 and input S(k, k) = 1. - Wolfdieter Lang, Aug 11 2017
The m-th row consecutive nonzero entries in order are (-1)^c*(c+b)!/c!b! with c = m/2, m/2-1, ..., 0 and b = m-2c if m is even and with c = (m-1)/2, (m-1)/2-1, ..., 0 with b = m-2c if m is odd. For the 8th row starting at a(36) the 5 consecutive nonzero entries in order are 1,-10,15,-7,1 given by c = 4,3,2,1,0 and b = 0,2,4,6,8. - Richard Turk, Aug 20 2017
O.g.f.: exp( Sum_{n >= 0} 2*T(n,x/2)*t^n/n ) = 1 + x*t + (-1 + x^2)*t^2 + (-2*x + x^3)*t^3 + (1 - 3*x^2 + x^4)*t^4 + ..., where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Aug 15 2022

A048994 Triangle of Stirling numbers of first kind, s(n,k), n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, -1, 1, 0, 2, -3, 1, 0, -6, 11, -6, 1, 0, 24, -50, 35, -10, 1, 0, -120, 274, -225, 85, -15, 1, 0, 720, -1764, 1624, -735, 175, -21, 1, 0, -5040, 13068, -13132, 6769, -1960, 322, -28, 1, 0, 40320, -109584, 118124, -67284, 22449, -4536, 546, -36, 1, 0, -362880, 1026576, -1172700, 723680, -269325, 63273, -9450, 870, -45, 1

Offset: 0

N. J. A. Sloane

The unsigned numbers are also called Stirling cycle numbers: |s(n,k)| = number of permutations of n objects with exactly k cycles.
Mirror image of the triangle A054654. - Philippe Deléham, Dec 30 2006
Also the triangle gives coefficients T(n,k) of x^k in the expansion of C(x,n) = (a(k)*x^k)/n!. - Mokhtar Mohamed, Dec 04 2012
From Wolfdieter Lang, Nov 14 2018: (Start)
This is the Sheffer triangle of Jabotinsky type (1, log(1 + x)). See the e.g.f. of the triangle below.
This is the inverse Sheffer triangle of the Stirling2 Sheffer triangle A008275.
The a-sequence of this Sheffer triangle (see a W. Lang link in A006232)
is from the e.g.f. A(x) = x/(exp(x) -1) a(n) = Bernoulli(n) = A027641(n)/A027642(n), for n >= 0. The z-sequence vanishes.
The Boas-Buck sequence for the recurrences of columns has o.g.f. B(x) = Sum_{n>=0} b(n)*x^n = 1/((1 + x)*log(1 + x)) - 1/x. b(n) = (-1)^(n+1)*A002208(n+1)/A002209(n+1), b = {-1/2, 5/12, -3/8, 251/720, -95/288, 19087/60480,...}. For the Boas-Buck recurrence of Riordan and Sheffer triangles see the Aug 10 2017 remark in A046521, adapted to the Sheffer case, also for two references. See the recurrence and example below. (End)
Let G(n,m,k) be the number of simple labeled graphs on [n] with m edges and k components. Then T(n,k) = Sum (-1)^m*G(n,m,k). See the Read link below. Equivalently, T(n,k) = Sum mu(0,p) where the sum is over all set partitions p of [n] containing k blocks and mu is the Moebius function in the incidence algebra associated to the set partition lattice on [n]. - Geoffrey Critzer, May 11 2024

			Triangle begins:
  n\k 0     1       2       3      4      5      6    7    8   9 ...
  0   1
  1   0     1
  2   0    -1       1
  3   0     2      -3       1
  4   0    -6      11      -6      1
  5   0    24     -50      35    -10      1
  6   0  -120     274    -225     85    -15      1
  7   0   720   -1764    1624   -735    175    -21    1
  8   0 -5040   13068  -13132   6769  -1960    322  -28    1
  9   0 40320 -109584  118124 -67284  22449  -4536  546  -36   1
  ... - _Wolfdieter Lang_, Aug 22 2012
------------------------------------------------------------------
From _Wolfdieter Lang_, Nov 14 2018: (Start)
Recurrence: s(5,2)= s(4, 1) - 4*s(4, 2) = -6 - 4*11 = -50.
Recurrence from the a- and z-sequences: s(6, 3) = 2*(1*1*(-50) + 3*(-1/2)*35 + 6*(1/6)*(-10) + 10*0*1) = -225.
Boas-Buck recurrence for column k = 3, with b = {-1/2, 5/12, -3/8, ...}:
s(6, 3) = 6!*((-3/8)*1/3! + (5/12)*(-6)/4! + (-1/2)*35/5!) = -225. (End)

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 833.
L. Comtet, Advanced Combinatorics, Reidel, 1974; Chapter V, also p. 310.
J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 93.
F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 226.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 245.
J. Riordan, An Introduction to Combinatorial Analysis, p. 48.

T. D. Noe, Rows n=0..100 of triangle, flattened
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].
Paul Barry, Generalized Stirling Numbers, Exponential Riordan Arrays, and Toda Chain Equations, Journal of Integer Sequences, 17 (2014), #14.2.3.
Fatima Zohra Bensaci, Rachid Boumahdi, and Laala Khaldi, Finite Sums Involving Fibonacci and Lucas Numbers, J. Int. Seq. (2024). See p. 9.
R. M. Dickau, Stirling numbers of the first kind. [Illustrates the unsigned Stirling cycle numbers A132393.]
Askar Dzhumadil'daev and Damir Yeliussizov, Walks, partitions, and normal ordering, Electronic Journal of Combinatorics, 22(4) (2015), #P4.10.
Bill Gosper, Colored illustrations of triangle of Stirling numbers of first kind read mod 2, 3, 4, 5, 6, 7.
Gergő Nemes, An Asymptotic Expansion for the Bernoulli Numbers of the Second Kind, J. Int. Seq. 14 (2011), #11.4.8.
A. Hennessy and Paul Barry, Generalized Stirling Numbers, Exponential Riordan Arrays, and Orthogonal Polynomials, J. Int. Seq. 14 (2011), #11.8.2 (A-number typo A048894).
NIST Digital Library of Mathematical Functions, Stirling Numbers
Ken Ono, Larry Rolen, and Florian Sprung, Zeta-Polynomials for modular form periods, p. 6, arXiv:1602.00752 [math.NT], 2016.
Ricardo A. Podestá, New identities for binary Krawtchouk polynomials, binomial coefficients and Catalan numbers, arXiv preprint arXiv:1603.09156 [math.CO], 2016.
Ronald Read, An Introduction to Chromatic Polynomials, Journal of Combinatorial Theory, 4(1968)52-71.
Wikipedia, Stirling numbers and exponential generating functions.

See especially A008275 which is the main entry for this triangle. A132393 is an unsigned version, and A008276 is another version.
Cf. A008277, A039814-A039817, A048993, A084938.
A000142(n) = Sum_{k=0..n} |s(n, k)| for n >= 0.
Row sums give A019590(n+1).
Cf. A002209, A027641/A027642, A216294, A271705.

a048994 n k = a048994_tabl !! n !! k
a048994_row n = a048994_tabl !! n
a048994_tabl = map fst $ iterate (\(row, i) ->
(zipWith (-) ([0] ++ row) $ map (* i) (row ++ [0]), i + 1)) ([1], 0)
-- Reinhard Zumkeller, Mar 18 2013

A048994:= proc(n,k) combinat[stirling1](n,k) end: # R. J. Mathar, Feb 23 2009
seq(print(seq(coeff(expand(k!*binomial(x,k)),x,i),i=0..k)),k=0..9); # Peter Luschny, Jul 13 2009
A048994_row := proc(n) local k; seq(coeff(expand(pochhammer(x-n+1,n)), x,k), k=0..n) end: # Peter Luschny, Dec 30 2010

Table[StirlingS1[n, m], {n, 0, 9}, {m, 0, n}] (* Peter Luschny, Dec 30 2010 *)

create_list(stirling1(n,k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */

a(n,k) = if(k<0 || k>n,0, if(n==0,1,(n-1)*a(n-1,k)+a(n-1,k-1)))

trg(nn)=for (n=0, nn-1, for (k=0, n, print1(stirling(n,k,1), ", ");); print();); \\ Michel Marcus, Jan 19 2015

s(n, k) = A008275(n,k) for n >= 1, k = 1..n; column k = 0 is {1, repeat(0)}.
s(n, k) = s(n-1, k-1) - (n-1)*s(n-1, k), n, k >= 1; s(n, 0) = s(0, k) = 0; s(0, 0) = 1.
The unsigned numbers a(n, k)=|s(n, k)| satisfy a(n, k)=a(n-1, k-1)+(n-1)*a(n-1, k), n, k >= 1; a(n, 0) = a(0, k) = 0; a(0, 0) = 1.
Triangle (signed) = [0, -1, -1, -2, -2, -3, -3, -4, -4, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...]; Triangle(unsigned) = [0, 1, 1, 2, 2, 3, 3, 4, 4, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...]; where DELTA is Deléham's operator defined in A084938.
Sum_{k=0..n} (-m)^(n-k)*s(n, k) = A000142(n), A001147(n), A007559(n), A007696(n), ... for m = 1, 2, 3, 4, ... .- Philippe Deléham, Oct 29 2005
A008275*A007318 as infinite lower triangular matrices. - Gerald McGarvey, Aug 20 2009
T(n,k) = n!*[x^k]([t^n]exp(x*log(1+t))). - Peter Luschny, Dec 30 2010, updated Jun 07 2020
From Wolfdieter Lang, Nov 14 2018: (Start)
Recurrence from the Sheffer a-sequence (see a comment above): s(n, k) = (n/k)*Sum_{j=0..n-k} binomial(k-1+j, j)*Bernoulli(j)*s(n-1, k-1+j), for n >= 1 and k >= 1, with s(n, 0) = 0 if n >= 1, and s(0,0) = 1.
Boas-Buck type recurrence for column k: s(n, k) = (n!*k/(n - k))*Sum_{j=k..n-1} b(n-1-j)*s(j, k)/j!, for n >= 1 and k = 0..n-1, with input s(n, n) = 1. For sequence b see the Boas-Buck comment above. (End)
T(n,k) = Sum_{j=k..n} (-1)^(n-j)*A271705(n,j)*A216294(j,k). - Mélika Tebni, Feb 23 2023

Offset corrected by R. J. Mathar, Feb 23 2009

Formula corrected by Philippe Deléham, Sep 10 2009

A002457 a(n) = (2n+1)!/n!^2.

Original entry on oeis.org

1, 6, 30, 140, 630, 2772, 12012, 51480, 218790, 923780, 3879876, 16224936, 67603900, 280816200, 1163381400, 4808643120, 19835652870, 81676217700, 335780006100, 1378465288200, 5651707681620, 23145088600920, 94684453367400, 386971244197200, 1580132580471900

Offset: 0

N. J. A. Sloane

Expected number of matches remaining in Banach's modified matchbox problem (counted when last match is drawn from one of the two boxes), multiplied by 4^(n-1). - Michael Steyer, Apr 13 2001
Hankel transform is (-1)^n*A014480(n). - Paul Barry, Apr 26 2009
Convolved with A000108: (1, 1, 1, 5, 14, 42, ...) = A000531: (1, 7, 38, 187, 874, ...). - Gary W. Adamson, May 14 2009
Convolution of A000302 and A000984. - Philippe Deléham, May 18 2009
1/a(n) is the integral of (x(1-x))^n on interval [0,1]. Apparently John Wallis computed these integrals for n=0,1,2,3,.... A004731, shifted left by one, gives numerators/denominators of related integrals (1-x^2)^n on interval [0,1]. - Marc van Leeuwen, Apr 14 2010
Extend the triangular peaks of Dyck paths of semilength n down to the baseline forming (possibly) larger and overlapping triangles. a(n) = sum of areas of these triangles. Also a(n) = triangular(n) * Catalan(n). - David Scambler, Nov 25 2010
Let H be the n X n Hilbert matrix H(i,j) = 1/(i+j-1) for 1 <= i,j <= n. Let B be the inverse matrix of H. The sum of the elements in row n of B equals a(n-1). - T. D. Noe, May 01 2011
Apparently the number of peaks in all symmetric Dyck paths with semilength 2n+1. - David Scambler, Apr 29 2013
Denominator of central elements of Leibniz's Harmonic Triangle A003506.
Central terms of triangle A116666. - Reinhard Zumkeller, Nov 02 2013
Number of distinct strings of length 2n+1 using n letters A, n letters B, and 1 letter C. - Hans Havermann, May 06 2014
Number of edges in the Hasse diagram of the poset of partitions in the n X n box ordered by containment (from Havermann's comment above, C represents the square added in the edge). - William J. Keith, Aug 18 2015
Let V(n, r) denote the volume of an n-dimensional sphere with radius r then V(n, 1/2^n) = V(n-1, 1/2^n) / a((n-1)/2) for all odd n. - Peter Luschny, Oct 12 2015
a(n) is the result of processing the n+1 row of Pascal's triangle A007318 with the method of A067056. Example: Let n=3. Given the 4th row of Pascal's triangle 1,4,6,4,1, we get 1*(4+6+4+1) + (1+4)*(6+4+1) + (1+4+6)*(4+1) + (1+4+6+4)*1 = 15+55+55+15 = 140 = a(3). - J. M. Bergot, May 26 2017
a(n) is the number of (n+1) X 2 Young tableaux with a two horizontal walls between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021] and A000984 for one horizontal wall. - Michael Wallner, Jan 31 2022
a(n) is the number of facets of the symmetric edge polytope of the cycle graph on 2n+1 vertices. - Mariel Supina, May 12 2022
Diagonal of the rational function 1 / (1 - x - y)^2. - Ilya Gutkovskiy, Apr 23 2025

			G.f. = 1 + 6*x + 30*x^2 + 140*x^3 + 630*x^4 + 2772*x^5 + 12012*x^6 + 51480*x^7 + ...

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 159.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 83, Problem 25; p. 168, #30.
W. Feller, An Introduction to Probability Theory and Its Applications, Vol. I.
C. Jordan, Calculus of Finite Differences. Röttig and Romwalter, Budapest, 1939; Chelsea, NY, 1965, p. 449.
M. Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see pp. 127-129.
C. Lanczos, Applied Analysis. Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 514.
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.
J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 92.
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).
J. Wallis, Operum Mathematicorum, pars altera, Oxford, 1656, pp 31,34 [Marc van Leeuwen, Apr 14 2010]

G. C. Greubel, Table of n, a(n) for n = 0..1000 [Terms 0 to 200 computed by T. D. Noe; terms 201 to 1000 by G. C. Greubel, Jan 14 2017]
Cyril Banderier and Michael Wallner, Young Tableaux with Periodic Walls: Counting with the Density Method, Séminaire Lotharingien de Combinatoire, 85B (2021), Art. 47, 12 pp.
Alexander Barg, Stolarsky's invariance principle for finite metric spaces, arXiv:2005.12995 [math.CO], 2020.
W. G. Bickley and J. C. P. Miller, Numerical differentiation near the limits of a difference table, Phil. Mag., 33 (1942), 1-12 (plus tables) [Annotated scanned copy]
Sara C. Billey, Matjaž Konvalinka, and Joshua P. Swanson, Asymptotic normality of the major index on standard tableaux, arXiv:1905.00975 [math.CO], 2019.See p. 15, Remark 4.2
R. Chapman, Moments of Dyck paths, Discrete Math., 204 (1999), 113-117.
Ömür Deveci and Anthony G. Shannon, Some aspects of Neyman triangles and Delannoy arrays, Mathematica Montisnigri (2021) Vol. L, 36-43.
F. Disanto, A. Frosini, R. Pinzani and S. Rinaldi, A closed formula for the number of convex permutominoes, arXiv:math/0702550 [math.CO], 2007.
Luca Ferrari and Emanuele Munarini, Enumeration of edges in some lattices of paths, arXiv preprint arXiv:1203.6792 [math.CO], 2012 and J. Int. Seq. 17 (2014) #14.1.5.
Nikita Gogin and Mika Hirvensalo, On the Moments of Squared Binomial Coefficients, (2020).
P.-Y. Huang, S.-C. Liu, and Y.-N. Yeh, Congruences of Finite Summations of the Coefficients in certain Generating Functions, The Electronic Journal of Combinatorics, 21 (2014), #P2.45.
Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.
C. Jordan, Calculus of Finite Differences, Budapest, 1939. [Annotated scans of pages 448-450 only]
Cemil Karaçam and Alper Vural, Enumerating 2D and 3D lattice paths with arbitrary steps, Mathematica Bohemica, pp. 1-13 (2025). See p. 4.
Bahar Kuloğlu, Engin Özkan, and Marin Marin, Fibonacci and Lucas Polynomials in n-gon, An. Şt. Univ. Ovidius Constanţa (Romania 2023) Vol. 31, No 2, 127-140.
C. Lanczos, Applied Analysis (Annotated scans of selected pages)
A. Petojevic and N. Dapic, The vAm(a,b,c;z) function, Preprint 2013.
H. E. Salzer, Coefficients for numerical differentiation with central differences, J. Math. Phys., 22 (1943), 115-135.
H. E. Salzer, Coefficients for numerical differentiation with central differences, J. Math. Phys., 22 (1943), 115-135. [Annotated scanned copy]
J. Ser, Les Calculs Formels des Séries de Factorielles, Gauthier-Villars, Paris, 1933 [Local copy].
J. Ser, Les Calculs Formels des Séries de Factorielles (Annotated scans of some selected pages)
L. W. Shapiro, W.-J. Woan and S. Getu, Runs, slides and moments, SIAM J. Alg. Discrete Methods, 4 (1983), 459-466.
Andrei K. Svinin, On some class of sums, arXiv:1610.05387 [math.CO], 2016. See p. 5.
T. R. Van Oppolzer, Lehrbuch zur Bahnbestimmung der Kometen und Planeten, Vol. 2, Engelmann, Leipzig, 1880, p. 21.
Eric Weisstein's World of Mathematics, Central Beta Function
Eric Weisstein's World of Mathematics, Pi Formulas
Y. Q. Zhao, Introduction to Probability with Applications

Cf. A000531 (Banach's original match problem).
Cf. A033876, A000984, A001803, A132818, A046521 (second column).
Cf. A000108, A000217.
A diagonal of A331430.
The rightmost diagonal of the triangle A331431.

a002457 n = a116666 (2 * n + 1) (n + 1)
-- Reinhard Zumkeller, Nov 02 2013

[Factorial(2*n+1)/Factorial(n)^2: n in [0..25]]; // Vincenzo Librandi, Oct 12 2015

A002457:=n->(n+1) * binomial(2*(n+1),(n+1)) / 2; seq(A002457(n), n=0..50);
seq((2*n)!*coeff(series(HeunC(0,0,-2,-1/4,7/4,4*x^2),x,2*n+1),x,2*n),n=0..22); # Peter Luschny, Nov 22 2013

a[n_]:=(2*n+1)!/n!^2; Array[f, 23, 0] (* Vladimir Joseph Stephan Orlovsky, Dec 13 2008 *)

{a(n) = if( n<0, 0, (2*n + 1)! / n!^2)}; /* Michael Somos, Dec 09 2002 */

a(n) = (2*n+1)*binomial(2*n, n); \\ Altug Alkan, Apr 16 2018

A002457 = lambda n: binomial(n+1/2,1/2)<<2*n
[A002457(n) for n in range(23)] # Peter Luschny, Sep 22 2014

G.f.: (1-4x)^(-3/2) = 1F0(3/2;;4x).
a(n-1) = binomial(2*n, n)*n/2 = binomial(2*n-1, n)*n.
a(n-1) = 4^(n-1)*Sum_{i=0..n-1} binomial(n-1+i, i)*(n-i)/2^(n-1+i).
a(n) ~ 2*Pi^(-1/2)*n^(1/2)*2^(2*n)*{1 + 3/8*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 21 2001
(2*n+2)!/(2*n!*(n+1)!) = (n+n+1)!/(n!*n!) = 1/beta(n+1, n+1) in A061928.
Sum_{i=0..n} i * binomial(n, i)^2 = n*binomial(2*n, n)/2. - Yong Kong (ykong(AT)curagen.com), Dec 26 2000
a(n) ~ 2*Pi^(-1/2)*n^(1/2)*2^(2*n). - Joe Keane (jgk(AT)jgk.org), Jun 07 2002
a(n) = 1/Integral_{x=0..1} x^n (1-x)^n dx. - Fred W. Helenius (fredh(AT)ix.netcom.com), Jun 10 2003
E.g.f.: exp(2*x)*((1+4*x)*BesselI(0, 2*x) + 4*x*BesselI(1, 2*x)). - Vladeta Jovovic, Sep 22 2003
a(n) = Sum_{i+j+k=n} binomial(2i, i)*binomial(2j, j)*binomial(2k, k). - Benoit Cloitre, Nov 09 2003
a(n) = (2*n+1)*A000984(n) = A005408(n)*A000984(n). - Zerinvary Lajos, Dec 12 2010
a(n-1) = Sum_{k=0..n} A039599(n,k)*A000217(k), for n >= 1. - Philippe Deléham, Jun 10 2007
Sum of (n+1)-th row terms of triangle A132818. - Gary W. Adamson, Sep 02 2007
Sum_{n>=0} 1/a(n) = 2*Pi/3^(3/2). - Jaume Oliver Lafont, Mar 07 2009
a(n) = Sum_{k=0..n} binomial(2k,k)*4^(n-k). - Paul Barry, Apr 26 2009
a(n) = A000217(n) * A000108(n). - David Scambler, Nov 25 2010
a(n) = f(n, n-3) where f is given in A034261.
a(n) = A005430(n+1)/2 = A002011(n)/4.
a(n) = binomial(2n+2, 2) * binomial(2n, n) / binomial(n+1, 1), a(n) = binomial(n+1, 1) * binomial(2n+2, n+1) / binomial(2, 1) = binomial(2n+2, n+1) * (n+1)/2. - Rui Duarte, Oct 08 2011
G.f.: (G(0) - 1)/(4*x) where G(k) = 1 + 2*x*((2*k + 3)*G(k+1) - 1)/(k + 1). - Sergei N. Gladkovskii, Dec 03 2011 [Edited by Michael Somos, Dec 06 2013]
G.f.: 1 - 6*x/(G(0)+6*x) where G(k) = 1 + (4*x+1)*k - 6*x - (k+1)*(4*k-2)/G(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Aug 13 2012
G.f.: Q(0), where Q(k) = 1 + 4*(2*k + 1)*x*(2*k + 2 + Q(k+1))/(k+1). - Sergei N. Gladkovskii, May 10 2013 [Edited by Michael Somos, Dec 06 2013]
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 4*x*(2*k+3)/(4*x*(2*k+3) + 2*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 06 2013
a(n) = 2^(4n)/Sum_{k=0..n} (-1)^k*C(2n+1,n-k)/(2k+1). - Mircea Merca, Nov 12 2013
a(n) = (2*n)!*[x^(2*n)] HeunC(0,0,-2,-1/4,7/4,4*x^2) where [x^n] f(x) is the coefficient of x^n in f(x) and HeunC is the Heun confluent function. - Peter Luschny, Nov 22 2013
0 = a(n) * (16*a(n+1) - 2*a(n+2)) + a(n+1) * (a(n+2) - 6*a(n+1)) for all n in Z. - Michael Somos, Dec 06 2013
a(n) = 4^n*binomial(n+1/2, 1/2). - Peter Luschny, Apr 24 2014
a(n) = 4^n*hypergeom([-2*n,-2*n-1,1/2],[-2*n-2,1],2)*(n+1)*(2*n+1). - Peter Luschny, Sep 22 2014
a(n) = 4^n*hypergeom([-n,-1/2],[1],1). - Peter Luschny, May 19 2015
a(n) = 2*4^n*Gamma(3/2+n)/(sqrt(Pi)*Gamma(1+n)). - Peter Luschny, Dec 14 2015
Sum_{n >= 0} 2^(n+1)/a(n) = Pi, related to Newton/Euler's Pi convergence transformation series. - Tony Foster III, Jul 28 2016. See the Weisstein Pi link, eq. (23). - Wolfdieter Lang, Aug 26 2016
Boas-Buck recurrence: a(n) = (6/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, and a(0) = 1. Proof from a(n) = A046521(n+1,1). See comment in A046521. - Wolfdieter Lang, Aug 10 2017
a(n) = (1/3)*Sum_{i = 0..n+1} C(n+1,i)*C(n+1,2*n+1-i)*C(3*n+2-i,n+1) = (1/3)*Sum_{i = 0..2*n+1} (-1)^(i+1)*C(2*n+1,i)*C(n+i+1,i)^2. - Peter Bala, Feb 07 2018
a(n) = (2*n+1)*binomial(2*n, n). - Kolosov Petro, Apr 16 2018
a(n) = (-4)^n*binomial(-3/2, n). - Peter Luschny, Oct 23 2018
a(n) = 1 / Sum_{s=0..n} (-1)^s * binomial(n, s) / (n+s+1). - Kolosov Petro, Jan 22 2019
a(n) = Sum_{k = 0..n} (2*k + 1)*binomial(2*n + 1, n - k). - Peter Bala, Feb 25 2019
4^n/a(n) = Integral_{x=0..1} (1 - x^2)^n. - Michael Somos, Jun 13 2019
D-finite with recurrence: 0 = a(n)*(6 + 4*n) - a(n+1)*(n + 1) for all n in Z. - Michael Somos, Jun 13 2019
Sum_{n>=0} (-1)^n/a(n) = 4*arcsinh(1/2)/sqrt(5). - Amiram Eldar, Sep 10 2020
From Jianing Song, Apr 10 2022: (Start)
G.f. for {1/a(n)}: 4*arcsin(sqrt(x)/2) / sqrt(x*(4-x)).
E.g.f. for {1/a(n)}: exp(x/4)*sqrt(Pi/x)*erf(sqrt(x)/2). (End)
G.f. for {1/a(n)}: 4*arctan(sqrt(x/(4-x))) / sqrt(x*(4-x)). - Michael Somos, Jun 17 2023
a(n) = Sum_{k = 0..n} (-1)^(n+k) * (n + 2*k + 1)*binomial(n+k, k).  This is the particular case m = 1 of the identity Sum_{k = 0..m*n} (-1)^k * (n + 2*k + 1) * binomial(n+k, k) = (-1)^(m*n) * (m*n + 1) * binomial((m+1)*n+1, n). Cf. A090816 and A306290. - Peter Bala, Nov 02 2024
a(n) = (1/Pi)*(2*n + 1)*(2^(2*n + 1))*Integral_{x=0..oo} 1/(x^2 + 1)^(n + 1) dx. - Velin Yanev, Jan 28 2025

A053121 Catalan triangle (with 0's) read by rows.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 2, 0, 1, 2, 0, 3, 0, 1, 0, 5, 0, 4, 0, 1, 5, 0, 9, 0, 5, 0, 1, 0, 14, 0, 14, 0, 6, 0, 1, 14, 0, 28, 0, 20, 0, 7, 0, 1, 0, 42, 0, 48, 0, 27, 0, 8, 0, 1, 42, 0, 90, 0, 75, 0, 35, 0, 9, 0, 1, 0, 132, 0, 165, 0, 110, 0, 44, 0, 10, 0, 1, 132, 0, 297, 0, 275, 0, 154, 0, 54, 0, 11, 0

Offset: 0

Wolfdieter Lang

Inverse lower triangular matrix of A049310(n,m) (coefficients of Chebyshev's S polynomials).
Walks with a wall: triangle of number of n-step walks from (0,0) to (n,m) where each step goes from (a,b) to (a+1,b+1) or (a+1,b-1) and the path stays in the nonnegative quadrant.
T(n,m) is the number of left factors of Dyck paths of length n ending at height m. Example: T(4,2)=3 because we have UDUU, UUDU, and UUUD, where U=(1,1) and D=(1,-1). (This is basically a different formulation of the previous - walks with a wall - property.) - Emeric Deutsch, Jun 16 2011
"The Catalan triangle is formed in the same manner as Pascal's triangle, except that no number may appear on the left of the vertical bar." [Conway and Smith]
G.f. for row polynomials p(n,x) := Sum_{m=0..n} (a(n,m)*x^m): c(z^2)/(1-x*z*c(z^2)). Row sums (x=1): A001405 (central binomial).
In the language of the Shapiro et al. reference such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group. The g.f. Ginv(x) of the m=0 column of the inverse of a given Bell-matrix (here A049310) is obtained from its g.f. of the m=0 column (here G(x)=1/(1+x^2)) by Ginv(x)=(f^{(-1)}(x))/x, with f(x) := x*G(x) and f^{(-1)}is the compositional inverse function of f (here one finds, with Ginv(0)=1, c(x^2)). See the Shapiro et al. reference.
Number of involutions of {1,2,...,n} that avoid the patterns 132 and have exactly k fixed points. Example: T(4,2)=3 because we have 2134, 4231 and 3214. Number of involutions of {1,2,...,n} that avoid the patterns 321 and have exactly k fixed points. Example: T(4,2)=3 because we have 1243, 1324 and 2134. Number of involutions of {1,2,...,n} that avoid the patterns 213 and have exactly k fixed points. Example: T(4,2)=3 because we have 1243, 1432 and 4231. - Emeric Deutsch, Oct 12 2006
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1 . Other triangles arise by choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. - Philippe Deléham, Sep 25 2007
Riordan array (c(x^2),xc(x^2)), where c(x) is the g.f. of Catalan numbers A000108. - Philippe Deléham, Nov 25 2007
A053121^2 = triangle A145973. Convolved with A001405 = triangle A153585. - Gary W. Adamson, Dec 28 2008
By columns without the zeros, n-th row = A000108 convolved with itself n times; equivalent to A = (1 + x + 2x^2 + 5x^3 + 14x^4 + ...), then n-th row = coefficients of A^(n+1). - Gary W. Adamson, May 13 2009
Triangle read by rows,product of A130595 and A064189 considered as infinite lower triangular arrays; A053121 = A130595*A064189 = B^(-1)*A097609*B where B = A007318. - Philippe Deléham, Dec 07 2009
From Mark Dols, Aug 17 2010: (Start)
As an upper right triangle, rows represent powers of 5-sqrt(24):
  5 - sqrt(24)^1 = 0.101020514...
  5 - sqrt(24)^2 = 0.010205144...
  5 - sqrt(24)^3 = 0.001030928...
(Divided by sqrt(96) these powers give a decimal representation of the columns of A007318, with 1/sqrt(96) being the middle column.) (End)
T(n,k) is the number of dispersed Dyck paths of length n (i.e., Motzkin paths of length n with no (1,0) steps at positive heights) having k (1,0)-steps. Example: T(5,3)=4 because, denoting U=(1,1), D=(1,-1), H=(1,0), we have HHHUD, HHUDH, HUDHH, and UDHHH. - Emeric Deutsch, Jun 01 2011
Let S(N,x) denote the N-th Chebyshev S-polynomial in x (see A049310, cf. [W. Lang]). Then x^n = sum_{k=0..n} T(n,k)*S(k,x). - L. Edson Jeffery, Sep 06 2012
This triangle a(n,m) appears also in the (unreduced) formula for the powers rho(N)^n for the algebraic number over the rationals rho(N) = 2*cos(Pi/N) = R(N, 2), the smallest diagonal/side ratio R in the regular N-gon:
  rho(N)^n = sum(a(n,m)*R(N,m+1),m=0..n), n>=0, identical in N >= 1. R(N,j) = S(j-1, x=rho(N)) (Chebyshev S (A049310)). See a comment on this under A039599 (even powers) and A039598 (odd powers). Proof: see the Sep 06 2012 comment by L. Edson Jeffery, which follows from T(n,k) (called here a(n,k)) being the inverse of the Riordan triangle A049310. - Wolfdieter Lang, Sep 21 2013
The so-called A-sequence for this Riordan triangle of the Bell type (c(x^2), x*c(x^2)) (see comments above) is A(x) = 1 + x^2. This proves the recurrence given in the formula section by Henry Bottomley for a(n, m) = a(n-1, m-1) + a(n-1, m+1) for n>=1 and m>=1, with inputs. The Z-sequence for this Riordan triangle is Z(x) = x which proves the recurrence a(n,0) = a(n-1,1), n>=1, a(0,0) = 1. For A- and Z-sequences for Riordan triangles see the W. Lang link under A006232. - Wolfdieter Lang, Sep 22 2013
Rows of the triangle describe decompositions of tensor powers of the standard (2-dimensional) representation of the Lie algebra sl(2) into irreducibles. Thus a(n,m) is the multiplicity of the m-th ((m+1)-dimensional) irreducible representation in the n-th tensor power of the standard one. - Mamuka Jibladze, May 26 2015
The Riordan row polynomials p(n, x) belong to the Boas-Buck class (see a comment and references in A046521), hence they satisfy the Boas-Buck identity: (E_x - n*1)*p(n, x) = (E_x + 1)*Sum_{j=0..n-1} (1/2)*(1 - (-1)^j)*binomial(j+1, (j+1)/2)*p(n-1-j, x), for n >= 0, where E_x = x*d/dx (Euler operator). For the triangle a(n, m) this entails a recurrence for the sequence of column m, given in the formula section. - Wolfdieter Lang, Aug 11 2017
From Roger Ford, Jan 22 2018: (Start)
For row n, the nonzero values represent the odd components (loops) formed by n+1 nonintersecting arches above and below the x-axis with the following constraints:  The top has floor((n+3)/2) starting arches at position 1 and the next consecutive odd positions. All other starting top arches are in even positions. The bottom arches are a rainbow of arches.  If the component=1 then the arch configuration is a semimeander solution.
Examples: For row 3 {0, 2, 0, 1} there are 3 arch configurations: 2 arch configurations have a component=1; 1 has a component=3.  c=components, U=top arch starting in odd position, u=top arch starting in an even position, d=ending top arch:
.
  top UuUdUddd c=3   top UdUuUddd c=1     top UdUdUudd c=1
       /\                    /\
      //\\                  /  \
     //  \\                / /\ \                    /\
    //    \\              / /  \ \                  /  \
   ///\  /\\\        /\  / / /\ \ \        /\  /\  / /\ \
   \\\ \/ ///        \ \ \ \/ / / /        \ \ \ \/ / / /
    \\\  ///          \ \ \  / / /          \ \ \  / / /
     \\\///            \ \ \/ / /            \ \ \/ / /
      \\//              \ \  / /              \ \  / /
       \/                \ \/ /                \ \/ /
                          \  /                  \  /
                           \/                    \/
For row 4 {2, 0, 3, 0, 1} there are 6 arch configurations: 2 have a component=1; 3 have a component=3: 1 has a component=1. (End)

			Triangle a(n,m) begins:
  n\m  0   1   2   3   4   5   6  7  8  9 10 ...
  0:   1
  1:   0   1
  2:   1   0   1
  3:   0   2   0   1
  4:   2   0   3   0   1
  5:   0   5   0   4   0   1
  6:   5   0   9   0   5   0   1
  7:   0  14   0  14   0   6   0  1
  8:  14   0  28   0  20   0   7  0  1
  9:   0  42   0  48   0  27   0  8  0  1
  10: 42   0  90   0  75   0  35  0  9  0  1
  ... (Reformatted by _Wolfdieter Lang_, Sep 20 2013)
E.g., the fourth row corresponds to the polynomial p(3,x)= 2*x + x^3.
From _Paul Barry_, May 29 2009: (Start)
Production matrix is
  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,
  0, 0, 0, 0, 0, 0, 0, 1, 0, 1 (End)
Boas-Buck recurrence for column k = 2, n = 6: a(6, 2) = (3/4)*(0 + 2*a(4 ,2) + 0 + 6*a(2, 2)) = (3/4)*(2*3 + 6) = 9. - _Wolfdieter Lang_, Aug 11 2017

J. H. Conway and D. A. Smith, On Quaternions and Octonions, A K Peters, Ltd., Natick, MA, 2003. See p. 60. MR1957212 (2004a:17002)
A. Nkwanta, Lattice paths and RNA secondary structures, in African Americans in Mathematics, ed. N. Dean, Amer. Math. Soc., 1997, pp. 137-147.

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
I. Bajunaid et al., Function series, Catalan numbers and random walks on trees, Amer. Math. Monthly 112 (2005), 765-785.
C. Banderier and D. Merlini, Lattice paths with an infinite set of jumps
Paul Barry, Riordan Arrays, Orthogonal Polynomials as Moments, and Hankel Transforms, J. Int. Seq. 14 (2011) # 11.2.2, example 3.
Paul Barry, On the inversion of Riordan arrays, arXiv:2101.06713 [math.CO], 2021.
Paul Barry and A. Hennessy, Meixner-Type Results for Riordan Arrays and Associated Integer Sequences, J. Int. Seq. 13 (2010) # 10.9.4, example 3.
Xiang-Ke Chang, X.-B. Hu, H. Lei, and Y.-N. Yeh, Combinatorial proofs of addition formulas, The Electronic Journal of Combinatorics, 23(1) (2016), #P1.8.
J. Cigler, Some q-analogues of Fibonacci, Lucas and Chebyshev polynomials with nice moments, 2013.
J. Cigler, Some remarks about q-Chebyshev polynomials and q-Catalan numbers and related results, 2013.
J. Cigler, Some notes on q-Gould polynomials, 2013.
Emeric Deutsch, A. Robertson and D. Saracino, Refined restricted involutions, European Journal of Combinatorics 28 (2007), 481-498 (see pp. 486 and 498).
J. East and R. D. Gray, Idempotent generators in finite partition monoids and related semigroups, arXiv preprint arXiv:1404.2359, 2014
D. Gouyou-Beauchamps, Chemins sous-diagonaux et tableau de Young, pp. 112-125 of "Combinatoire Enumerative (Montreal 1985)", Lect. Notes Math. 1234, 1986 (see |F_{l,p}| on page 114). - _N. J. A. Sloane_, Jan 29 2011
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
V. E. Hoggatt, Jr. and M. Bicknell, Catalan and related sequences arising from inverses of Pascal's triangle matrices, Fib. Quart., 14 (1976), 395-405.
W. F. Klostermeyer, M. E. Mays, L. Soltes and G. Trapp, A Pascal rhombus, Fibonacci Quarterly, 35 (1997), 318-328.
Wolfdieter Lang, Chebyshev S-polynomials: ten applications.
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38,5 (2000) 408-419; Note 4, pp. 414-415.
MathOverflow, Catalan numbers as sums of squares of numbers in the rows of the Catalan triangle - is there a combinatorial explanation?
A. Nkwanta and A. Tefera, Curious Relations and Identities Involving the Catalan Generating Function and Numbers, Journal of Integer Sequences, 16 (2013), #13.9.5.
Karim Ritter von Merkl, Computing colored Khovanov homology, arXiv:2505.03916 [math.QA], 2025. See p. 2.
Frank Ruskey and Mark Weston, Spherical Venn Diagrams with Involutory Isometries, Electronic Journal of Combinatorics, 18 (2011), #P191.
L. W. Shapiro, S. Getu, Wen-Jin Woan and L. C. Woodson, The Riordan Group, Discrete Appl. Maths. 34 (1991) 229-239.
Yidong Sun and Luping Ma, Minors of a class of Riordan arrays related to weighted partial Motzkin paths. Eur. J. Comb. 39, 157-169 (2014).
Mark C. Wilson, Diagonal asymptotics for products of combinatorial classes.
W.-J. Woan, Area of Catalan Paths, Discrete Math., 226 (2001), 439-444.
Index entries for sequences related to Chebyshev polynomials.

Cf. A008315, A049310, A000108, A001405 (row sums), A145973, A153585, A108786, A037952. Another version: A008313. A039598 and A039599 without zeros, and odd and even numbered rows.

Variant without zero-diagonals: A033184 and with rows reversed: A009766.

a053121 n k = a053121_tabl !! n !! k
a053121_row n = a053121_tabl !! n
a053121_tabl = iterate
   (\row -> zipWith (+) ([0] ++ row) (tail row ++ [0,0])) [1]
-- Reinhard Zumkeller, Feb 24 2012

T:=proc(n,k): if n+k mod 2 = 0 then (k+1)*binomial(n+1,(n-k)/2)/(n+1) else 0 fi end: for n from 0 to 13 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form; Emeric Deutsch, Oct 12 2006
F:=proc(l,p) if ((l-p) mod 2) = 1 then 0 else (p+1)*l!/( ( (l-p)/2 )! * ( (l+p)/2 +1)! ); fi; end;
r:=n->[seq( F(n,p),p=0..n)]; [seq(r(n),n=0..15)]; # N. J. A. Sloane, Jan 29 2011
A053121 := proc(n,k) option remember; `if`(k>n or k<0,0,`if`(n=k,1,
procname(n-1,k-1)+procname(n-1,k+1))) end proc:
seq(print(seq(A053121(n,k), k=0..n)),n=0..12); # Peter Luschny, May 01 2011

a[n_, m_] /; n < m || OddQ[n-m] = 0; a[n_, m_] = (m+1) Binomial[n+1, (n-m)/2]/(n+1); Flatten[Table[a[n, m], {n, 0, 12}, {m, 0, n}]] [[1 ;; 90]] (* Jean-François Alcover, May 18 2011 *)
T[0, 0] := 1; T[n_, k_]/;0<=k<=n := T[n, k] = T[n-1, k-1]+T[n-1, k+1]; T[n_, k_] := 0; Flatten@Table[T[n, k], {n, 0, 12}, {k, 0, n}] (* Oliver Seipel, Dec 31 2024 *)

T(n, m)=if(nCharles R Greathouse IV, Mar 09 2016

def A053121_triangle(dim):
    M = matrix(ZZ,dim,dim)
    for n in (0..dim-1): M[n,n] = 1
    for n in (1..dim-1):
        for k in (0..n-1):
            M[n,k] = M[n-1,k-1] + M[n-1,k+1]
    return M
A053121_triangle(13) # Peter Luschny, Sep 19 2012

a(n, m) := 0 if n

a(n, m) = (4*(n-1)*a(n-2, m) + 2*(m+1)*a(n-1, m-1))/(n+m+2), a(n, m)=0 if n

G.f. for m-th column: c(x^2)*(x*c(x^2))^m, where c(x) = g.f. for Catalan numbers A000108.

G.f.: G(t,z) = c(z^2)/(1 - t*z*c(z^2)), where c(z) = (1 - sqrt(1-4*z))/(2*z) is the g.f. for the Catalan numbers (A000108). - Emeric Deutsch, Jun 16 2011

a(n, m) = a(n-1, m-1) + a(n-1, m+1) if n > 0 and m >= 0, a(0, 0)=1, a(0, m)=0 if m > 0, a(n, m)=0 if m < 0. - Henry Bottomley, Jan 25 2001

Sum_{k>=0} T(m,k)^2 = A000108(m). - Paul D. Hanna, Apr 23 2005

Sum_{k>=0} T(m, k)*T(n, k) = 0 if m+n is odd; Sum_{k>=0} T(m, k)*T(n, k) = A000108((m+n)/2) if m+n is even. - Philippe Deléham, May 26 2005

T(n,k)=sum{i=0..n, (-1)^(n-i)*C(n,i)*sum{j=0..i, C(i,j)*(C(i-j,j+k)-C(i-j,j+k+2))}}; Column k has e.g.f. BesselI(k,2x)-BesselI(k+2,2x). - Paul Barry, Feb 16 2006

Sum_{k=0..n} T(n,k)*(k+1) = 2^n. - Philippe Deléham, Mar 22 2007

Sum_{j>=0} T(n,j)*binomial(j,k) = A054336(n,k). - Philippe Deléham, Mar 30 2007

T(2*n+1,2*k+1) = A039598(n,k), T(2*n,2*k) = A039599(n,k). - Philippe Deléham, Apr 16 2007

Sum_{k=0..n} T(n,k)^x = A000027(n+1), A001405(n), A000108(n), A003161(n), A129123(n) for x = 0,1,2,3,4 respectively. - Philippe Deléham, Nov 22 2009

Sum_{k=0..n} T(n,k)*x^k = A126930(n), A126120(n), A001405(n), A054341(n), A126931(n) for x = -1, 0, 1, 2, 3 respectively. - Philippe Deléham, Nov 28 2009

Sum_{k=0..n} T(n,k)*A000045(k+1) = A098615(n). - Philippe Deléham, Feb 03 2012

Recurrence for row polynomials C(n, x) := Sum_{m=0..n} a(n, m)*x^m = x*Sum_{k=0..n} Chat(k)*C(n-1-k, x), n >= 0, with C(-1, 1/x) = 1/x and Chat(k) = A000108(k/2) if n is even and 0 otherwise. From the o.g.f. of the row polynomials: G(z; x) := Sum_{n >= 0} C(n, x)*z^n = c(z^2)*(1 + x*z*G(z, x)), with the o.g.f. c of A000108. - Ahmet Zahid KÜÇÜK and Wolfdieter Lang, Aug 23 2015

The Boas-Buck recurrence (see a comment above) for the sequence of column m is: a(n, m) = ((m+1)/(n-m))*Sum_{j=0..n-1-m} (1/2)*(1 - (-1)^j)*binomial(j+1, (j+1)/2)* a(n-1-j, k), for n > m >= 0 and input a(m, m) = 1. - Wolfdieter Lang, Aug 11 2017

Sum_{m=1..n} a(n,m) = A037952(n). - R. J. Mathar, Sep 23 2021

Edited by N. J. A. Sloane, Jan 29 2011

A002802 a(n) = (2n+3)!/(6n!*(n+1)!).

Original entry on oeis.org

1, 10, 70, 420, 2310, 12012, 60060, 291720, 1385670, 6466460, 29745716, 135207800, 608435100, 2714556600, 12021607800, 52895074320, 231415950150, 1007340018300, 4365140079300, 18839025605400, 81007810103220, 347176329013800, 1483389769422600

Offset: 0

N. J. A. Sloane

For n >= 1 a(n) is also the number of rooted bicolored unicellular maps of genus 1 on n+2 edges. - Ahmed Fares (ahmedfares(AT)my-deja.com), Aug 20 2001
a(n) is half the number of (n+2) X 2 Young tableaux with a three horizontal walls between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021], A000984 for one horizontal wall, and A002457 for two. - Michael Wallner, Jan 31 2022
From Robert Coquereaux, Feb 12 2024: (Start)
Call B(p,g) the number of genus g partitions of a set with p elements (genus-dependent Bell number). Up to an appropriate shift the given sequence counts the genus 1 partitions of a set: we have a(n) = B(n+4,1), with a(0)= B(4,1)=1.
When shifted with an offset 4 (i.e., defining b(p)=a(p-4), which starts with 0,0,0,1,10,70, etc., and b(4)=1), the given sequence reads b(p) = (1/( 2^4 3 )) * (1/( (2 p - 1) (2 p - 3))) * (1/(p - 4)!) * (2p)!/p!. In this form it appears as a generalization of Catalan numbers (that indeed count the genus 0 partitions).
Call C[p, [alpha], g] the number of partitions of a set with p elements, of cyclic type [alpha], and of genus g (genus g Faa di Bruno coefficients of type [alpha]). Up to an appropriate shift the given sequence also counts the genus 1 partitions of p=2k into k parts of length 2, which is then called C[2k, [2^k], 1], and we have a(n) = C[2k, [2^k], 1] for k=n+2.
The two previous interpretations of this sequence, leading to a(n) = B(n+4, 1) and to a(n) = C[2(n+2), [2^(n+2)], 1] are not related in any obvious way. (End)

			G.f. = 1 + 10*x + 70*x^2 + 420*x^3 + 2310*x^4 + 12012*x^5 + 60060*x^6 + ...

C. Jordan, Calculus of Finite Differences. Röttig and Romwalter, Budapest, 1939; Chelsea, NY, 1965, p. 449.
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).

T. D. Noe, Table of n, a(n) for n = 0..200
Cyril Banderier and Michael Wallner, Young Tableaux with Periodic Walls: Counting with the Density Method, Séminaire Lotharingien de Combinatoire, 85B (2021), Art. 47, 12 pp.
Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus. A compendium of results, arXiv:2305.01100 [math.CO], 2023. See pp. 4, 12.
Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus: a compendium of results, Journal of Integer Sequences, Vol. 27 (2024), Article 24.2.6. See p. 9. See also arXiv:2305.01100, 2023. See pp. 9, 19.
R. Cori and G. Hetyei, Counting genus one partitions and permutations, arXiv preprint arXiv:1306.4628 [math.CO], 2013.
R. Cori and G. Hetyei, How to count genus one partitions, FPSAC 2014, Chicago, Discrete Mathematics and Theoretical Computer Science (DMTCS), Nancy, France, 2014, 333-344.
Alain Goupil and Gilles Schaeffer, Factoring N-Cycles and Counting Maps of Given Genus, Europ. J. Combinatorics (1998) 19 819-834.
T. R. S. Walsh and A. B. Lehman, Counting rooted maps by genus. I, J. Comb. Theory B 13 (1972), 192-218 (Tab.1).
Liang Zhao and Fengyao Yan, Note on Total Positivity for a Class of Recursive Matrices, Journal of Integer Sequences, Vol. 19 (2016), Article 16.6.5.
Notes.

Cf. A035309, A000108 (for genus 0 maps), A046521 (third column).
Cf. A000984, A000911, A002457, A002697, A051133.
Column g=1 of A370235.

F:=Factorial;; List([0..25], n-> F(2*n+3)/(6*F(n)*F(n+1)) ); # G. C. Greubel, Jul 20 2019

F:=Factorial; [F(2*n+3)/(6*F(n)*F(n+1)): n in [0..25]]; // G. C. Greubel, Jul 20 2019

seq(simplify(4^n*hypergeom([-n,-3/2], [1], 1)),n=0..25); # Peter Luschny, Apr 26 2016

Table[(2*n+3)!/(6*n!*(n+1)!), {n, 0, 25}] (* Vladimir Joseph Stephan Orlovsky, Dec 13 2008 *)

{a(n) = if( n<0, 0, (2*n + 3)! / (6 * n! * (n+1)!))}; /* Michael Somos, Sep 16 2013 */

{a(n) = 2^(n+3) * polcoeff( pollegendre(n+4), n) / 3}; /* Michael Somos, Sep 16 2013 */

f=factorial; [f(2*n+3)/(6*f(n)*f(n+1)) for n in (0..25)] # G. C. Greubel, Jul 20 2019

G.f.: (1 - 4*x)^(-5/2) = 1F0(5/2;;4x).
Asymptotic expression for a(n) is a(n) ~ (n+2)^(3/2) * 4^(n+2) / (sqrt(Pi) * 48).
a(n) = Sum_{a+b+c+d+e=n} f(a)*f(b)*f(c)*f(d)*f(e) with f(n) = binomial(2n, n) = A000984(n). - Philippe Deléham, Jan 22 2004
a(n-1) = (1/4)*Sum_{k=1..n} k*(k+1)*binomial(2*k, k). - Benoit Cloitre, Mar 20 2004
a(n) = A051133(n+1)/3 = A000911(n)/6. - Zerinvary Lajos, Jun 02 2007
From Rui Duarte, Oct 08 2011: (Start)
Also convolution of A000984 with A002697, also convolution of A000302 with A002457.
a(n) = ((2n+3)(2n+1)/(3*1)) * binomial(2n, n).
a(n) = binomial(2n+4, 4) * binomial(2n, n) / binomial(n+2, 2).
a(n) = binomial(n+2, 2) * binomial(2n+4, n+2) / binomial(4, 2).
a(n) = binomial(2n+4, n+2) * (n+2)*(n+1) / 12. (End)
D-finite with recurrence: n*a(n) - 2*(2*n+3)*a(n-1) = 0. - R. J. Mathar, Jan 31 2014
a(n) = 4^n*hypergeom([-n,-3/2], [1], 1). - Peter Luschny, Apr 26 2016
Boas-Buck recurrence: a(n) = (10/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+2, 2). See a comment there. - Wolfdieter Lang, Aug 10 2017
a(n) = (-4)^n*binomial(-5/2, n). - Peter Luschny, Oct 23 2018
Sum_{n>=0} 1/a(n) = 12 - 2*sqrt(3)*Pi. - Amiram Eldar, Oct 13 2020
E.g.f.: (1/12) exp(2 x) x^2 BesselI[2, 2 x]. - Robert Coquereaux, Feb 12 2024

A053122 Triangle of coefficients of Chebyshev's S(n,x-2) = U(n,x/2-1) polynomials (exponents of x in increasing order).

Original entry on oeis.org

1, -2, 1, 3, -4, 1, -4, 10, -6, 1, 5, -20, 21, -8, 1, -6, 35, -56, 36, -10, 1, 7, -56, 126, -120, 55, -12, 1, -8, 84, -252, 330, -220, 78, -14, 1, 9, -120, 462, -792, 715, -364, 105, -16, 1, -10, 165, -792, 1716, -2002, 1365, -560, 136, -18, 1, 11, -220, 1287, -3432, 5005, -4368, 2380, -816, 171, -20

Offset: 0

Wolfdieter Lang

Apart from signs, identical to A078812.
Another version with row-leading 0's and differing signs is given by A285072.
G.f. for row polynomials S(n,x-2) (signed triangle): 1/(1+(2-x)*z+z^2). Unsigned triangle |a(n,m)| has g.f. 1/(1-(2+x)*z+z^2) for row polynomials.
Row sums (signed triangle) A049347(n) (periodic(1,-1,0)). Row sums (unsigned triangle) A001906(n+1)=F(2*(n+1)) (even-indexed Fibonacci).
In the language of Shapiro et al. (see A053121 for the reference) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group.
The (unsigned) column sequences are A000027, A000292, A000389, A000580, A000582, A001288 for m=0..5, resp. For m=6..23 they are A010966..(+2)..A011000 and for m=24..49 they are A017713..(+2)..A017763.
Riordan array (1/(1+x)^2,x/(1+x)^2). Inverse array is A039598. Diagonal sums have g.f. 1/(1+x^2). - Paul Barry, Mar 17 2005. Corrected by Wolfdieter Lang, Nov 13 2012.
Unsigned version is in A078812. - Philippe Deléham, Nov 05 2006
Also row n gives (except for an overall sign) coefficients of characteristic polynomial of the Cartan matrix for the root system A_n. - Roger L. Bagula, May 23 2007
From Wolfdieter Lang, Nov 13 2012: (Start)
The A-sequence for this Riordan triangle is A115141, and the Z-sequence is A115141(n+1), n>=0. For A- and Z-sequences for Riordan matrices see the W. Lang link under A006232 with details and references.
S(n,x^2-2) = sum(r(j,x^2),j=0..n) with Chebyshev's S-polynomials and r(j,x^2) := R(2*j+1,x)/x, where R(n,x) are the monic integer Chebyshv T-polynomials with coefficients given in A127672. Proof from comparing the o.g.f. of the partial sum of the r(j,x^2) polynomials (see a comment on the signed Riordan triangle A111125) with the present Riordan type o.g.f. for the row polynomials with x -> x^2.  (End)
S(n,x^2-2) = S(2*n+1,x)/x, n >= 0, from the odd part of the bisection of the o.g.f. - Wolfdieter Lang, Dec 17 2012
For a relation to a generator for the Narayana numbers A001263, see A119900, whose columns are unsigned shifted rows (or antidiagonals) of this array, referring to the tables in the example sections. - Tom Copeland, Oct 29 2014
The unsigned rows of this array are alternating rows of a mirrored A011973 and alternating shifted rows of A030528 for the Fibonacci polynomials. - Tom Copeland, Nov 04 2014
Boas-Buck type recurrence for column k >= 0 (see Aug 10 2017 comment in A046521 with references): a(n, m) =  (2*(m + 1)/(n - m))*Sum_{k = m..n-1} (-1)^(n-k)*a(k, m), with input a(n, n) = 1, and a(n,k) = 0 for n < k. - Wolfdieter Lang, Jun 03 2020
Row n gives the characteristic polynomial of the (n X n)-matrix M where M[i,j] = 2 if i = j, -1 if |i-j| = 1 and 0 otherwise. The matrix M is positive definite and has 2-condition number (cot(Pi/(2*n+2)))^2. - Jianing Song, Jun 21 2022
Also the convolution triangle of (-1)^(n+1)*n. - Peter Luschny, Oct 07 2022

			The triangle a(n,m) begins:
n\m   0    1    2     3     4     5     6    7    8  9
0:    1
1:   -2    1
2:    3   -4    1
3:   -4   10   -6     1
4:    5  -20   21    -8     1
5:   -6   35  -56    36   -10     1
6:    7  -56  126  -120    55   -12     1
7:   -8   84 -252   330  -220    78   -14    1
8:    9 -120  462  -792   715  -364   105  -16    1
9:  -10  165 -792  1716 -2002  1365  -560  136  -18  1
... Reformatted and extended by _Wolfdieter Lang_, Nov 13 2012
E.g., fourth row (n=3) {-4,10,-6,1} corresponds to the polynomial S(3,x-2) = -4+10*x-6*x^2+x^3.
From _Wolfdieter Lang_, Nov 13 2012: (Start)
Recurrence: a(5,1) = 35 = 1*5 + (-2)*(-20) -1*(10).
Recurrence from Z-sequence [-2,-1,-2,-5,...]: a(5,0) = -6 = (-2)*5 + (-1)*(-20) + (-2)*21 + (-5)*(-8) + (-14)*1.
Recurrence from A-sequence [1,-2,-1,-2,-5,...]: a(5,1) = 35 = 1*5  + (-2)*(-20) + (-1)*21 + (-2)*(-8) + (-5)*1.
(End)
E.g., the fourth row (n=3) {-4,10,-6,1} corresponds also to the polynomial S(7,x)/x = -4 + 10*x^2 - 6*x^4 + x^6. - _Wolfdieter Lang_, Dec 17 2012
Boas-Buck type recurrence: -56 = a(5, 2) = 2*(-1*1 + 1*(-6) - 1*21) = -2*28 = -56. - _Wolfdieter Lang_, Jun 03 2020

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.
Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990.
R. N. Cahn, Semi-Simple Lie Algebras and Their Representations, Dover, NY, 2006, ISBN 0-486-44999-8, p. 62.
Sigurdur Helgasson, Differential Geometry, Lie Groups and Symmetric Spaces, Graduate Studies in Mathematics, volume 34. A. M. S.: ISBN 0-8218-2848-7, 1978, p. 463.

T. D. Noe, Rows n=0..50 of triangle, flattened
Wolfdieter Lang, First rows of the triangle.
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].
P. Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
Naiomi T. Cameron and Asamoah Nkwanta, On some (pseudo) involutions in the Riordan group, J. of Integer Sequences, 8(2005), 1-16.
P. Damianou, On the characteristic polynomials of Cartan matrices and Chebyshev polynomials, arXiv preprint arXiv:1110.6620 [math.RT], 2014 (p. 10). - _Tom Copeland_, Oct 11 2014
Pentti Haukkanen, Jorma Merikoski, Seppo Mustonen, Some polynomials associated with regular polygons, Acta Univ. Sapientiae, Mathematica, 6, 2 (2014) 178-193.
Eric Weisstein's World of Mathematics, Cartan Matrix
Eric Weisstein's World of Mathematics, Dynkin Diagram
Index entries for sequences related to Chebyshev polynomials.

Cf. A005248, A127677, A053123, A049310.
Cf. A011973, A030528, A034867, A119900.
Cf. A285072 (version with row-leading 0's and differing signs). - Eric W. Weisstein, Apr 09 2017

seq(seq((-1)^(n+m)*binomial(n+m+1,2*m+1),m=0..n),n=0..10); # Robert Israel, Oct 15 2014
# Uses function PMatrix from A357368. Adds a row above and a column to the left.
PMatrix(10, n -> -(-1)^n*n); # Peter Luschny, Oct 07 2022

T[n_, m_, d_] := If[ n == m, 2, If[n == m - 1 || n == m + 1, -1, 0]]; M[d_] := Table[T[n, m, d], {n, 1, d}, {m, 1, d}]; a = Join[M[1], Table[CoefficientList[Det[M[d] - x*IdentityMatrix[d]], x], {d, 1, 10}]]; Flatten[a] (* Roger L. Bagula, May 23 2007 *)
(* Alternative code for the matrices from MathWorld: *)
sln[n_] := 2IdentityMatrix[n] - PadLeft[PadRight[IdentityMatrix[n - 1], {n, n - 1}], {n, n}] - PadLeft[PadRight[IdentityMatrix[n - 1], {n - 1, n}], {n, n}] (* Roger L. Bagula, May 23 2007 *)

@CachedFunction
def A053122(n,k):
    if n< 0: return 0
    if n==0: return 1 if k == 0 else 0
    return A053122(n-1,k-1)-A053122(n-2,k)-2*A053122(n-1,k)
for n in (0..9): [A053122(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

a(n, m) := 0 if n

a(n, m) = -2*a(n-1, m) + a(n-1, m-1) - a(n-2, m), a(n, -1) := 0 =: a(-1, m), a(0, 0)=1, a(n, m) := 0 if n

O.g.f. for m-th column (signed triangle): ((x/(1+x)^2)^m)/(1+x)^2.

From Jianing Song, Jun 21 2022: (Start)

T(n,k) = [x^k]f_n(x), where f_{-1}(x) = 0, f_0(x) = 1, f_n(x) = (x-2)*f_{n-1}(x) - f_{n-2}(x) for n >= 2.

f_n(x) = (((x-2+sqrt(x^2-4*x))/2)^(n+1) - ((x-2-sqrt(x^2-4*x))/2)^(n+1))/sqrt(x^2-4x).

The roots of f_n(x) are 2 + 2*cos(k*Pi/(n+1)) = 4*(cos(k*Pi/(2*n+2)))^2 for 1 <= k <= n. (End)

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cp's OEIS Frontend

A046748 Row sums of triangle A046521.

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A176280 Diagonal sums of number triangle A046521.

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A007318 Pascal's triangle read by rows: C(n,k) = binomial(n,k) = n!/(k!*(n-k)!), 0 <= k <= n.

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A000984 Central binomial coefficients: binomial(2*n,n) = (2*n)!/(n!)^2.

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A049310 Triangle of coefficients of Chebyshev's S(n,x) := U(n,x/2) polynomials (exponents in increasing order).

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A000984 Central binomial coefficients: binomial(2n,n) = (2n)!/(n!)^2.

A002802 a(n) = (2n+3)!/(6n!*(n+1)!).