A068555 -id:A068555 - 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

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
Amulya Kumar Bag, Binomial theorem in ancient India, Indian Journal of History of Science, vol. 1 (1966), pp. 68-74.
Arthur T. Benjamin and Jennifer Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 63ff.
Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 306.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 68-74.
Paul Curtz, Intégration numérique des systèmes différentiels à conditions initiales, Centre de Calcul Scientifique de l'Armement, Arcueil, 1969.
A. W. F. Edwards, Pascal's Arithmetical Triangle, 2002.
William Feller, An Introduction to Probability Theory and Its Application, Vol. 1, 2nd ed. New York: Wiley, p. 36, 1968.
Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, p. 155.
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §4.4 Powers and Roots, pp. 140-141.
David Hök, Parvisa mönster i permutationer [Swedish], 2007.
Donald E. Knuth, The Art of Computer Programming, Vol. 1, 2nd ed., p. 52.
Sergei K. Lando, Lecture on Generating Functions, Amer. Math. Soc., Providence, R.I., 2003, pp. 60-61.
Blaise Pascal, Traité du triangle arithmétique, avec quelques autres petits traitez sur la mesme matière, Desprez, Paris, 1665.
Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 71.
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 271-275.
A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 6.
John Riordan, Combinatorial Identities, Wiley, 1968, p. 2.
Robert Sedgewick and Philippe Flajolet, An Introduction to the Analysis of Algorithms, Addison-Wesley, Reading, MA, 1996, p. 143.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 6, pages 43-52.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 13, 30-33.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 115-118.
Douglas B. West, Combinatorial Mathematics, Cambridge, 2021, p. 25.

N. J. A. Sloane, First 141 rows of Pascal's triangle, formatted as a simple linear sequence: (n, a(n)), n=0..10152.
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].
Tewodros Amdeberhan, Moa Apagodu, and Doron Zeilberger, Wilf's "Snake Oil" Method Proves an Identity in The Motzkin Triangle, arXiv:1507.07660 [math.CO], 2015.
Said Amrouche and Hacène Belbachir, Asymmetric extension of Pascal-Dellanoy triangles, arXiv:2001.11665 [math.CO], 2020.
Shaun V. Ault and Charles Kicey, Counting paths in corridors using circular Pascal arrays, Discrete Mathematics, Vol. 332, No. 6 (2014), pp. 45-54.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38 (2012), pp. 1871-1876.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42 (2012), pp. 2053-2059.
Amulya Kumar Bag, Binomial theorem in ancient India, Indian Journal of History of Science, Vol. 1 (1966), pp. 68-74.
Armen G. Bagdasaryan and Ovidiu Bagdasar, On some results concerning generalized arithmetic triangles, Electronic Notes in Discrete Mathematics, Vol. 67 (2018), pp. 71-77.
Peter Bala, A combinatorial interpretation for the binomial coefficients, 2013.
Cyril Banderier and Donatella Merlini, Lattice paths with an infinite set of jumps, Proceedings of the 14th International Conference on Formal Power Series and Algebraic Combinatorics, Melbourne, Australia. 2002.
J. Fernando Barbero G., Jesús Salas, and Eduardo J. S. Villaseñor, Bivariate Generating Functions for a Class of Linear Recurrences. I. General Structure, arXiv:1307.2010 [math.CO], 2013.
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays , JIS, Vol. 12 (2009) Article 09.8.6.
Paul Barry, Eulerian polynomials as moments, via exponential Riordan arrays, arXiv:1105.3043 [math.CO], 2011.
Paul Barry, Combinatorial polynomials as moments, Hankel transforms and exponential Riordan arrays, arXiv:1105.3044 [math.CO], 2011.
Paul Barry, On the Central Coefficients of Bell Matrices, J. Int. Seq., Vol. 14 (2011) Article 11.4.3, example 2.
Paul Barry, Riordan-Bernstein Polynomials, Hankel Transforms and Somos Sequences, Journal of Integer Sequences, Vol. 15 (2012), Article 12.8.2.
Paul Barry, On the Central Coefficients of Riordan Matrices, Journal of Integer Sequences, Vol. 16 (2013), Article 13.5.1.
Paul Barry, A Note on a Family of Generalized Pascal Matrices Defined by Riordan Arrays, Journal of Integer Sequences, Vol. 16 (2013), Article 13.5.4.
Paul Barry, On the Inverses of a Family of Pascal-Like Matrices Defined by Riordan Arrays, Journal of Integer Sequences, Vol. 16 (2013), Article 13.5.6.
Paul Barry, On the Connection Coefficients of the Chebyshev-Boubaker polynomials, The Scientific World Journal, Vol. 2013 (2013), Article ID 657806, 10 pages.
Paul Barry, General Eulerian Polynomials as Moments Using Exponential Riordan Arrays, Journal of Integer Sequences, Vol. 16 (2013), Article 13.9.6.
Paul Barry, Riordan arrays, generalized Narayana triangles, and series reversion, Linear Algebra and its Applications, Vol. 491 (2016), pp. 343-385.
Paul Barry, The Gamma-Vectors of Pascal-like Triangles Defined by Riordan Arrays, arXiv:1804.05027 [math.CO], 2018.
Paul Barry, On the f-Matrices of Pascal-like Triangles Defined by Riordan Arrays, arXiv:1805.02274 [math.CO], 2018.
Paul Barry, The Central Coefficients of a Family of Pascal-like Triangles and Colored Lattice Paths, J. Int. Seq., Vol. 22 (2019), Article 19.1.3.
Paul Barry, On the halves of a Riordan array and their antecedents, arXiv:1906.06373 [math.CO], 2019.
Paul Barry, On the r-shifted central triangles of a Riordan array, arXiv:1906.01328 [math.CO], 2019.
Paul Barry, Generalized Catalan Numbers Associated with a Family of Pascal-like Triangles, J. Int. Seq., Vol. 22 (2019), Article 19.5.8.
Paul Barry, A Note on Riordan Arrays with Catalan Halves, arXiv:1912.01124 [math.CO], 2019.
Paul Barry, Chebyshev moments and Riordan involutions, arXiv:1912.11845 [math.CO], 2019.
Paul Barry, Characterizations of the Borel triangle and Borel polynomials, arXiv:2001.08799 [math.CO], 2020.
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Paul Barry, Extensions of Riordan Arrays and Their Applications, Mathematics (2025) Vol. 13, No. 2, 242. See p. 13.
Paul Barry, Notes on Riordan arrays and lattice paths, arXiv:2504.09719 [math.CO], 2025. See p. 2.
Paul Barry and Aoife Hennessy, Four-term Recurrences, Orthogonal Polynomials and Riordan Arrays, Journal of Integer Sequences, Vol. 15 (2012), Article 12.4.2.
Jonathan W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977v1 [math.NT], J. London Math. Soc. (2), Vol. 79 (2009), pp. 422-444.
Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids, English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see p. 4.
Michael Bukata, Ryan Kulwicki, Nicholas Lewandowski, Lara Pudwell, Jacob Roth and Teresa Wheeland, Distributions of Statistics over Pattern-Avoiding Permutations, arXiv preprint arXiv:1812.07112 [math.CO], 2018.
Douglas Butler, Pascal's Triangle.
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, and Graça Tomaz, Intrinsic Properties of a Non-Symmetric Number Triangle, J. Int. Seq., Vol. 26 (2023), Article 23.4.8.
Naiomi T. Cameron and Asamoah Nkwanta, On Some (Pseudo) Involutions in the Riordan Group, Journal of Integer Sequences, Vol. 8 (2005), Article 05.3.7.
Dario T. de Castro, p-adic Order of Positive Integers via Binomial Coefficients, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 22, Paper A61, 2022.
Ji Young Choi, Digit Sums Generalizing Binomial Coefficients, J. Int. Seq., Vol. 22 (2019), Article 19.8.3.
Cristian Cobeli and Alexandru Zaharescu, Promenade around Pascal Triangle - Number Motives, Bull. Math. Soc. Sci. Math. Roumanie, Tome 56(104) No. 1 (2013), pp. 73-98.
CombOS - Combinatorial Object Server, Generate combinations.
J. H. Conway and N. J. A. Sloane, Low-dimensional lattices. VII Coordination sequences, Proc. R. Soc. Lond. A, Vo. 453, No. 1966 (1997), pp. 2369-2389.
Tom Copeland, Infinigens, the Pascal Triangle, and the Witt and Virasoro Algebras.
Persi Diaconis, The distribution of leading digits and uniform distribution mod 1, Ann. Probability, Vol. 5 (1977), pp. 72-81.
Karl Dilcher and Kenneth B. Stolarsky, A Pascal-Type Triangle Characterizing Twin Primes, The American Mathematical Monthly, Vol. 112, No. 8 (Oct 2005), pp. 673-681.
Tomislav Došlic and Darko Veljan, Logarithmic behavior of some combinatorial sequences, Discrete Math., Vol. 308, No. 11 (2008), pp. 2182-2212. MR2404544 (2009j:05019).
Steffen Eger, Some Elementary Congruences for the Number of Weighted Integer Compositions, J. Int. Seq., Vol. 18 (2015), Article 15.4.1.
Leonhard Euler, On the expansion of the power of any polynomial (1+x+x^2+x^3+x^4+etc.)^n, arXiv:math/0505425 [math.HO], 2005. See also The Euler Archive, item E709.
Jackson Evoniuk, Steven Klee, and Van Magnan, Enumerating Minimal Length Lattice Paths, J. Int. Seq., Vol. 21 (2018), Article 18.3.6.
A. Farina, S. Giompapa, A. Graziano, A. Liburdi, M. Ravanelli, and F. Zirilli, Tartaglia-Pascal's triangle: a historical perspective with applications, Signal, Image and Video Processing, Vol. 7, No. 1 (January 2013), pp. 173-188.
Steven Finch, Pascal Sebah, and Zai-Qiao Bai, Odd Entries in Pascal's Trinomial Triangle, arXiv:0802.2654 [math.NT], 2008.
David Fowler, The binomial coefficient function, Amer. Math. Monthly, Vol. 103, No. 1 (1996), pp. 1-17.
Shishuo Fu and Yaling Wang, Bijective recurrences concerning two Schröder triangles, arXiv:1908.03912 [math.CO], 2019.
Tom Halverson and Theodore N. Jacobson, Set-partition tableaux and representations of diagram algebras, arXiv:1808.08118 [math.RT], 2018.
T. Han and S. Kitaev, Joint distributions of statistics over permutations avoiding two patterns of length 3, arXiv:2311.02974 [math.CO], 2023
Brady Haran and Casandra Monroe, Pascal's Triangle, Numberphile video (2017).
Tian-Xiao He and Renzo Sprugnoli, Sequence characterization of Riordan arrays, Discrete Math., Vol. 309, No. 12 (2009), pp. 3962-3974.
Nick Hobson, Python program for A007318.
V. E. Hoggatt, Jr. and Marjorie Bicknell, Catalan and related sequences arising from inverses of Pascal's triangle matrices, Fib. Quart., Vol. 14, No. 5 (1976), pp. 395-405.
Matthew Hubbard and Tom Roby, Pascal's Triangle From Top to Bottom. [archived page]
Charles Jordan, Calculus of Finite Differences (p. 65).
Subhash Kak, The golden mean and the physics of aesthetics, in: B. Yadav and M. Mohan (eds.), Ancient Indian Leaps into Mathematics, Birkhäuser, Boston, MA, 2009, pp. 111-119; arXiv preprint, arXiv:physics/0411195 [physics.hist-ph], 2004.
Petro Kolosov, Polynomial identities involving Pascal's triangle rows, 2022.
Wolfdieter Lang, On generalizations of Stirling number triangles, J. Integer Seq., Vol. 3 (2000), Article 00.2.4.
Eitan Y. Levine, GCD formula proof.
Meng Li and Ron Goldman, Limits of sums for binomial and Eulerian numbers and their associated distributions, Discrete mathematics, Vol. 343, No. 7 (2020), 111870.
P. A. MacMahon, Memoir on the Theory of the Compositions of Numbers, Phil. Trans. Royal Soc. London A, Vol. 184 (1893), pp. 835-901.
Mathforum, Pascal's Triangle
Carl McTague, On the Greatest Common Divisor of C(q*n,n), C(q*n,2*n), ...C(q*n,q*n-q), arXiv:1510.06696 [math.CO], 2015.
D. Merlini, R. Sprugnoli, and M. C. Verri, An algebra for proper generating trees, in: D. Gardy and A. Mokkadem (eds.), Mathematics and Computer Science, Trends in Mathematics, Birkhäuser, Basel, 2000, pp. 127-139; alternative link.
Donatella Merlini, Francesca Uncini, and M. Cecilia Verri, A unified approach to the study of general and palindromic compositions, Integers, Vol. 4 (2004), A23, 26 pp.
Ângela Mestre and José Agapito, A Family of Riordan Group Automorphisms, J. Int. Seq., Vol. 22 (2019), Article 19.8.5.
Pierre Remond de Montmort, Essay d'analyse sur les jeux de hazard, Paris: Chez Jacque Quillau, 1708, p. 80.
Yossi Moshe, The density of 0's in recurrence double sequences, J. Number Theory, Vol. 103 (2003), pp. 109-121.
Lili Mu and Sai-nan Zheng, On the Total Positivity of Delannoy-Like Triangles, Journal of Integer Sequences, Vol. 20 (2017), Article 17.1.6.
Abdelkader Necer, Séries formelles et produit de Hadamard, Journal de théorie des nombres de Bordeaux, Vol. 9, No. 2 (1997), pp. 319-335.
Asamoah Nkwanta and Earl R. Barnes, Two Catalan-type Riordan Arrays and their Connections to the Chebyshev Polynomials of the First Kind, Journal of Integer Sequences, Vol. 15 (2012), Article 12.3.3.
Asamoah Nkwanta and Akalu Tefera, Curious Relations and Identities Involving the Catalan Generating Function and Numbers, Journal of Integer Sequences, Vol. 16 (2013), Article 13.9.5.
Mustafa A. A. Obaid, S. Khalid Nauman, Wafaa M. Fakieh, and Claus Michael Ringel, The numbers of support-tilting modules for a Dynkin algebra, 2014.
OEIS Wiki, Binomial coefficients
Richard L. Ollerton and Anthony G. Shannon, Some properties of generalized Pascal squares and triangles, Fib. Q., Vol. 36, No. 2 (1998), pp. 98-109.
Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003.
Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003. [Cached copy, with permission (pdf only)]
Balak Ram, Common factors of n!/(m!(n-m)!), (m = 1, 2, ... n-1), Journal of the Indian Mathematical Club (Madras) 1 (1909), pp. 39-43.
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
Franck Ramaharo, A generating polynomial for the pretzel knot, arXiv:1805.10680 [math.CO], 2018.
Franck Ramaharo, A generating polynomial for the two-bridge knot with Conway's notation C(n,r), arXiv:1902.08989 [math.CO], 2019.
Franck Ramaharo, A bracket polynomial for 2-tangle shadows, arXiv:2002.06672 [math.CO], 2020.
Jack Ramsay, On Arithmetical Triangles, The Pulse of Long Island, June 1965 [Mentions application to design of antenna arrays. Annotated scan.]
Thomas M. Richardson, The Reciprocal Pascal Matrix, arXiv preprint arXiv:1405.6315 [math.CO], 2014.
Yuriy Shablya, Dmitry Kruchinin, and Vladimir Kruchinin, Method for Developing Combinatorial Generation Algorithms Based on AND/OR Trees and Its Application, Mathematics, Vol. 8, No. 6 (2020), 962.
Louis W. Shapiro, Seyoum Getu, Wen-Jin Woan, and Leon C. Woodson, The Riordan group, Discrete Applied Math., Vol. 34 (1991), pp. 229-239.
N. J. A. Sloane, My favorite integer sequences, in Sequences and their Applications (Proceedings of SETA '98).
N. J. A. Sloane, Triangle showing silhouette of first 30 rows of Pascal's triangle (after Cobeli and Zaharescu)
N. J. A. Sloane, The OEIS: A Fingerprint File for Mathematics, arXiv:2105.05111 [math.HO], 2021.
N. J. A. Sloane, "A Handbook of Integer Sequences" Fifty Years Later, arXiv:2301.03149 [math.NT], 2023, p. 5.
Hermann Stamm-Wilbrandt, Compute C(n+m,...) based on C(n,...) and C(m,...) values animation.
Igor Victorovich Statsenko, On the ordinal numbers of triangles of generalized special numbers, Innovation science No 2-2, State Ufa, Aeterna Publishing House, 2024, pp. 15-19. In Russian.
Christopher Stover and Eric W. Weisstein, Composition. From MathWorld - A Wolfram Web Resource.
Gérard Villemin's Almanach of Numbers, Triangle de Pascal.
Eric Weisstein's World of Mathematics, Pascal's Triangle.
Wikipedia, Pascal's triangle.
Herbert S. Wilf, Generatingfunctionology, 2nd edn., Academic Press, NY, 1994, pp. 12ff.
Ken Williams, Mathforum, Interactive Pascal's Triangle.
Doron Zeilberger, The Combinatorial Astrology of Rabbi Abraham Ibn Ezra, arXiv:math/9809136 [math.CO], 1998.
Chris Zheng and Jeffrey Zheng, Triangular Numbers and Their Inherent Properties, Variant Construction from Theoretical Foundation to Applications, Springer, Singapore, 51-65.
Index entries for triangles and arrays related to Pascal's triangle.
Index entries for "core" sequences.
Index entries for sequences related to Benford's law.

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

A046521 Array T(i,j) = binomial(-1/2-i,j)*(-4)^j, i,j >= 0 read by antidiagonals going down.

Original entry on oeis.org

1, 2, 1, 6, 6, 1, 20, 30, 10, 1, 70, 140, 70, 14, 1, 252, 630, 420, 126, 18, 1, 924, 2772, 2310, 924, 198, 22, 1, 3432, 12012, 12012, 6006, 1716, 286, 26, 1, 12870, 51480, 60060, 36036, 12870, 2860, 390, 30, 1, 48620, 218790, 291720, 204204, 87516, 24310

Offset: 0

Wolfdieter Lang

Or, a triangle related to A000984 (central binomial) and A000302 (powers of 4).
This is an example of a Riordan matrix. See the Shapiro et al. reference quoted under A053121 and Notes 1 and 2 of the Wolfdieter Lang reference, p. 306.
As a number triangle, this is the Riordan array (1/sqrt(1-4x),x/(1-4x)). - Paul Barry, May 30 2005
The A- and Z- sequences for this Riordan matrix are (see the Wolfdieter Lang link under A006232 for the D. G. Rogers, D. Merlini et al. and R. Sprugnoli references on Riordan A- and Z-sequences with a summary): A-sequence [1,4,0,0,0,...] and Z-sequence 4+2*A000108(n)*(-1)^(n+1)=[2, 2, -4, 10, -28, 84, -264, 858, -2860, 9724, -33592, 117572, -416024, 1485800, -5348880, 19389690, -70715340, 259289580, -955277400, 3534526380], n >= 0. The o.g.f. for the Z-sequence is 4-2*c(-x) with the Catalan number o.g.f. c(x). - Wolfdieter Lang, Jun 01 2007
As a triangle, T(2n,n) is A001448. Row sums are A046748. Diagonal sums are A176280. - Paul Barry, Apr 14 2010
From Wolfdieter Lang, Aug 10 2017: (Start)
The row polynomials R(n, x) of Riordan triangles R = (G(x), F(x)), with F(x)= x*Fhat(x), belong to the class of Boas-Buck polynomials (see the reference). Hence they satisfy the Boas-Buck identity (we use the notation of Rainville, Theorem 50, p. 141):
  (E_x - n*1)*R(n, x) = -Sum_{k=0..n-1} (alpha(k)*1 + beta(k)*E_x)*R(n-1.k, x), for n >= 0, where E_x = x*d/dx (Euler operator). The Boas-Buck sequences are given by alpha(k) := [x^k] ((d/dx)log(G(x))) and beta(k) := [x^k] (d/dx)log(Fhat(x)).
This entails a recurrence for the sequence of column m of the Riordan triangle T, n > m >= 0: T(n, m) = (1/(n-m))*Sum_{k=m..n-1} (alpha(n-1-k) + m*beta(n-1-k))*T(k, m), with input T(m,m).
For the present case the Boas-Buck identity for the row polynomials is (E_x - n*1)*R(n, x) = -Sum_{k=0..n-1} 2^(2*k+1)*(1 + 2*E_x)*R(n-1-k, x), for n >= 0. For the ensuing recurrence for the columns m of the triangle T see the formula and example section. (End)
From Peter Bala, Mar 04 2018: (Start)
The following two remarks are particular cases of more general results for Riordan arrays of the form (f(x), x/(1 - k*x)).
1) Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,4*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(4*x)/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(x) * the e.g.f. for the polynomial R(n,4*x). For example, when n = 3 we have exp(x)*(20 + 30*(4*x) + 10*(4*x)^2/2! + (4*x)^3/3!) = 20 + 140*x + 420*x^2/2! + 924*x^3/3! + 1716*x^4/4! + ....
2) Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. P(n,x) is the n-th degree Taylor polynomial of (1 + 4*x)^(n-1/2) about 0. For example, for n = 4 we have (1 + 4*x)^(7/2) = 70*x^4 + 140*x^3 + 70*x^2 + 14*x + 1 + O(x^5).
Let C(x) = (1 - sqrt(1 - 4*x))/(2*x) denote the o.g.f. of the Catalan numbers A000108. The derivatives of C(x) are determined by the identity (-1)^n * x^n/n! * (d/dx)^n(C(x)) = 1/(2*x)*( 1 - P(n,-x)/(1 - 4*x)^(n-1/2) ), n = 0,1,2,.... See Lang 2002. Cf. A283150 and A283151. (End)

			Array begins:
  1,  2,   6,  20,   70, ...
  1,  6,  30, 140,  630, ...
  1, 10,  70, 420, 2310, ...
  1, 14, 126, 924, 6006, ...
Recurrence from A-sequence: 140 = a(4,1) = 20 + 4*30.
Recurrence from Z-sequence: 252 = a(5,0) = 2*70 + 2*140 - 4*70 + 10*14 - 28*1.
From _Paul Barry_, Apr 14 2010: (Start)
As a number triangle, T(n, m) begins:
n\k       0      1       2       3      4      5     6    7   8  9 10 ...
0:        1
1:        2      1
2:        6      6       1
3:       20     30      10       1
4:       70    140      70      14      1
5:      252    630     420     126     18      1
6:      924   2772    2310     924    198     22     1
7:     3432  12012   12012    6006   1716    286    26    1
8:    12870  51480   60060   36036  12870   2860   390   30   1
9:    48620 218790  291720  204204  87516  24310  4420  510  34  1
10:  184756 923780 1385670 1108536 554268 184756 41990 6460 646 38  1
... [Reformatted and extended by _Wolfdieter Lang_, Aug 10 2017]
Production matrix begins
      2, 1,
      2, 4, 1,
     -4, 0, 4, 1,
     10, 0, 0, 4, 1,
    -28, 0, 0, 0, 4, 1,
     84, 0, 0, 0, 0, 4, 1,
   -264, 0, 0, 0, 0, 0, 4, 1,
    858, 0, 0, 0, 0, 0, 0, 4, 1,
  -2860, 0, 0, 0, 0, 0, 0, 0, 4, 1 (End)
Boas-Buck recurrence for column m = 2, and n = 4: T(4, 2) = (2*(2*2+1)/2) * Sum_{k=2..3} 4^(3-k)*T(k, 2) = 5*(4*1 + 1*10) = 70. - _Wolfdieter Lang_, Aug 10 2017
From _Peter Bala_, Feb 15 2018: (Start)
With C(x) = (1 - sqrt( 1 - 4*x))/(2*x),
-x^3/3! * (d/dx)^3(C(x)) = 1/(2*x)*( 1 - (1 - 10*x + 30*x^2 - 20*x^3)/(1 - 4*x)^(5/2) ).
x^4/4! * (d/dx)^4(C(x)) = 1/(2*x)*( 1 - (1 - 14*x + 70*x^2 - 140*x^3 + 70*x^4 )/(1 - 4*x)^(7/2) ). (End)

Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.

Muniru A Asiru, Antidiagonals n = 0..50, flattened
Peter Bala, A 4-parameter family of embedded Riordan arrays
Peter Bala, A note on the diagonals of a proper Riordan Array
Paul Barry, Embedding structures associated with Riordan arrays and moment matrices, arXiv preprint arXiv:1312.0583 [math.CO], 2013.
Jonathan W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc. (2) 79 2009, 422-444.
Wolfdieter Lang, First 10 rows.
Wolfdieter Lang, On polynomials related to derivatives of the generating function of Catalan numbers, Fib. Quart. 40,4 (2002) 299-313; T(n,m) is called B(n,m) there.
Benjamin Lovitz and Nathaniel Johnston, A hierarchy of eigencomputations for polynomial optimization on the sphere, arXiv:2310.17827 [math.OC], 2023. See pp. 35, 39. See also Math. Prog. (2025) Series A.
Helmut Prodinger, Some information about the binomial transform, The Fibonacci Quarterly, 32, 1994, 412-415.

Columns include: A000984 (m=0), A002457 (m=1), A002802 (m=2), A020918 (m=3), A020920 (m=4), A020922 (m=5), A020924 (m=6), A020926 (m=7), A020928 (m=8), A020930 (m=9), A020932 (m=10).
Row sums: A046748.
Cf. A001147, A006882, A007318, A068555, A111959, A283150, A283151.

Flat(List([0..9],n->List([0..n],m->Binomial(2*n,n)*Binomial(n,m)/Binomial(2*m,m)))); # Muniru A Asiru, Jul 19 2018

[Binomial(n+1,k+1)*Catalan(n)/Catalan(k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 28 2024

t[i_, j_] := If[i < 0 || j < 0, 0, (2*i + 2*j)!*i!/(2*i)!/(i + j)!/j!]; Flatten[Reverse /@ Table[t[n, k - n] , {k, 0, 9}, {n, k, 0, -1}]][[1 ;; 51]] (* Jean-François Alcover, Jun 01 2011, after PARI prog. *)

T(i,j)=if(i<0 || j<0,0,(2*i+2*j)!*i!/(2*i)!/(i+j)!/j!)

def A046521(n,k): return binomial(n+1, k+1)*catalan_number(n)/catalan_number(k)
flatten([[A046521(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 28 2024

T(n, m) = binomial(2*n, n)*binomial(n, m)/binomial(2*m, m), n >= m >= 0.
G.f. for column m: ((x/(1-4*x))^m)/sqrt(1-4*x).
Recurrence from the A-sequence given above: a(n,m) = a(n-1,m-1) + 4*a(n-1,m), for n >= m >= 1.
Recurrence from the Z-sequence given above: a(n,0) = Sum_{j=0..n-1} Z(j)*a(n-1,j), n >= 1; a(0,0)=1.
As a number triangle, T(n,k) = C(2*n,n)*C(n,k)/C(2*k,k) = C(n-1/2,n-k)*4^(n-k). - Paul Barry, Apr 14 2010
From Peter Bala, Apr 11 2012: (Start):
One of three infinite families of integral factorial ratio sequences of height 1 (see Bober, Theorem 1.2). The other two are A007318 and A068555.
The triangular array equals exp(S), where the infinitesimal generator S has [2,6,10,14,18,...] on the main subdiagonal and zeros elsewhere.
Recurrence equation for the square array: T(n+1,k) = (k+1)/(4*n+2)*T(n,k+1). (End)
T(n,k) = 4^(n-k)*A006882(2*n - 1)/(A006882(2*n - 2*k)*A006882(2*k - 1)) = 4^(n-k)*(2*n - 1)!!/((2*n - 2*k)!*(2*k - 1)!!). - Peter Bala, Nov 07 2016
Boas-Buck recurrence for column m, m > n >= 0: T(n, m) = (2*(2*m+1)/(n-m))*Sum_{k=m..n-1} 4^(n-1-k)*T(k, m), with input T(n, n) = 1. See a comment above. - Wolfdieter Lang, Aug 10 2017
From Peter Bala, Aug 13 2021: (Start)
Analogous to the binomial transform we have the following sequence transformation formula: g(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*f(k) iff f(n) = Sum_{k = 0..n} (-1)^(n-k)*T(n,k)*b^(n-k)*g(k). See Prodinger, bottom of p. 413, with b replaced with 4*b, c = 1 and d = 1/2.
Equivalently, if F(x) = Sum_{n >= 0} f(n)*x^n and G(x) = Sum_{n >= 0} g(n)*x^n are a pair of  formal power series then
G(x) = 1/sqrt(1 - 4*b*x) * F(x/(1 - 4*b*x)) iff F(x) = 1/sqrt(1 + 4*b*x) * G(x/(1 + 4*b*x)).
The m-th power of this array has entries m^(n-k)*T(n,k). (End)

A082590 Expansion of 1/((1 - 2x)sqrt(1 - 4*x)).

Original entry on oeis.org

1, 4, 14, 48, 166, 584, 2092, 7616, 28102, 104824, 394404, 1494240, 5692636, 21785872, 83688344, 322494208, 1246068806, 4825743832, 18726622964, 72798509728, 283443548276, 1105144970992, 4314388905704, 16862208539008, 65972020761116, 258354647959984, 1012627828868072

Offset: 0

Vladeta Jovovic, May 13 2003

nonn

Row sums of A068555 and A112336. - Paul Barry, Sep 04 2005
Hankel transform is 2^n*(-1)^C(n+1,2) (A120617). - Paul Barry, Apr 26 2009
Number of n-lettered words in the alphabet {1, 2, 3, 4} with as many occurrences of the substring (consecutive subword) [1, 2] as of [1, 3]. - N. J. A. Sloane, Apr 08 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Shalosh B. Ekhad and Doron Zeilberger, Automatic Solution of Richard Stanley's Amer. Math. Monthly Problem #11610 and ANY Problem of That Type, arXiv:1112.6207 [math.CO], 2011. See subpages for rigorous derivations of the g.f., the recurrence, asymptotics for this sequence.
Alejandro Erickson and Frank Ruskey, Enumerating maximal tatami mat coverings of square grids with v vertical dominoes, arXiv:1304.0070 [math.CO], 2013.
Y. Kamiyama, On the middle dimensional homology classes of equilateral polygon spaces, arXiv:1507.03161 [math.AT], 2015.

Bisection of A226302.

Cf. A034430, A068555, A076729, A112336.

A082590 := proc(n)
    coeftayl( 1/(1-2*x)/sqrt(1-4*x),x=0,n) ;
end proc: # R. J. Mathar, Nov 06 2013
A082590 := n -> 2^n*JacobiP(n, 1/2, -1 - n, 3):
seq(simplify(A082590(n)), n = 0..26);  # Peter Luschny, Jan 22 2025

CoefficientList[ Series[ 1/((1 - 2*x)*Sqrt[1 - 4*x]), {x, 0, 25}], x] (* Jean-François Alcover, Mar 26 2013 *)
Table[2^(n) JacobiP[n, 1/2, -1-n, 3], {n, 0, 30}] (* Vincenzo Librandi, May 26 2013 *)

a(n) = 2^n*JacobiP(n, 1/2, -1-n, 3).
A034430(n) = (n!/2^n)*a(n). A076729(n) = n!*a(n).
a(n) = Sum_{k=0..n+1} binomial(2*n+2, k) * sin((n - k + 1)*Pi/2). - Paul Barry, Nov 02 2004
From Paul Barry, Sep 04 2005: (Start)
a(n) = Sum_{k=0..n} 2^(n-k)*binomial(2*k, k).
a(n) = Sum_{k=0..n} (2*k)! * (2*(n-k))!/(n!*k!*(n-k)!). (End)
a(n) = Sum_{k=0..n} C(2*n, n)*C(n, k)/C(2*n, 2*k). - Paul Barry, Mar 18 2007
G.f.: 1/(1 - 4*x + 2*x^2/(1 + x^2/(1 - 4*x + x^2/(1 + x^2/(1 - 4*x + x^2/(1 + ... (continued fraction). - Paul Barry, Apr 26 2009
D-finite with recurrence: n*a(n) + 2*(-3*n+1)*a(n-1) + 4*(2*n-1)*a(n-2) = 0. - R. J. Mathar, Dec 03 2012
a(n) ~ 2^(2*n + 1)/sqrt(Pi*n). - Vaclav Kotesovec, Aug 15 2013
a(n) = 2^(n + 1)*Pochhammer(1/2, n+1)*hyper2F1([1/2,-n], [3/2], -1)/n!. - Peter Luschny, Aug 02 2014
a(n) - 2*a(n-1) = A000984(n). - R. J. Mathar, Apr 24 2024
a(n) = 2^n*JacobiP(n, 1/2, -1 - n, 3). - Peter Luschny, Jan 22 2025

A248324 Square array read by antidiagonals downwards: super Patalan numbers of order 3.

Original entry on oeis.org

1, 3, 6, 18, 9, 45, 126, 36, 45, 360, 945, 189, 135, 270, 2970, 7371, 1134, 567, 648, 1782, 24948, 58968, 7371, 2835, 2268, 3564, 12474, 212058, 480168, 50544, 15795, 9720, 10692, 21384, 90882, 1817640, 3961386, 360126, 94770, 47385, 40095, 56133, 136323, 681615, 15677145, 33011550, 2640924, 600210, 252720, 173745, 187110, 318087, 908820, 5225715, 135868590

Offset: 0

Tom Richardson, Oct 04 2014

Generalization of super Catalan numbers of Gessel, A068555, based on Patalan numbers of order 3, A097188.

			T(0..4,0..4) is:
  1    3    18   126   945
  6    9    36   189   1134
  45   45   135  567   2835
  360  270  648  2268  9720
  2970 1782 3564 10692 40095

Thomas M. Richardson, The Super Patalan Numbers, arXiv:1410.5880 [math.CO], 2014.
Thomas M. Richardson, The Super Patalan Numbers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.3.3.

Cf. A068555, A248325. First column is A004988, first row is A004987. a(n,1) = -A004990(n+1) = 3*A097188(n). a(1,k) = -A004989(k+1).

T(0,0)=1, T(n,k) = T(n-1,k)*(9*n-3)/(n+k), T(n,k) = T(n,k-1)*(9*k-6)/(n+k).

G.f.: (x/(1-9*x)^(2/3)+y/(1-9*y)^(1/3))/(x+y-9*x*y).

A248325 Square array read by antidiagonals downwards: super Patalan numbers of order 4.

Original entry on oeis.org

1, 4, 12, 40, 24, 168, 480, 160, 224, 2464, 6240, 1440, 1120, 2464, 36960, 84864, 14976, 8064, 9856, 29568, 561792, 1188096, 169728, 69888, 59136, 98560, 374528, 8614144, 16972800, 2036736, 678912, 439296, 506880, 1070080, 4922368, 132903936, 246105600, 25459200, 7128576, 3734016, 3294720, 4815360

Offset: 0

Tom Richardson, Oct 04 2014

Generalization of super Catalan numbers of Gessel, A068555, based on Patalan numbers of order 4, A025749.

			T(0..4, 0..4) is:
  1      4      40     480    6240
  12     24     160    1440   14976
  168    224    1120   8064   69888
  2464   2464   9856   59136  439296
  36960  29568  98560  506880 3294720

Thomas M. Richardson, The Super Patalan Numbers, arXiv:1410.5880 [math.CO], 2014.
Thomas M. Richardson, The Super Patalan Numbers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.3.3.

Cf. A068555, A025749, A216702 (first column), A034385 (first row), A248324.

T(0,0)=1, T(n,k) = T(n-1,k)*(16*n-4)/(n+k), T(n,k) = T(n,k-1)*(16*k-12)/(n+k).

G.f.: (x/(1-16*x)^(3/4)+y/(1-16*y)^(1/4))/(x+y-16*x*y).

A248326 Square array read by downward antidiagonals: super Patalan numbers of order 5.

Original entry on oeis.org

1, 5, 20, 75, 50, 450, 1375, 500, 750, 10500, 27500, 6875, 5625, 13125, 249375, 577500, 110000, 61875, 78750, 249375, 5985000, 12512500, 1925000, 825000, 721875, 1246875, 4987500, 144637500, 277062500, 35750000, 12375000, 8250000, 9796875, 21375000, 103312500, 3512625000, 6233906250, 692656250

Offset: 0

Tom Richardson, Oct 04 2014

Generalization of super Catalan numbers of Gessel, A068555, based on Patalan numbers of order 5, A025750.

			T(0..4,0..4) is
  1       5       75      1375    27500
  20      50      500     6875    110000
  450     750     5625    61875   825000
  10500   13125   78750   721875  8250000
  249375  249375  1246875 9796875 97968750

Thomas M. Richardson, The Super Patalan Numbers, arXiv:1410.5880 [math.CO], 2014.
Thomas M. Richardson, The Super Patalan Numbers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.3.3.

Cf. A068555, A025750, A034688 (first row), A049382 (first column), A248324, A248325, A248328, A248329, A248332.

matrix(5, 5, nn, kk, n=nn-1;k=kk-1;(-1)^k*25^(n+k)*binomial(n-1/5,n+k)) \\ Michel Marcus, Oct 09 2014

T(0,0)=1, T(n,k) = T(n-1,k)*(25*n-5)/(n+k), T(n,k) = T(n,k-1)*(25*k-20)/(n+k).
G.f.: (x/(1-25*x)^(4/5)+y/(1-25*y)^(1/5))/(x+y-25*x*y).
T(n,k) = (-1)^k*25^(n+k)*binomial(n-1/5,n+k).

A248328 Square array read by antidiagonals downwards: super Patalan numbers of order 6.

Original entry on oeis.org

1, 6, 30, 126, 90, 990, 3276, 1260, 1980, 33660, 93366, 24570, 20790, 50490, 1161270, 2800980, 560196, 324324, 424116, 1393524, 40412196, 86830380, 14004900, 6162156, 5513508, 9754668, 40412196, 1414426860, 2753763480, 372130200, 132046200, 89791416, 108694872, 242473176, 1212365880

Offset: 0

Tom Richardson, Oct 04 2014

Generalization of super Catalan numbers, A068555, based on Patalan numbers of order 6, A025751.

			T(0..4,0..4) is
  1          6         126       3276      93366
  30         90        1260      24570     560196
  990        1980      20790     324324    6162156
  33660      50490     424116    5513508   89791416
  1161270    1393524   9754668   108694872 1548901926

Thomas M. Richardson, The Super Patalan Numbers, arXiv:1410.5880 [math.CO], 2014.
Thomas M. Richardson, The Super Patalan Numbers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.3.3.

Cf. A068555, A025751, A004993 (first row), A004994 (first column), A004995 (second row), A004996 (second column), A248324, A248325, A248326, A248329, A248332.

matrix(5, 5, nn, kk, n=nn-1;k=kk-1;(-1)^k*36^(n+k)*binomial(n-1/6,n+k)) \\ Michel Marcus, Oct 09 2014

T(0,0)=1, T(n,k) = T(n-1,k)*(36*n-6)/(n+k), T(n,k) = T(n,k-1)*(36*k-30)/(n+k).
G.f.: (x/(1-36*x)^(5/6)+y/(1-36*y)^(1/6))/(x+y-36*x*y).
T(n,k) = (-1)^k*36^(n+k)*binomial(n-1/6,n+k).

A248329 Square array read by antidiagonals downwards: super Patalan numbers of order 7.

Original entry on oeis.org

1, 7, 42, 196, 147, 1911, 6860, 2744, 4459, 89180, 264110, 72030, 62426, 156065, 4213755, 10722866, 2218524, 1310946, 1747928, 5899257, 200574738, 450360372, 75060062, 33647614, 30588740, 55059732, 234003861, 9594158301, 19365495996, 2702162232, 975780806, 672952280, 825895980, 1872030888

Offset: 0

Tom Richardson, Oct 04 2014

Generalization of super Catalan numbers, A068555, based on Patalan numbers of order 7, A025752.

			T(0..4,0..4) is
  1           7          196        6860       264110
  42          147        2744       72030      2218524
  1911        4459       62426      1310946    33647614
  89180       156065     1747928    30588740   672952280
  4213755     5899257    55059732   825895980  15898497615

Thomas M. Richardson, The Super Patalan Numbers, arXiv:1410.5880 [math.CO], 2014.
Thomas M. Richardson, The Super Patalan Numbers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.3.3.

Cf. A068555, A025752, A034835 (first row), A216703 (first column), A248324, A248325, A248326, A248328, A248332.

matrix(5, 5, nn, kk, n=nn-1;k=kk-1;(-1)^k*49^(n+k)*binomial(n-1/7,n+k)) \\ Michel Marcus, Oct 09 2014

T(0,0)=1, T(n,k) = T(n-1,k)*(49*n-7)/(n+k), T(n,k) = T(n,k-1)*(49*k-42)/(n+k).
G.f.: (x/(1-49*x)^(6/7)+y/(1-49*y)^(1/7))/(x+y-49*x*y).
T(n,k) = (-1)^k*49^(n+k)*binomial(n-1/7,n+k).

A248332 Square array read by antidiagonals downwards: super Patalan numbers of order 8.

Original entry on oeis.org

1, 8, 56, 288, 224, 3360, 13056, 5376, 8960, 206080, 652800, 182784, 161280, 412160, 12776960, 34467840, 7311360, 4386816, 5935104, 20443136, 797282304, 1884241920, 321699840, 146227200, 134529024, 245317632, 1063043072, 49963024384, 105517547520, 15073935360, 5514854400, 3843686400

Offset: 0

Tom Richardson, Oct 04 2014

Generalization of super Catalan numbers, A068555, based on Patalan numbers of order 8, A025753.

			T(0..4,0..4) is
  1           8           288         13056       652800
  56          224         5376        182784      7311360
  3360        8960        161280      4386816     146227200
  206080      412160      5935104     134529024   3843686400
  12776960    20443136    245317632   4766171136  119154278400

Thomas M. Richardson, The Super Patalan Numbers, arXiv:1410.5880 [math.CO], 2014.
Thomas M. Richardson, The Super Patalan Numbers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.3.3.

Cf. A068555, A025753, A034977 (first row), A216704 (first column), A248324, A248325, A248326, A248328, A248329.

matrix(5, 5, nn, kk, n=nn-1;k=kk-1;(-1)^k*64^(n+k)*binomial(n-1/8,n+k)) \\ Michel Marcus, Oct 09 2014

T(0,0)=1, T(n,k) = T(n-1,k)*(64*n-8)/(n+k), T(n,k) = T(n,k-1)*(64*k-56)/(n+k).
G.f.: (x/(1-64*x)^(7/8)+y/(1-64*y)^(1/8))/(x+y-64*x*y).
T(n,k) = (-1)^k*64^(n+k)*binomial(n-1/8,n+k).

A078718 a(n) = (-1)^n(2n - 1)*CatalanNumber(n - 2) for n >= 2, a(n) = n for n = 0, 1.

Original entry on oeis.org

0, 1, 3, -5, 14, -45, 154, -546, 1980, -7293, 27170, -102102, 386308, -1469650, 5616324, -21544100, 82907640, -319929885, 1237518450, -4796857230, 18627909300, -72457790790, 282257178060, -1100982015900, 4299680491080, -16809921068850, 65785111513524, -257683159276956

Offset: 0

N. J. A. Sloane, Dec 20 2002

sign

Original definition: Let f(i, j) = Sum_{k=0..2*i} binomial(2*i, k)*binomial(2*j, i+j-k)*(-1)^(i+j-k) (this is essentially the same as the triangle in A068555); then a(n) = f(n, n-2)/2. Apart perhaps from signs, f(n, 0) and f(n, n) give A000984, f(n, 1) gives A002420, f(n, n-1) gives 2*A000984.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Guo-Niu Han, Enumeration of Standard Puzzles [Cached copy]

Cf. A000108, A068555, A000984, A002420.

s = (Sqrt[4*x+1]*(4*x^2+x) + 6*x^2 + x)/(Sqrt[4*x+1] + 4*x+1) + O[x]^28; CoefficientList[s, x] (* Jean-François Alcover, Dec 17 2016, after Vladimir Kruchinin *)
Table[(1/2)*Sum[ Binomial[2*n, k]*Binomial[2*(n - 2), 2*n - 2 - k]*(-1)^(2*n - 2 - k), {k, 0, 2*n}], {n, 0, 50}] (* G. C. Greubel, Feb 16 2017 *)
(* Mathematica returns CatalanNumber[-2] = 0 and CatalanNumber[-1] = -1. This extended definition is in accordance with the alternative definition of the Catalan numbers C(n) = binomial(2*n, n) - binomial(2*n, n-1). *)
a[n_] := (-1)^n (2 n - 1) CatalanNumber[n - 2];
Table[a[n], {n, 0, 21}] (* Peter Luschny, Nov 28 2021 *)

A(x) := x*(1 + sqrt(1+4*x))/2;
G(x) := (2*cosh(asinh((3^(3/2)*sqrt(x))/2)/3)*sinh(asinh((3^(3/2)*sqrt(x))/2)/3)* sqrt(x))/(sqrt(3)*sqrt((27*x)/4+1))+(4*sinh(asinh((3^(3/2)*sqrt(x))/2)/3)^2)/3  + 1;
taylor(x^2*diff(A(x),x)*G(A(x))/A(x),x,0,20); /* Vladimir Kruchinin, Dec 16 2016 */

concat([0,1], for(n=2,25, print1(sum(k=0,2*n, (1/2)* binomial(2*n,k)* binomial(2*( n-2),2*n-k-2)*(-1)^(2*n-k-2)), ", "))) \\ G. C. Greubel, Feb 16 2017

G.f.: x^2*A'(x)*G(A(x))/A(x), where A(x) = x*(1+sqrt(1+4*x))/2, G(x) =(2*cosh(asinh((3^(3/2)*sqrt(x))/2)/3)*sinh(asinh((3^(3/2)*sqrt(x))/2)/3)*sqrt(x))/(sqrt(3)*sqrt((27*x)/4+1))+(4*sinh(asinh((3^(3/2)*sqrt(x))/2)/3)^2)/3+1. - Vladimir Kruchinin, Dec 16 2016
G.f.: (sqrt(4*x+1)*(4*x^2+x)+6*x^2+x)/(sqrt(4*x+1)+4*x+1). - Vladimir Kruchinin, Dec 17 2016
a(n) ~ (-1)^n * 2^(2*n-3) / sqrt(Pi*n). - Vaclav Kotesovec, Dec 17 2016
a(n) = (1/2)*Sum_{k=0,..,2*n} ( binomial(2*n, k)*binomial(2*(n - 2), 2*n - 2 - k)*(-1)^(2*n - 2 - k) ), with a(0)=0, a(1)=1. - G. C. Greubel, Feb 16 2017
D-finite with recurrence (-n+1)*a(n) +6*(-2*n+5)*a(n-1) +16*(-2*n+7)*a(n-2)=0. - R. J. Mathar, Nov 22 2024

New name by Peter Luschny, Nov 28 2021

Showing 1-10 of 12 results. Next

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Axiom

GAP

Haskell

Magma

Maple

Mathematica

Maxima

PARI

Python

Python

Sage

Formula

Extensions

A046521 Array T(i,j) = binomial(-1/2-i,j)*(-4)^j, i,j >= 0 read by antidiagonals going down.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

GAP

Magma

Mathematica

PARI

SageMath

Formula

A082590 Expansion of 1/((1 - 2*x)*sqrt(1 - 4*x)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A248324 Square array read by antidiagonals downwards: super Patalan numbers of order 3.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A248325 Square array read by antidiagonals downwards: super Patalan numbers of order 4.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A248326 Square array read by downward antidiagonals: super Patalan numbers of order 5.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A082590 Expansion of 1/((1 - 2x)sqrt(1 - 4*x)).

A078718 a(n) = (-1)^n(2n - 1)*CatalanNumber(n - 2) for n >= 2, a(n) = n for n = 0, 1.