A132311 -id:A132311 - cp's OEIS frontend

A000041 a(n) is the number of partitions of n (the partition numbers).

1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718, 4565, 5604, 6842, 8349, 10143, 12310, 14883, 17977, 21637, 26015, 31185, 37338, 44583, 53174, 63261, 75175, 89134, 105558, 124754, 147273, 173525

Offset: 0

N. J. A. Sloane

Also number of nonnegative solutions to b + 2c + 3d + 4e + ... = n and the number of nonnegative solutions to 2c + 3d + 4e + ... <= n. - Henry Bottomley, Apr 17 2001
a(n) is also the number of conjugacy classes in the symmetric group S_n (and the number of irreducible representations of S_n).
Also the number of rooted trees with n+1 nodes and height at most 2.
Coincides with the sequence of numbers of nilpotent conjugacy classes in the Lie algebras gl(n). A006950, A015128 and this sequence together cover the nilpotent conjugacy classes in the classical A,B,C,D series of Lie algebras. - Alexander Elashvili, Sep 08 2003
Number of distinct Abelian groups of order p^n, where p is prime (the number is independent of p). - Lekraj Beedassy, Oct 16 2004
Number of graphs on n vertices that do not contain P3 as an induced subgraph. - Washington Bomfim, May 10 2005
Numbers of terms to be added when expanding the n-th derivative of 1/f(x). - Thomas Baruchel, Nov 07 2005
Sequence agrees with expansion of Molien series for symmetric group S_n up to the term in x^n. - Maurice D. Craig (towenaar(AT)optusnet.com.au), Oct 30 2006
Also the number of nonnegative integer solutions to x_1 + x_2 + x_3 + ... + x_n = n such that n >= x_1 >= x_2 >= x_3 >= ... >= x_n >= 0, because by letting y_k = x_k - x_(k+1) >= 0 (where 0 < k < n) we get y_1 + 2y_2 + 3y_3 + ... + (n-1)y_(n-1) + nx_n = n. - Werner Grundlingh (wgrundlingh(AT)gmail.com), Mar 14 2007
Let P(z) := Sum_{j>=0} b_j z^j, b_0 != 0. Then 1/P(z) = Sum_{j>=0} c_j z^j, where the c_j must be computed from the infinite triangular system b_0 c_0 = 1, b_0 c_1 + b_1 c_0 = 0 and so on (Cauchy products of the coefficients set to zero). The n-th partition number arises as the number of terms in the numerator of the expression for c_n: The coefficient c_n of the inverted power series is a fraction with b_0^(n+1) in the denominator and in its numerator having a(n) products of n coefficients b_i each. The partitions may be read off from the indices of the b_i. - Peter C. Heinig (algorithms(AT)gmx.de), Apr 09 2007
A sequence of positive integers p = p_1 ... p_k is a descending partition of the positive integer n if p_1 + ... + p_k = n and p_1 >= ... >= p_k. If formally needed p_j = 0 is appended to p for j > k. Let P_n denote the set of these partition for some n >= 1. Then a(n) = 1 + Sum_{p in P_n} floor((p_1-1)/(p_2+1)). (Cf. A000065, where the formula reduces to the sum.) Proof in Kelleher and O'Sullivan (2009). For example a(6) = 1 + 0 + 0 + 0 + 0 + 1 + 0 + 0 + 1 + 1 + 2 + 5 = 11. - Peter Luschny, Oct 24 2010
Let n = Sum( k_(p_m) p_m ) = k_1 + 2k_2 + 5k_5 + 7k_7 + ..., where p_m is the m-th generalized pentagonal number (A001318). Then a(n) is the sum over all such pentagonal partitions of n of (-1)^(k_5+k_7 + k_22 + ...) ( k_1 + k_2 + k_5 + ...)! /( k_1! k_2! k_5! ...), where the exponent of (-1) is the sum of all the k's corresponding to even-indexed GPN's. - Jerome Malenfant, Feb 14 2011
From Jerome Malenfant, Feb 14 2011: (Start)
The matrix of a(n) values
  a(0)
  a(1) a(0)
  a(2) a(1) a(0)
  a(3) a(2) a(1) a(0)
  ....
  a(n) a(n-1) a(n-2) ... a(0)
is the inverse of the matrix
   1
  -1  1
  -1 -1  1
   0 -1 -1  1
  ....
  -d_n  -d_(n-1) -d_(n-2) ... -d_1  1
where d_q = (-1)^(m+1) if q = m(3m-1)/2 = the m-th generalized pentagonal number (A001318), = 0 otherwise. (End)
Let k > 0 be an integer, and let i_1, i_2, ..., i_k be distinct integers such that 1 <= i_1 < i_2 < ... < i_k. Then, equivalently, a(n) equals the number of partitions of N = n + i_1 + i_2 + ... + i_k in which each i_j (1 <= j <= k) appears as a part at least once. To see this, note that the partitions of N of this class must be in 1-to-1 correspondence with the partitions of n, since N - i_1 - i_2 - ... - i_k = n. - L. Edson Jeffery, Apr 16 2011
a(n) is the number of distinct degree sequences over all free trees having n + 2 nodes.  Take a partition of the integer n, add 1 to each part and append as many 1's as needed so that the total is 2n + 2.  Now we have a degree sequence of a tree with n + 2 nodes.  Example: The partition 3 + 2 + 1 = 6 corresponds to the degree sequence {4, 3, 2, 1, 1, 1, 1, 1} of a tree with 8 vertices. - Geoffrey Critzer, Apr 16 2011
a(n) is number of distinct characteristic polynomials among  n! of permutations matrices size n X n. - Artur Jasinski, Oct 24 2011
Conjecture: starting with offset 1 represents the numbers of ordered compositions of n using the signed (++--++...) terms of A001318 starting (1, 2, -5, -7, 12, 15, ...). - Gary W. Adamson, Apr 04 2013 (this is true by the pentagonal number theorem, Joerg Arndt, Apr 08 2013)
a(n) is also number of terms in expansion of the n-th derivative of log(f(x)). In Mathematica notation: Table[Length[Together[f[x]^n * D[Log[f[x]], {x, n}]]], {n, 1, 20}]. - Vaclav Kotesovec, Jun 21 2013
Conjecture: No a(n) has the form x^m with m > 1 and x > 1. - Zhi-Wei Sun, Dec 02 2013
Partitions of n that contain a part p are the partitions of n - p. Thus, number of partitions of m*n - r that include k*n as a part is A000041(h*n-r), where h = m - k >= 0, n >= 2, 0 <= r < n; see A111295 as an example.  - Clark Kimberling, Mar 03 2014
a(n) is the number of compositions of n into positive parts avoiding the pattern [1, 2]. - Bob Selcoe, Jul 08 2014
Conjecture: For any j there exists k such that all primes p <= A000040(j) are factors of one or more a(n) <= a(k). Growth of this coverage is slow and irregular. k = 1067 covers the first 102 primes, thus slower than A000027. - Richard R. Forberg, Dec 08 2014
a(n) is the number of nilpotent conjugacy classes in the order-preserving, order-decreasing and (order-preserving and order-decreasing) injective transformation semigroups. - Ugbene Ifeanyichukwu, Jun 03 2015
Define a segmented partition a(n,k, ) to be a partition of n with exactly k parts, with s(j) parts t(j) identical to each other and distinct from all the other parts. Note that n >= k, j <= k, 0 <= s(j) <= k, s(1)t(1) + ... + s(j)t(j) = n and s(1) + ... + s(j) = k. Then there are up to a(k) segmented partitions of n with exactly k parts. - Gregory L. Simay, Nov 08 2015
(End)
From Gregory L. Simay, Nov 09 2015: (Start)
The polynomials for a(n, k, ) have degree j-1.
  a(n, k, ) = 1 if n = 0 mod k, = 0 otherwise
  a(rn, rk, ) = a(n, k, )
  a(n odd, k, ) = 0
Established results can be recast in terms of segmented partitions:
  For j(j+1)/2 <= n < (j+1)(j+2)/2, A000009(n) = a(n, 1, <1>) + ... + a(n, j, ), j < n
  a(n, k, ) = a(n - j(j-1)/2, k)
(End)
a(10^20) was computed using the NIST Arb package. It has 11140086260 digits and its head and tail sections are 18381765...88091448. See the Johansson 2015 link. - Stanislav Sykora, Feb 01 2016
Satisfies Benford's law [Anderson-Rolen-Stoehr, 2011]. - N. J. A. Sloane, Feb 08 2017
The partition function p(n) is log-concave for all n>25 [DeSalvo-Pak, 2014]. - Michel Marcus, Apr 30 2019
a(n) is also the dimension of the n-th cohomology of the infinite real Grassmannian with coefficients in Z/2. - Luuk Stehouwer, Jun 06 2021
Number of equivalence relations on n unlabeled nodes. - Lorenzo Sauras Altuzarra, Jun 13 2022
Equivalently, number of idempotent mappings f from a set X of n elements into itself (i.e., satisfying f o f = f) up to permutation (i.e., f~f' :<=> There is a permutation sigma in Sym(X) such that f' o sigma = sigma o f). - Philip Turecek, Apr 17 2023
Conjecture: Each integer n > 2 different from 6 can be written as a sum of finitely many numbers of the form a(k) + 2 (k > 0) with no summand dividing another. This has been verified for n <= 7140. - Zhi-Wei Sun, May 16 2023
a(n) is also the number of partitions of n*(n+3)/2 into n distinct parts. - David García Herrero, Aug 20 2024
a(n) is also the number of non-isomorphic sigma algebras on {1,...,n}. A000110(n) counts all sigma algebras on {1,...,n}. Every sigma algebra on a finite set X is exactly the collection of all unions of its atoms (its minimal nonempty members), and those atoms partition X. An isomorphism of sigma algebras must map atoms to atoms, so the isomorphism class of a sigma algebra is determined by the multiset of its atom-sizes, which is an integer partition of n. - Matthew Azar, Jul 18 2025

			a(5) = 7 because there are seven partitions of 5, namely: {1, 1, 1, 1, 1}, {2, 1, 1, 1}, {2, 2, 1}, {3, 1, 1}, {3, 2}, {4, 1}, {5}. - _Bob Selcoe_, Jul 08 2014
G.f. = 1 + x + 2*x^2 + 3*x^3 + 5*x^4 + 7*x^5 + 11*x^6 + 15*x^7 + 22*x^8 + ...
G.f. = 1/q + q^23 + 2*q^47 + 3*q^71 + 5*q^95 + 7*q^119 + 11*q^143 + 15*q^167 + ...
From _Gregory L. Simay_, Nov 08 2015: (Start)
There are up to a(4)=5 segmented partitions of the partitions of n with exactly 4 parts. They are a(n,4, <4>), a(n,4,<3,1>), a(n,4,<2,2>), a(n,4,<2,1,1>), a(n,4,<1,1,1,1>).
The partition 8,8,8,8 is counted in a(32,4,<4>).
The partition 9,9,9,5 is counted in a(32,4,<3,1>).
The partition 11,11,5,5 is counted in a(32,4,<2,2>).
The partition 13,13,5,1 is counted in a(32,4,<2,1,1>).
The partition 14,9,6,3 is counted in a(32,4,<1,1,1,1>).
a(n odd,4,<2,2>) = 0.
a(12, 6, <2,2,2>) = a(6,3,<1,1,1>) = a(6-3,3) = a(3,3) = 1. The lone partition is 3,3,2,2,1,1.
(End)

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J. D. Rosenhouse, Solutions to Problems
Kate Rudolph, Pattern Popularity in 132-Avoiding Permutations, The Electronic Journal of Combinatorics 20(1)(2013), #P8. [Cited by Shalosh B. Ekhad and Doron Zeilberger, 2014] - _N. J. A. Sloane_, Mar 31 2014
F. Ruskey, Generate Numerical Partitions
F. Ruskey, The first 284547 partition numbers (52MB compressed file, archived link)
M. Savic, The Partition Function and Ramanujan's 5k+4 Congruence
Zhumagali Shomanov, Combinatorial formula for the partition function, arXiv:1508.03173 [math.CO], 2015.
T. Sillke, Number of integer partitions
R. P. Stanley, A combinatorial miscellany
Cormac O'Sullivan, Detailed asymptotic expansions for partitions into powers, arXiv:2205.13468 [math.NT], 2022-3.
Yi Wang and Bao-Xuan Zhu, Proofs of some conjectures on monotonicity of number-theoretic and combinatorial sequences, arXiv preprint arXiv:1303.5595 [math.CO], 2013.
R. L. Weaver, New Congruences for the Partition Function, The Ramanujan Journal 5(1) 2001.
Eric Weisstein's World of Mathematics, Partition, Partition Function P, q-Pochhammer Symbol, and Ramanujan's Identity
West Sussex Grid for Learning, Multicultural Mathematics, Ramanujan's Partition of Numbers
Thomas Wieder, Comment on A000041
Wikipedia, Partition (number theory)
H. S. Wilf, Lectures on Integer Partitions
Wolfram Research, Generating functions of p(n)
D. J. Wright, Partitions [broken link]
Doron Zeilberger, Noam Zeilberger, Two Questions about the Fractional Counting of Partitions, arXiv:1810.12701 [math.CO], 2018.
Robert M. Ziff, On Cardy's formula for the critical crossing probability in 2d percolation, J. Phys. A. 28, 1249-1255 (1995).
Index entries for "core" sequences
Index entries for related partition-counting sequences
Index entries for expansions of Product_{k >= 1} (1-x^k)^m
Index entries for sequences related to rooted trees
Index entries for sequences related to Benford's law

Cf. A000009, A000079, A000203, A001318, A008284, A026820, A065446, A078506, A113685, A132311, A000248, A000110.
Partial sums give A000070.
For successive differences see A002865, A053445, A072380, A081094, A081095.
Antidiagonal sums of triangle A092905. a(n) = A054225(n,0).
Boustrophedon transforms: A000733, A000751.
Cf. A167376 (complement), A061260 (multisets), A000700 (self-conjug), A330644 (not self-conj).

List([1..10],n->Size(OrbitsDomain(SymmetricGroup(IsPermGroup,n),SymmetricGroup(IsPermGroup,n),\^))); # Attila Egri-Nagy, Aug 15 2014

import Data.MemoCombinators (memo2, integral)
a000041 n = a000041_list !! n
a000041_list = map (p' 1) [0..] where
   p' = memo2 integral integral p
   p _ 0 = 1
   p k m = if m < k then 0 else p' k (m - k) + p' (k + 1) m
-- Reinhard Zumkeller, Nov 03 2015, Nov 04 2013

# DedekindEta is defined in A000594
A000041List(len) = DedekindEta(len, -1)
A000041List(50) |> println # Peter Luschny, Mar 09 2018

a:= func< n | NumberOfPartitions(n) >; [ a(n) : n in [0..10]];

A000041 := n -> combinat:-numbpart(n): [seq(A000041(n), n=0..50)]; # Warning: Maple 10 and 11 give incorrect answers in some cases: A110375.
spec := [B, {B=Set(Set(Z,card>=1))}, unlabeled ];
[seq(combstruct[count](spec, size=n), n=0..50)];
with(combstruct):ZL0:=[S,{S=Set(Cycle(Z,card>0))}, unlabeled]: seq(count(ZL0,size=n),n=0..45); # Zerinvary Lajos, Sep 24 2007
G:={P=Set(Set(Atom,card>0))}: combstruct[gfsolve](G,labeled,x); seq(combstruct[count]([P,G,unlabeled],size=i),i=0..45); # Zerinvary Lajos, Dec 16 2007
# Using the function EULER from Transforms (see link at the bottom of the page).
1,op(EULER([seq(1,n=1..49)])); # Peter Luschny, Aug 19 2020

Table[ PartitionsP[n], {n, 0, 45}]
a[ n_] := SeriesCoefficient[ q^(1/24) / DedekindEta[ Log[q] / (2 Pi I)], {q, 0, n}]; (* Michael Somos, Jul 11 2011 *)
a[ n_] := SeriesCoefficient[ 1 / Product[ 1 - x^k, {k, n}], {x, 0, n}]; (* Michael Somos, Jul 11 2011 *)
CoefficientList[1/QPochhammer[q] + O[q]^100, q] (* Jean-François Alcover, Nov 25 2015 *)
a[0] := 1; a[n_] := a[n] = Block[{k=1, s=0, i=n-1}, While[i >= 0, s=s-(-1)^k (a[i]+a[i-k]); k=k+1; i=i-(3 k-2)]; s]; Map[a, Range[0, 49]] (* Oliver Seipel, Jun 01 2024 after Euler *)

num_partitions(60,list); /* Emanuele Munarini, Feb 24 2014 */

combinat::partitions::count(i) $i=0..54 // Zerinvary Lajos, Apr 16 2007

{a(n) = if( n<0, 0, polcoeff( 1 / eta(x + x * O(x^n)), n))};

/* The Hardy-Ramanujan-Rademacher exact formula in PARI is as follows (this is no longer necessary since it is now built in to the numbpart command): */
Psi(n, q) = local(a, b, c); a=sqrt(2/3)*Pi/q; b=n-1/24; c=sqrt(b); (sqrt(q)/(2*sqrt(2)*b*Pi))*(a*cosh(a*c)-(sinh(a*c)/c))
L(n, q) = if(q==1,1,sum(h=1,q-1,if(gcd(h,q)>1,0,cos((g(h,q)-2*h*n)*Pi/q))))
g(h, q) = if(q<3,0,sum(k=1,q-1,k*(frac(h*k/q)-1/2)))
part(n) = round(sum(q=1,max(5,0.5*sqrt(n)),L(n,q)*Psi(n,q)))
/* Ralf Stephan, Nov 30 2002, fixed by Vaclav Kotesovec, Apr 09 2018 */

PARI
```
{a(n) = numbpart(n)};
```

{a(n) = if( n<0, 0, polcoeff( sum( k=1, sqrtint(n), x^k^2 / prod( i=1, k, 1 - x^i, 1 + x * O(x^n))^2, 1), n))};

f(n)= my(v,i,k,s,t);v=vector(n,k,0);v[n]=2;t=0;while(v[1]1,i--;s+=i*(v[i]=(n-s)\i));t++);t \\ Thomas Baruchel, Nov 07 2005

a(n)=if(n<0, 0, polcoeff(exp(sum(k=1, n, x^k/(1-x^k)/k, x*O(x^n))), n)) \\ Joerg Arndt, Apr 16 2010

use ntheory ":all"; my @p = map { partitions($) } 0..100; say "[@p]"; # _Dana Jacobsen, Sep 06 2015

from sympy.functions.combinatorial.numbers import partition
print([partition(i) for i in range(101)]) # Joan Ludevid, May 25 2025

#lang racket
; SUM(k,-inf,+inf) (-1)^k p(n-k(3k-1)/2)
; For k outside the range (1-(sqrt(1-24n))/6 to (1+sqrt(1-24n))/6) argument n-k(3k-1)/2 < 0.
; Therefore the loops below are finite. The hash avoids repeated identical computations.
(define (p n) ; Nr of partitions of n.
(hash-ref h n
  (λ ()
   (define r
    (+
     (let loop ((k 1) (n (sub1 n)) (s 0))
      (if (< n 0) s
       (loop (add1 k) (- n (* 3 k) 1) (if (odd? k) (+ s (p n)) (- s (p n))))))
     (let loop ((k -1) (n (- n 2)) (s 0))
      (if (< n 0) s
       (loop (sub1 k) (+ n (* 3 k) -2) (if (odd? k) (+ s (p n)) (- s (p n))))))))
   (hash-set! h n r)
   r)))
(define h (make-hash '((0 . 1))))
; (for ((k (in-range 0 50))) (printf "~s, " (p k))) runs in a moment.
; Jos Koot, Jun 01 2016

[number_of_partitions(n) for n in range(46)]  # Zerinvary Lajos, May 24 2009

@CachedFunction
def A000041(n):
    if n == 0: return 1
    S = 0; J = n-1; k = 2
    while 0 <= J:
        T = A000041(J)
        S = S+T if is_odd(k//2) else S-T
        J -= k if is_odd(k) else k//2
        k += 1
    return S
[A000041(n) for n in range(50)]  # Peter Luschny, Oct 13 2012

# uses[EulerTransform from A166861]
a = BinaryRecurrenceSequence(1, 0)
b = EulerTransform(a)
print([b(n) for n in range(50)]) # Peter Luschny, Nov 11 2020

G.f.: Product_{k>0} 1/(1-x^k) = Sum_{k>= 0} x^k Product_{i = 1..k} 1/(1-x^i) = 1 + Sum_{k>0} x^(k^2)/(Product_{i = 1..k} (1-x^i))^2.
G.f.: 1 + Sum_{n>=1} x^n/(Product_{k>=n} 1-x^k). - Joerg Arndt, Jan 29 2011
a(n) - a(n-1) - a(n-2) + a(n-5) + a(n-7) - a(n-12) - a(n-15) + ... = 0, where the sum is over n-k and k is a generalized pentagonal number (A001318) <= n and the sign of the k-th term is (-1)^([(k+1)/2]). See A001318 for a good way to remember this!
a(n) = (1/n) * Sum_{k=0..n-1} sigma(n-k)*a(k), where sigma(k) is the sum of divisors of k (A000203).
a(n) ~ 1/(4*n*sqrt(3)) * e^(Pi * sqrt(2n/3)) as n -> infinity (Hardy and Ramanujan). See A050811.
a(n) = a(0)*b(n) + a(1)*b(n-2) + a(2)*b(n-4) + ... where b = A000009.
From Jon E. Schoenfield, Aug 17 2014: (Start)
It appears that the above approximation from Hardy and Ramanujan can be refined as
a(n) ~ 1/(4*n*sqrt(3)) * e^(Pi * sqrt(2n/3 + c0 + c1/n^(1/2) + c2/n + c3/n^(3/2) + c4/n^2 + ...)), where the coefficients c0 through c4 are approximately
     c0 = -0.230420145062453320665537
     c1 = -0.0178416569128570889793
     c2 =  0.0051329911273
     c3 = -0.0011129404
     c4 =  0.0009573,
as n -> infinity. (End)
From Vaclav Kotesovec, May 29 2016 (c4 added Nov 07 2016): (Start)
c0 = -0.230420145062453320665536704197233... = -1/36 - 2/Pi^2
c1 = -0.017841656912857088979502135349949... = 1/(6*sqrt(6)*Pi) - sqrt(3/2)/Pi^3
c2 =  0.005132991127342167594576391633559... = 1/(2*Pi^4)
c3 = -0.001112940489559760908236602843497... = 3*sqrt(3/2)/(4*Pi^5) - 5/(16*sqrt(6)*Pi^3)
c4 =  0.000957343284806972958968694349196... = 1/(576*Pi^2) - 1/(24*Pi^4) + 93/(80*Pi^6)
a(n) ~ exp(Pi*sqrt(2*n/3))/(4*sqrt(3)*n) * (1 - (sqrt(3/2)/Pi + Pi/(24*sqrt(6)))/sqrt(n) + (1/16 + Pi^2/6912)/n).
a(n) ~ exp(Pi*sqrt(2*n/3) - (sqrt(3/2)/Pi + Pi/(24*sqrt(6)))/sqrt(n) + (1/24 - 3/(4*Pi^2))/n) / (4*sqrt(3)*n).
(End)
a(n) < exp( (2/3)^(1/2) Pi sqrt(n) ) (Ayoub, p. 197).
G.f.: Product_{m>=1} (1+x^m)^A001511(m). - Vladeta Jovovic, Mar 26 2004
a(n) = Sum_{i=0..n-1} P(i, n-i), where P(x, y) is the number of partitions of x into at most y parts and P(0, y)=1. - Jon Perry, Jun 16 2003
G.f.: Product_{i>=1} Product_{j>=0} (1+x^((2i-1)*2^j))^(j+1). - Jon Perry, Jun 06 2004
G.f. e^(Sum_{k>0} (x^k/(1-x^k)/k)). - Franklin T. Adams-Watters, Feb 08 2006
a(n) = A114099(9*n). - Reinhard Zumkeller, Feb 15 2006
Euler transform of all 1's sequence (A000012). Weighout transform of A001511. - Franklin T. Adams-Watters, Mar 15 2006
a(n) = A027187(n) + A027193(n) = A000701(n) + A046682(n). - Reinhard Zumkeller, Apr 22 2006
A026820(a(n),n) = A134737(n) for n > 0. - Reinhard Zumkeller, Nov 07 2007
Convolved with A152537 gives A000079, powers of 2. - Gary W. Adamson, Dec 06 2008
a(n) = A026820(n, n); a(n) = A108949(n) + A045931(n) + A108950(n) = A130780(n) + A171966(n) - A045931(n) = A045931(n) + A171967(n). - Reinhard Zumkeller, Jan 21 2010
a(n) = Tr(n)/(24*n-1) = A183011(n)/A183010(n), n>=1. See the Bruinier-Ono paper in the Links. - Omar E. Pol, Jan 23 2011
From Jerome Malenfant, Feb 14 2011: (Start)
a(n) = determinant of the n X n Toeplitz matrix:
   1  -1
   1   1  -1
   0   1   1  -1
   0   0   1   1  -1
  -1   0   0   1   1  -1
   . . .
  d_n  d_(n-1) d_(n-2)...1
where d_q = (-1)^(m+1) if q = m(3m-1)/2 = p_m, the m-th generalized pentagonal number (A001318), otherwise d_q = 0. Note that the 1's run along the diagonal and the -1's are on the superdiagonal. The (n-1) row (not written) would end with ... 1 -1. (End)
Empirical: let F*(x) = Sum_{n=0..infinity} p(n)*exp(-Pi*x*(n+1)), then F*(2/5) = 1/sqrt(5) to a precision of 13 digits.
F*(4/5) = 1/2+3/2/sqrt(5)-sqrt(1/2*(1+3/sqrt(5))) to a precision of 28 digits. These are the only values found for a/b when a/b is from F60, Farey fractions up to 60. The number for F*(4/5) is one of the real roots of 25*x^4 - 50*x^3 - 10*x^2 - 10*x + 1. Note here the exponent (n+1) compared to the standard notation with n starting at 0. - Simon Plouffe, Feb 23 2011
The constant (2^(7/8)*GAMMA(3/4))/(exp(Pi/6)*Pi^(1/4)) = 1.0000034873... when expanded in base exp(4*Pi) will give the first 52 terms of a(n), n>0, the precision needed is 300 decimal digits. - Simon Plouffe, Mar 02 2011
a(n) = A035363(2n). - Omar E. Pol, Nov 20 2009
G.f.: A(x)=1+x/(G(0)-x); G(k) = 1 + x - x^(k+1) - x*(1-x^(k+1))/G(k+1); (continued fraction Euler's kind, 1-step ). - Sergei N. Gladkovskii, Jan 25 2012
Convolution of A010815 with A000712. - Gary W. Adamson, Jul 20 2012
G.f.: 1 + x*(1 - G(0))/(1-x) where G(k) = 1 - 1/(1-x^(k+1))/(1-x/(x-1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 22 2013
G.f.: Q(0) where Q(k) = 1 + x^(4*k+1)/( (x^(2*k+1)-1)^2 - x^(4*k+3)*(x^(2*k+1)-1)^2/( x^(4*k+3) + (x^(2*k+2)-1)^2/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 16 2013
a(n) = 24*spt(n) + 12*N_2(n) - Tr(n) = 24*A092269(n) + 12*A220908(n) - A183011(n), n >= 1. - Omar E. Pol, Feb 17 2013
G.f.: 1/(x; x){inf} where (a; q)_k is the q-Pochhammer symbol. - _Vladimir Reshetnikov, Apr 24 2013
a(n) = A066186(n)/n, n >= 1. - Omar E. Pol, Aug 16 2013
From Peter Bala, Dec 23 2013: (Start)
a(n-1) = Sum_{parts k in all partitions of n} mu(k), where mu(k) is the arithmetical Möbius function (see A008683).
Let P(2,n) denote the set of partitions of n into parts k >= 2. Then a(n-2) = -Sum_{parts k in all partitions in P(2,n)} mu(k).
n*( a(n) - a(n-1) ) = Sum_{parts k in all partitions in P(2,n)} k (see A138880).
Let P(3,n) denote the set of partitions of n into parts k >= 3. Then
a(n-3) = (1/2)*Sum_{parts k in all partitions in P(3,n)} phi(k), where phi(k) is the Euler totient function (see A000010). Using this result and Mertens's theorem on the average order of the phi function, we can find an approximate 3-term recurrence for the partition function: a(n) ~ a(n-1) + a(n-2) + (Pi^2/(3*n) - 1)*a(n-3). For example, substituting the values a(47) = 124754, a(48) = 147273 and a(49) = 173525 into the recurrence gives the approximation a(50) ~ 204252.48... compared with the true value a(50) = 204226. (End)
a(n) = Sum_{k=1..n+1} (-1)^(n+1-k)*A000203(k)*A002040(n+1-k). - Mircea Merca, Feb 27 2014
a(n) = A240690(n) + A240690(n+1), n >= 1. - Omar E. Pol, Mar 16 2015
From Gary W. Adamson, Jun 22 2015: (Start)
A production matrix for the sequence with offset 1 is M, an infinite n x n matrix of the following form:
  a, 1, 0, 0, 0, 0, ...
  b, 0, 1, 0, 0, 0, ...
  c, 0, 0, 1, 0, 0, ...
  d, 0, 0, 0, 1, 0, ...
.
.
... such that (a, b, c, d, ...) is the signed version of A080995 with offset 1: (1,1,0,0,-1,0,-1,...)
and a(n) is the upper left term of M^n.
This operation is equivalent to the g.f. (1 + x + 2x^2 + 3x^3 + 5x^4 + ...) = 1/(1 - x - x^2 + x^5 + x^7 - x^12 - x^15 + x^22 + ...). (End)
G.f.: x^(1/24)/eta(log(x)/(2 Pi i)). - Thomas Baruchel, Jan 09 2016, after Michael Somos (after Richard Dedekind).
a(n) = Sum_{k=-inf..+inf} (-1)^k a(n-k(3k-1)/2) with a(0)=1 and a(negative)=0. The sum can be restricted to the (finite) range from k = (1-sqrt(1-24n))/6 to (1+sqrt(1-24n))/6, since all terms outside this range are zero. - Jos Koot, Jun 01 2016
G.f.: (conjecture) (r(x) * r(x^2) * r(x^4) * r(x^8) * ...) where r(x) is A000009: (1, 1, 1, 2, 2, 3, 4, ...). - Gary W. Adamson, Sep 18 2016; Doron Zeilberger observed today that "This follows immediately from Euler's formula 1/(1-z) = (1+z)*(1+z^2)*(1+z^4)*(1+z^8)*..." Gary W. Adamson, Sep 20 2016
a(n) ~ 2*Pi * BesselI(3/2, sqrt(24*n-1)*Pi/6) / (24*n-1)^(3/4). - Vaclav Kotesovec, Jan 11 2017
G.f.: Product_{k>=1} (1 + x^k)/(1 - x^(2*k)). - Ilya Gutkovskiy, Jan 23 2018
a(n) = p(1, n) where p(k, n) = p(k+1, n) + p(k, n-k) if k < n, 1 if k = n, and 0 if k > n. p(k, n) is the number of partitions of n into parts >= k. - Lorraine Lee, Jan 28 2020
Sum_{n>=1} 1/a(n) = A078506. - Amiram Eldar, Nov 01 2020
Sum_{n>=0} a(n)/2^n = A065446. - Amiram Eldar, Jan 19 2021
From Simon Plouffe, Mar 12 2021: (Start)
Sum_{n>=0} a(n)/exp(Pi*n) = 2^(3/8)*Gamma(3/4)/(Pi^(1/4)*exp(Pi/24)).
Sum_{n>=0} a(n)/exp(2*Pi*n) = 2^(1/2)*Gamma(3/4)/(Pi^(1/4)*exp(Pi/12)).
[corrected by Vaclav Kotesovec, May 12 2023] (End)
[These are the reciprocals of phi(exp(-Pi)) (A259148) and phi(exp(-2*Pi)) (A259149), where phi(q) is the Euler modular function. See B. C. Berndt (RLN, Vol. V, p. 326), and formulas (13) and (14) in I. Mező, 2013. - Peter Luschny, Mar 13 2021]
a(n) = A000009(n) + A035363(n) + A006477(n). - R. J. Mathar, Feb 01 2022
a(n) = A008284(2*n,n) is also the number of partitions of 2n into n parts. - Ryan Brooks, Jun 11 2022
a(n) = A000700(n) + A330644(n). - R. J. Mathar, Jun 15 2022
a(n) ~ exp(Pi*sqrt(2*n/3)) / (4*n*sqrt(3)) * (1 + Sum_{r>=1} w(r)/n^(r/2)), where w(r) = 1/(-4*sqrt(6))^r * Sum_{k=0..(r+1)/2} binomial(r+1,k) * (r+1-k) / (r+1-2*k)! * (Pi/6)^(r-2*k) [Cormac O'Sullivan, 2023, pp. 2-3]. - Vaclav Kotesovec, Mar 15 2023

Additional comments from Ola Veshta (olaveshta(AT)my-deja.com), Feb 28 2001

Additional comments from Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001

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

Original entry on oeis.org

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

A132312 Triangle read by rows: T(n,k) = number of partitions of binomial(n,k) into distinct parts of the first n rows of Pascal's triangle, 0<=k<=n.

Original entry on oeis.org

0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 3, 2, 1, 1, 4, 7, 6, 7, 4, 1, 1, 4, 11, 14, 14, 11, 4, 1, 1, 5, 28, 57, 56, 57, 28, 5, 1, 1, 7, 73, 273, 434, 434, 273, 73, 7, 1, 1, 10, 189, 1411, 3479, 3980, 3479, 1411, 189, 10, 1, 1, 11, 300, 4138, 16293, 26555, 26555, 16293, 4138, 300, 11, 1

Offset: 0

Reinhard Zumkeller, Aug 18 2007

T(n,k) = T(n,n-k);
T(n,0) = 1 for n>0;
A000009(n) - 1 <= T(n,1) <= A000009(n) for n>1;

			T(9,1) = A000009(9)-1 = 7;
A007318(5,2) = A007318(10,1) = 10:
T(5,2) = #{6+4, 6+3+1, 4+3+2+1} = 3,
but T(10,1) = A000009(10) = 10.

Alois P. Heinz, Rows n = 0..28, flattened
Index entries for triangles and arrays related to Pascal's triangle

Cf. A132311, A007318, A126257, A014631.

T[n_] := T[n] = Table[Binomial[m, k], {m, 0, n-1}, {k, 0, m}] // Flatten // Union;
T[n_, k_] /; k <= n/2 := T[n, k] = Select[ IntegerPartitions[ Binomial[n, k], Length[T[n]], T[n]], Length[#] == Length[Union[#]]&] // Length;
T[n_, k_] /; k > n/2 := T[n, k] = T[n, n-k];
Table[Print["T[", n, ",", k, "] = ", T[n, k]]; T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 02 2020 *)

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cp's OEIS Frontend

A000041 a(n) is the number of partitions of n (the partition numbers).

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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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A132312 Triangle read by rows: T(n,k) = number of partitions of binomial(n,k) into distinct parts of the first n rows of Pascal's triangle, 0<=k<=n.

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