A107430 -id:A107430 - 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

A126869 a(n) = Sum_{k = 0..n} binomial(n,floor(k/2))*(-1)^(n-k).

Original entry on oeis.org

1, 0, 2, 0, 6, 0, 20, 0, 70, 0, 252, 0, 924, 0, 3432, 0, 12870, 0, 48620, 0, 184756, 0, 705432, 0, 2704156, 0, 10400600, 0, 40116600, 0, 155117520, 0, 601080390, 0, 2333606220, 0, 9075135300, 0, 35345263800, 0, 137846528820, 0, 538257874440, 0, 2104098963720, 0, 8233430727600, 0, 32247603683100, 0, 126410606437752, 0

Offset: 0

Philippe Deléham, Mar 16 2007

Hankel transform is 2^n. Successive binomial transforms are A002426, A000984, A026375, A081671, A098409, A098410.
From Andrew V. Sutherland, Feb 29 2008: (Start)
Counts returning walks of length n on a 1-d integer lattice with step set {-1,+1}.
Moment sequence of the trace of a random matrix in G = SO(2). If X = tr(A) is a random variable (A distributed with Haar measure on G), then a(n) = E[X^n].
Also the moment sequence of the trace of the k-th power of a random matrix in USp(2) = SU(2), for all k > 2.
(End)
From Paul Barry, Aug 10 2009: (Start)
The Hankel transform of 0,1,0,2,0,6,... is 0,-1,0,4,0,-16,0,... with general term I*(-4)^(n/2)(1 - (-1)^n)/4, I = sqrt(-1).
The Hankel transform of 1,1,0,2,0,6,... (which has g.f. 1 + x/sqrt(1 - 4*x^2)) is A164111. (End)
a(n) = A204293(2*n,n): central terms of the triangle in A204293. - Reinhard Zumkeller, Jan 14 2012
a(n) is the total number of closed walks (round trips) of length n on the graph P_N (a line with N nodes and N-1 edges), divided by N, in the limit N -> infinity. See a comment on A198632 and a link under A201198. - Wolfdieter Lang, Oct 10 2012

			a(4) = 6 {UUDD,UDUD,UDDU,DUUD,DUDU,DDUU}.

Lin Yang and S.-L. Yang, The parametric Pascal rhombus. Fib. Q., 57:4 (2019), 337-346.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Francesc Fite, Kiran S. Kedlaya, Victor Rotger and Andrew V. Sutherland, Sato-Tate distributions and Galois endomorphism modules in genus 2, arXiv preprint arXiv:1110.6638 [math.NT], 2011.
Francesc Fite and Andrew V. Sutherland, Sato-Tate distributions of twists of y^2= x^5-x and y^2= x^6+1, arXiv preprint arXiv:1203.1476 [math.NT], 2012. - From _N. J. A. Sloane_, Sep 14 2012
Nikita Gogin and Mika Hirvensalo, On the Moments of Squared Binomial Coefficients, (2020).
Kiran S. Kedlaya and Andrew V. Sutherland, Hyperelliptic curves, L-polynomials and random matrices, arXiv:0803.4462 [math.NT], 2008-2010.
Paveł Szabłowski, Beta distributions whose moment sequences are related to integer sequences listed in the OEIS, Contrib. Disc. Math. (2024) Vol. 19, No. 4, 85-109. See p. 99.

This is A000984 with interspersed zeros. m-th binomial transforms of A000984: A126869 (m = -2), A002426 (m = -1 and m = -3 for signed version), A000984 (m = 0 and m = -4 for signed version), A026375 (m = 1 and m = -5 for signed version), A081671 (m = 2 and m = -6 for signed version), A098409 (m = 3 and m = -7 for signed version), A098410 (m = 4 and m = -8 for signed version), A104454 (m = 5 and m = -9 for signed version).

Cf. A107430, A061554, A126120.

a126869 n = a204293_row (2*n) !! n  -- Reinhard Zumkeller, Jan 14 2012

seq((-1)^(n/2)*pochhammer(-n,n/2)/(n/2)!, n=0..43); # Peter Luschny, May 17 2013
seq(n!*coeff(series(hypergeom([],[1],x^2),x,n+1),x,n),n=0..42); # Peter Luschny, Jan 31 2015

Table[(-1)^Floor[n/2] HypergeometricPFQ[{-n,-n},{1},-1],{n,0,30}] (* Peter Luschny, Nov 01 2011 *)

A126869 = lambda n: (2^(n-1)*((-1)^n+1)*gamma((n+1)/2))/(sqrt(pi)*gamma((n+2)/2))
[A126869(n) for n in range(44)] # Peter Luschny, Sep 10 2014

From Andrew V. Sutherland, Feb 29 2008: (Start)
a(2*n) = binomial(2*n,n) = A000984(n); a(2*n+1) = 0.
a(n) = Sum_{k = 0..n} A107430(n,k)*(-1)^(n-k).
a(n) = Sum_{k = 0..n} A061554(n,k)*(-1)^k.
a(n) = (1/Pi)*Integral_{t = 0..Pi} cos^n(t) dt. (End)
E.g.f.: I_0(2*x) where I_n(x) is the modified Bessel function as a function of x. - Benjamin Phillabaum, Mar 10 2011
G.f.: A(x) = 1/sqrt(1 - 4*x^2). - Vladimir Kruchinin, Apr 16 2011
a(n) = (1/Pi)*Integral{x = -2..2} x^n/sqrt((2 - x)*(2 + x)). - Peter Luschny, Sep 12 2011
a(n) = (-1)^floor(n/2) * Hypergeometric([-n,-n],[1], -1). - Peter Luschny, Nov 01 2011
E.g.f.: E(0)/(1 - x) where E(k) = 1 - x/(1 - x/(x - (k+1)^2/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Apr 05 2013
E.g.f.: 1 + x^2/(Q(0) - x^2), where Q(k) = x^2 + (k+1)^2 - x^2*(k+1)^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Apr 28 2013
G.f.: 1/(1 - 2*x^2*Q(0)), where Q(k) = 1 + (4*k+1)*x^2/(k+1 - x^2*(2*k+2)*(4*k+3)/(2*x^2*(4*k+3) + (2*k+3)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 15 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 2*x/(2*x + (k+1)/(x*(2*k+1))/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
G.f.: G(0)/(1+x), where G(k) = 1 + x*(2+5*x)*(4*k+1)/((4*k+2)*(1+x)^2 - 2*(2*k+1)*(4*k+3)*x*(2+5*x)*(1+x)^2/((4*k+3)*x*(2+5*x) + 4*(k+1)*(1+x)^2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 19 2014
a(n) = 2^n*JacobiP(n,0,-1/2-n,-3). - Peter Luschny, Aug 02 2014
a(n) = (2^(n-1)*((-1)^n+1)*Gamma((n+1)/2))/(sqrt(Pi)*Gamma((n+2)/2)). - Peter Luschny, Sep 10 2014
a(n) = n!*[x^n]hypergeom([],[1],x^2). - Peter Luschny, Jan 31 2015
a(n) = 2^n*hypergeom([1/2,-n],[1],2). - Peter Luschny, Feb 03 2015
From Peter Bala, Jul 25 2016: (Start)
a(n) = (-1)^floor(n/2)*Sum_{k = 0..n} (-1)^k*binomial(n,k)^2.
D-finite with recurrence: a(n) = 4*(n - 1)/n * a(n-2) with a(0) = 1, a(1) = 0. (End)
From Ilya Gutkovskiy, Jul 25 2016: (Start)
Inverse binomial transform of A002426.
a(n) = Sum_{k=0..n} (-1)^k*A128014(k).
a(n) ~ 2^n*((-1)^n + 1)/sqrt(2*Pi*n). (End)

A061554 Square table read by antidiagonals: a(n,k) = binomial(n+k, floor(k/2)).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 6, 4, 4, 1, 1, 10, 10, 5, 5, 1, 1, 20, 15, 15, 6, 6, 1, 1, 35, 35, 21, 21, 7, 7, 1, 1, 70, 56, 56, 28, 28, 8, 8, 1, 1, 126, 126, 84, 84, 36, 36, 9, 9, 1, 1, 252, 210, 210, 120, 120, 45, 45, 10, 10, 1, 1, 462, 462, 330, 330, 165, 165, 55, 55, 11, 11, 1, 1

Offset: 0

Henry Bottomley, May 17 2001

Equivalently, a triangle read by rows, where the rows are obtained by sorting the elements of the rows of Pascal's triangle (A007318) into descending order. - Philippe Deléham, May 21 2005
Equivalently, as a triangle read by rows, this is T(n,k)=binomial(n,floor((n-k)/2)); column k then has e.g.f. Bessel_I(k,2x)+Bessel_I(k+1,2x). - Paul Barry, Feb 28 2006
Antidiagonal sums are A037952(n+1) = C(n+1,[n/2]). Matrix inverse is the row reversal of triangle A066170. Eigensequence is A125094(n) = Sum_{k=0..n-1} A125093(n-1,k)*A125094(k). - Paul D. Hanna, Nov 20 2006
Riordan array (1/(1-x-x^2*c(x^2)),x*c(x^2)); where c(x)=g.f.for Catalan numbers A000108. - Philippe Deléham, Mar 17 2007
Triangle T(n,k), 0<=k<=n, read by rows given by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+T(n-1,k+1) for k>=1. - Philippe Deléham, Mar 27 2007
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
T(n,k) is the number of paths from (0,k) to some (n,m) which never dip below y=0, touch y=0 at least once and are made up only of the steps (1,1) and (1,-1). This can be proved using the recurrence supplied by Deléham. - Gerald McGarvey, Oct 15 2008
Triangle read by rows = partial sums of A053121 terms starting from the right. - Gary W. Adamson, Oct 24 2008
As a subset of the "family of triangles" (Deleham comment of Sep 25 2007), beginning with a signed variant of A061554, M = (-1,0) =  (1; -1, 1; 2, -1, 1; -3, 3, -1, 1; ...) successive binomial transforms of M yield (0,1) - A089942; (1,2) - A039599; (2,3) - A124733; (3,4) - A124574; (4,5) - A126331; ... such that the binomial transform of the triangle generated from (n,n+1) = the triangle generated from (n+1,n+2). Similarly, another subset beginning with A053121 - (0,0), and taking successive binomial transforms yields (1,1) - A064189; (2,2) - A039598; (3,3) - A091965, ... By rows, the triangle generated from (n,n) can be obtained by taking pairwise sums from the (n-1,n) triangle starting from the right. For example, row 2 of (1,2) - A039599 = (2, 3, 1); and taking pairwise sums from the right we obtain (5, 4, 1) = row 2 of (2,2) - A039598. - Gary W. Adamson, Aug 04 2011
The triangle by rows (n) with alternating signs (+-+...) from the top as a set of simultaneous equations solves for diagonal lengths of odd N (N = 2n+1) regular polygons. The constants in each case are powers of c = 2*cos(2*Pi/N). By way of example, the first 3 rows relate to the heptagon and the simultaneous equations are (1,0,0) = 1; (-1,1,0) = c = 1.24697...; and (2,-1,1) = c^2. The answers are 1, 2.24697..., and 1.801...; the 3 distinct diagonal lengths of the heptagon with edge = 1. - Gary W. Adamson, Sep 07 2011

			The array starts:
   1, 1,  2,  3,  6, 10,  20,  35,   70,  126, ...
   1, 1,  3,  4, 10, 15,  35,  56,  126,  210, ...
   1, 1,  4,  5, 15, 21,  56,  84,  210,  330, ...
   1, 1,  5,  6, 21, 28,  84, 120,  330,  495, ...
   1, 1,  6,  7, 28, 36, 120, 165,  495,  715, ...
   1, 1,  7,  8, 36, 45, 165, 220,  715, 1001, ...
   1, 1,  8,  9, 45, 55, 220, 286, 1001, 1365, ...
   1, 1,  9, 10, 55, 66, 286, 364, 1365, 1820, ...
   1, 1, 10, 11, 66, 78, 364, 455, 1820, 2380, ...
   1, 1, 11, 12, 78, 91, 455, 560, 2380, 3060, ...
Triangle (antidiagonal) version begins:
    1;
    1,   1;
    2,   1,   1;
    3,   3,   1,   1;
    6,   4,   4,   1,   1;
   10,  10,   5,   5,   1,   1;
   20,  15,  15,   6,   6,   1,  1;
   35,  35,  21,  21,   7,   7,  1,  1;
   70,  56,  56,  28,  28,   8,  8,  1,  1;
  126, 126,  84,  84,  36,  36,  9,  9,  1,  1;
  252, 210, 210, 120, 120,  45, 45, 10, 10,  1, 1;
  462, 462, 330, 330, 165, 165, 55, 55, 11, 11, 1, 1; ...
Matrix inverse begins:
   1;
  -1,  1;
  -1, -1,   1;
   1, -2,  -1,   1;
   1,  2,  -3,  -1,  1;
  -1,  3,   3,  -4, -1,  1;
  -1, -3,   6,   4, -5, -1,  1;
   1, -4,  -6,  10,  5, -6, -1,  1;
   1,  4, -10, -10, 15,  6, -7, -1, 1; ...
From _Paul Barry_, May 21 2009: (Start)
Production matrix is
  1, 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 (End)

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Reed Acton, T. Kyle Petersen, Blake Shirman, and Bridget Eileen Tenner, The clairvoyant maître d', arXiv:2401.11680 [math.CO], 2024. See p. 15.
Johann Cigler, Some remarks and conjectures related to lattice paths in strips along the x-axis, arXiv:1501.04750 [math.CO], 2015.
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.
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), Table 2.2.

Rows are A001405, A037952, A037955, A037951, A037956, A037953, A037957 etc. Columns are truncated pairs of A000012, A000027, A000217, A000292, A000332, A000389, A000579, etc. Main diagonal is alternate values of A051036.

Cf. A007318, A107430, A125094, A037952, A066170.

T := proc(n, k) option remember;
if n = k then 1 elif k < 0 or n < 0 or k > n then 0
elif k = 0 then T(n-1, 0) + T(n-1, 1) else T(n-1, k-1) + T(n-1, k+1) fi end:
for n from 0 to 9 do seq(T(n, k), k = 0..n) od; # Peter Luschny, May 25 2021

t[n_, k_] = Binomial[n, Floor[(n+1)/2 - (-1)^(n-k)*(k+1)/2]]; Flatten[Table[t[n, k], {n, 0, 11}, {k, 0, n}]] (* Jean-François Alcover, May 31 2011 *)

T(n,k)=binomial(n,(n+1)\2-(-1)^(n-k)*((k+1)\2))

As a triangle: T(n,k) = binomial(n,m) where m = floor((n+1)/2 - (-1)^(n-k)*(k+1)/2).
a(0, k) = binomial(k, floor(k/2)) = A001405(k); for n>0 T(n, k) = T(n+1, k-2) + T(n-1, k).
n-th row = M^n * V, where M = the infinite tridiagonal matrix with all 1's in the super and subdiagonals and (1,0,0,0,...) in the main diagonal. V = the infinite vector [1,0,0,0,...]. Example: (3,3,1,1,0,0,0,...) = M^3 * V. - Gary W. Adamson, Nov 04 2006
Sum_{k=0..n} T(m,k)*T(n,k) = T(m+n,0) = A001405(m+n). - Philippe Deléham, Feb 26 2007
Sum_{k=0..n} T(n,k)=2^n. - Philippe Deléham, Mar 27 2007
Sum_{k=0..n} T(n,k)*x^k = A127361(n), A126869(n), A001405(n), A000079(n), A127358(n), A127359(n), A127360(n) for x = -2, -1, 0, 1, 2, 3, 4 respectively. - Philippe Deléham, Dec 04 2009

Entry revised by N. J. A. Sloane, Nov 22 2006

A034868 Left half of Pascal's triangle.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 1, 4, 6, 1, 5, 10, 1, 6, 15, 20, 1, 7, 21, 35, 1, 8, 28, 56, 70, 1, 9, 36, 84, 126, 1, 10, 45, 120, 210, 252, 1, 11, 55, 165, 330, 462, 1, 12, 66, 220, 495, 792, 924, 1, 13, 78, 286, 715, 1287, 1716, 1, 14, 91, 364, 1001, 2002, 3003, 3432, 1, 15

Offset: 0

N. J. A. Sloane

			1;
1;
1, 2;
1, 3;
1, 4,  6;
1, 5, 10;
1, 6, 15, 20;
...

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318, A107430, A062344, A122366, A027306 (row sums).
Cf. A008619.
Cf. A225860.
Cf. A126257.
Cf. A034869 (right half), A014413, A014462, A265848.

a034868 n k = a034868_tabf !! n !! k
a034868_row n = a034868_tabf !! n
a034868_tabf = map reverse a034869_tabf
-- Reinhard Zumkeller, improved Dec 20 2015, Jul 27 2012

Flatten[ Table[ Binomial[n, k], {n, 0, 15}, {k, 0, Floor[n/2]}]] (* Robert G. Wilson v, May 28 2005 *)

for(n=0, 14, for(k=0, floor(n/2), print1(binomial(n, k),", ");); print();) \\ Indranil Ghosh, Mar 31 2017

import math
from sympy import binomial
for n in range(15):
    print([binomial(n, k) for k in range(int(math.floor(n/2)) + 1)]) # Indranil Ghosh, Mar 31 2017

from itertools import count, islice
def A034868_gen(): # generator of terms
    yield from (s:=(1,))
    for i in count(0):
        yield from (s:=(1,)+tuple(s[j]+s[j+1] for j in range(len(s)-1)) + ((s[-1]<<1,) if i&1 else ()))
A034868_list = list(islice(A034868_gen(),30)) # Chai Wah Wu, Oct 17 2023

T(n,k) = A034869(n,floor(n/2)-k), k = 0..floor(n/2). - Reinhard Zumkeller, Jul 27 2012

A103284 Triangle, read by rows, where row n+1 is formed by sorting, in ascending order, the result of the convolution of row n with {1,1}.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 2, 2, 3, 1, 3, 3, 4, 5, 1, 4, 5, 6, 7, 9, 1, 5, 9, 9, 11, 13, 16, 1, 6, 14, 16, 18, 20, 24, 29, 1, 7, 20, 29, 30, 34, 38, 44, 53, 1, 8, 27, 49, 53, 59, 64, 72, 82, 97, 1, 9, 35, 76, 97, 102, 112, 123, 136, 154, 179, 1, 10, 44, 111, 173, 179, 199, 214, 235, 259

Offset: 0

Paul D. Hanna, Jan 28 2005

Main diagonal is A103285.

			Convolution of row 5 {1,4,5,6,7,9} with {1,1} = {1,5,9,11,13,16,9}; sort to obtain row 6: {1,5,9,9,11,13,16}.
Rows begin:
1,
1,1,
1,1,2,
1,2,2,3,
1,3,3,4,5,
1,4,5,6,7,9,
1,5,9,9,11,13,16,
1,6,14,16,18,20,24,29,
1,7,20,29,30,34,38,44,53,
1,8,27,49,53,59,64,72,82,97,
1,9,35,76,97,102,112,123,136,154,179,...

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened

Cf. A103285.

Cf. A000079 (row sums), A257641 (central terms), A257646, A107430.

import Data.List (sort)
a103284 n k = a103284_tabl !! n !! k
a103284_row n = a103284_tabl !! n
a103284_tabl = iterate (\xs -> sort $ zipWith (+) (xs++[0]) ([0]++xs)) [1]
-- Reinhard Zumkeller, Nov 19 2015

{T(n,k)=local(A=vector(n+1,i,vector(i)),B);A[1][1]=1; for(k=1,n,B=vector(k+1);B[1]=1;B[k+1]=A[k][k]; for(i=2,k,B[i]=A[k][i]+A[k][i-1]); A[k+1]=vecsort(B));return(A[n+1][k+1])}

Row(n+1) = union of {1} and {T(n,k-1) + T(n,k): k=1..n}, sorted in ascending order. - Reinhard Zumkeller, Nov 19 2015

A127358 a(n) = Sum_{k=0..n} binomial(n, floor(k/2))*2^(n-k).

Original entry on oeis.org

1, 3, 8, 21, 54, 138, 350, 885, 2230, 5610, 14088, 35346, 88596, 221952, 555738, 1391061, 3480870, 8708610, 21783680, 54483510, 136254964, 340729788, 852000828, 2130354786, 5326563004, 13317759588, 33296999120, 83247698100, 208129274400, 520343244300

Offset: 0

Paul Barry, Jan 11 2007

Hankel transform is (-1)^n. In general, given r >= 0, the sequence given by Sum_{k=0..n} binomial(n, floor(k/2))*r^(n-k) has Hankel transform (1-r)^n. The sequence is the image of the sequence with g.f. (1+x)/(1-2*x) under the Chebyshev mapping g(x) -> (1/sqrt(1-4*x^2))*g(x*c(x^2)), where c(x) is the g.f. of the Catalan numbers A000108.

			a(3) = 21 = (12 + 6 + 2 + 1), where the top row of M^3 = (12, 6, 2, 1).

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Isaac DeJager, Madeleine Naquin, Frank Seidl, Colored Motzkin Paths of Higher Order, VERUM 2019.

Cf. A107430. - Philippe Deléham, Sep 16 2009

Table[Sum[Binomial[n,Floor[k/2]]2^(n-k),{k,0,n}],{n,0,30}] (* Harvey P. Dale, Jun 03 2012 *)
CoefficientList[Series[(1 + 2*x - Sqrt[1 - 4*x^2])/(2*Sqrt[1 - 4*x^2]*(x - 1 + Sqrt[1 - 4*x^2])), {x, 0, 50}], x] (* G. C. Greubel, May 22 2017 *)

my(x='x+O('x^50)); Vec((1 + 2*x - sqrt(1 - 4*x^2))/(2*sqrt(1 - 4*x^2)*(x - 1 + sqrt(1 - 4*x^2)))) \\ G. C. Greubel, May 22 2017

G.f.: (1/sqrt(1 - 4x^2))(1 + x*c(x^2))/(1 - 2*x*c(x^2)).
a(n) = 2*a(n-1) + A054341(n-1). a(n) = Sum_{k=0..n} A126075(n,k). - Philippe Deléham, Mar 03 2007
a(n) = Sum_{k=0..n} A061554(n,k)*2^k. - Philippe Deléham, Dec 04 2009
From Gary W. Adamson, Sep 07 2011: (Start)
a(n) is the sum of top row terms of M^n, M is an infinite square production matrix as follows:
  2, 1, 0, 0, 0, ...
  1, 0, 1, 0, 0, ...
  0, 1, 0, 1, 0, ...
  0, 0, 1, 0, 1, ...
  0, 0, 0, 1, 0, ...
  ... (End)
D-finite with recurrence 2*n*a(n) + (-5*n-4)*a(n-1) + 2*(-4*n+13)*a(n-2) + 20*(n-2)*a(n-3) = 0. - R. J. Mathar, Nov 30 2012
a(n) ~ 3 * 5^n / 2^(n+1). - Vaclav Kotesovec, Feb 13 2014

A189390 The maximum possible value for the apex of a triangle of numbers whose base consists of a permutation of the numbers 0 to n, and each number in a higher row is the sum of the two numbers directly below it.

Original entry on oeis.org

0, 1, 5, 16, 45, 116, 286, 680, 1581, 3604, 8106, 18008, 39650, 86568, 187804, 404944, 868989, 1856180, 3950194, 8376056, 17708310, 37329016, 78499620, 164682416, 344789970, 720430216, 1502768996, 3129355120, 6507087396, 13510929104

Offset: 0

Nathaniel Johnston, Apr 20 2011

The maximal value is reached when the largest numbers are placed in the middle and the smallest numbers at the border of the first row, i.e., [0,2,...,n,...,3,1]. Since the value of the apex is given as sum(c_k binomial(n,k)), one can compute this maximal value directly.

			For n = 4 consider the triangle:
         45
       21  24
      8  13  11
    2   6   7   4
  0   2   4   3   1
This triangle has 45 at its apex and no other such triangle with the numbers 0 through 4 on its base has a larger apex value, so a(4) = 45.

Nathaniel Johnston, Table of n, a(n) for n = 0..1000
Steven Finch, How far might we walk at random?, arXiv:1802.04615 [math.HO], 2018.

Cf. A066411, A099325, A189162, A189391.

a:= proc(n) return add((4*k+1)*binomial(n,k), k=0..floor((n-1)/2)) + `if`(n mod 2=0, n*binomial(n,n/2), 0):end:
seq(a(n), n=0..50);

a[n_] := Sum[(4k+1)*Binomial[n, k], {k, 0, Floor[(n-1)/2]}] + If[EvenQ[n], n*Binomial[n, n/2], 0]; Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Feb 18 2017, translated from Maple *)

A189390(n)=sum(i=0, (n-1)\2, (4*i+1)*binomial(n, i), if(!bittest(n,0),n*binomial(n, n\2)))  \\ - M. F. Hasler, Jan 24 2012

from math import comb
def A189390(n): return sum(((k<<2)|1)*comb(n,k) for k in range(n+1>>1))+(0 if n&1 else n*comb(n,n>>1)) # Chai Wah Wu, Oct 28 2024

a(n) = Sum_{k=0..floor((n-1)/2)} (4*k+1)*C(n,k) + (n+1 mod 2)*n*C(n,n/2).
a(n) = n*2^n-A189391(n). - M. F. Hasler, Jan 24 2012
a(n) = Sum_{k=0..n} k * C(n,floor(k/2)) = Sum_{k=0..n} k*A107430(n,k). - Alois P. Heinz, Feb 02 2012
G.f.: (2*x-sqrt(1-4*x^2)+1) / (2*(2*x-1)^2). - Alois P. Heinz, Feb 09 2012
D-finite with recurrence n*a(n) -4*n*a(n-1) +12*a(n-2) +16*(n-3)*a(n-3) +16*(-n+3)*a(n-4)=0. - R. J. Mathar, Jul 28 2016
D-finite with recurrence n*(2*n-3)*a(n) +2*(-2*n^2-n+5)*a(n-1) +4*(-2*n^2+9*n-5)*a(n-2) +8*(2*n-1)*(n-2)*a(n-3)=0. - R. J. Mathar, Jul 28 2016
a(n) = Sum_{k=1..n} Sum_{i=1..k} C(n,floor((n-k)/2)+i). - Stefano Spezia, Aug 20 2019

A127359 a(n) = Sum_{k=0..n} binomial(n, floor(k/2))*3^(n-k).

Original entry on oeis.org

1, 4, 14, 48, 162, 544, 1820, 6080, 20290, 67680, 225684, 752448, 2508468, 8362176, 27875064, 92919168, 309734850, 1032458080, 3441543140, 11471842880, 38239537852, 127465249344, 424884399624, 1416281802368, 4720940242612, 15736469278144, 52454901060680

Offset: 0

Paul Barry, Jan 11 2007

Hankel transform is (-2)^n. In general, given r>=0, the sequence given by Sum_{k=0..n} C(n,floor(k/2))*r^(n-k) has Hankel transform (1-r)^n. The sequence is the image of the sequence with g.f. (1+x)/(1-3x) under the Chebyshev mapping g(x)->(1/sqrt(1-4x^2))*g(xc(x^2)), where c(x) is the g.f. of the Catalan numbers A000108.

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Isaac DeJager, Madeleine Naquin, and Frank Seidl, Colored Motzkin Paths of Higher Order, VERUM 2019.

Cf. A107430. - Philippe Deléham, Sep 16 2009

Cf. A000108 (Catalan numbers).

a:=[1, 4, 14];; for n in [4..30] do a[n]:=(2*(5*n-2)*a[n-1] +4*(3*n-14)*a[n-2] -40*(n-3)*a[n-3])/(3*(n-1)); od; a; # G. C. Greubel, Dec 15 2019

I:=[1, 4, 14]; [n le 3 select I[n] else (2*(5*n-2)*Self(n-1) + 4*(3*n - 14)*Self(n-2) -40*(n-3)*Self(n-3))/(3*(n-1)): n in [1..30]]; // G. C. Greubel, Dec 15 2019

A127359:=n->sum(binomial(n,floor(k/2))*3^(n-k), k=0..n): seq(A127359(n), n=0..30); # Wesley Ivan Hurt, Mar 14 2015

Table[Sum[Binomial[n, Floor[k/2]]*3^(n-k), {k,0,n}], {n,0,30}] (* Vaclav Kotesovec, Oct 19 2012 *)

a(n) = sum(j=0, n, binomial(n, j\2)*3^(n-j));
vector(31, n, a(n-1)) \\ G. C. Greubel, Dec 15 2019

[sum(binomial(n, floor(j/2))*3^(n-j) for j in (0..n)) for n in (0..30)] # G. C. Greubel, Dec 15 2019

G.f.: (1/sqrt(1-4*x^2))*(1+x*c(x^2))/(1-3*x*c(x^2)), where c(x) = (1 - sqrt(1 - 4*x))/(2*x).
a(n) = Sum_{k=0..n} A061554(n,k)*3^k. - Philippe Deléham, Dec 04 2009
Recurrence: 3*n*a(n) = 2*(5*n + 3)*a(n-1) + 4*(3*n - 11)*a(n-2) - 40*(n-2)*a(n-3). - Vaclav Kotesovec, Oct 19 2012
a(n) ~ 4*10^n/3^(n+1). - Vaclav Kotesovec, Oct 19 2012

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A126136 Binomial transform of A107430.

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A127159 Triangle T(n,k) with T(n,k) = A061554(n,k) + A107430(n,k).

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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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A126869 a(n) = Sum_{k = 0..n} binomial(n,floor(k/2))*(-1)^(n-k).

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A061554 Square table read by antidiagonals: a(n,k) = binomial(n+k, floor(k/2)).

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A034868 Left half of Pascal's triangle.