A001297 -id:A001297 - cp's OEIS frontend

A008277 Triangle of Stirling numbers of the second kind, S2(n,k), n >= 1, 1 <= k <= n.

1, 1, 1, 1, 3, 1, 1, 7, 6, 1, 1, 15, 25, 10, 1, 1, 31, 90, 65, 15, 1, 1, 63, 301, 350, 140, 21, 1, 1, 127, 966, 1701, 1050, 266, 28, 1, 1, 255, 3025, 7770, 6951, 2646, 462, 36, 1, 1, 511, 9330, 34105, 42525, 22827, 5880, 750, 45, 1, 1, 1023, 28501, 145750, 246730, 179487, 63987, 11880, 1155, 55, 1

Offset: 1

N. J. A. Sloane

Also known as Stirling set numbers and written {n, k}.
S2(n,k) counts partitions of an n-set into k nonempty subsets.
From Manfred Boergens, Apr 07 2025: (Start)
With regard to the preceding comment:
For disjoint collections of subsets see A256894.
For arbitrary collections of subsets see A163353.
For arbitrary collections of nonempty subsets see A055154. (End)
Triangle S2(n,k), 1 <= k <= n, read by rows, given by [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is Deléham's operator defined in A084938.
Number of partitions of {1, ..., n+1} into k+1 nonempty subsets of nonconsecutive integers, including the partition 1|2|...|n+1 if n=k. E.g., S2(3,2)=3 since the number of partitions of {1,2,3,4} into three subsets of nonconsecutive integers is 3, i.e., 13|2|4, 14|2|3, 1|24|3. - Augustine O. Munagi, Mar 20 2005
Draw n cards (with replacement) from a deck of k cards. Let prob(n,k) be the probability that each card was drawn at least once. Then prob(n,k) = S2(n,k)*k!/k^n (see A090582). - Rainer Rosenthal, Oct 22 2005
Define f_1(x), f_2(x), ..., such that f_1(x)=e^x and for n = 2, 3, ..., f_{n+1}(x) = (d/dx)(x*f_n(x)). Then f_n(x) = e^x*Sum_{k=1..n} S2(n,k)*x^(k-1). - Milan Janjic, May 30 2008
From Peter Bala, Oct 03 2008: (Start)
For tables of restricted Stirling numbers of the second kind see A143494 - A143496.
S2(n,k) gives the number of 'patterns' of words of length n using k distinct symbols - see [Cooper & Kennedy] for an exact definition of the term 'pattern'. As an example, the words AADCBB and XXEGTT, both of length 6, have the same pattern of letters. The five patterns of words of length 3 are AAA, AAB, ABA, BAA and ABC giving row 3 of this table as (1,3,1).
Equivalently, S2(n,k) gives the number of sequences of positive integers (N_1,...,N_n) of length n, with k distinct entries, such that N_1 = 1 and N_(i+1) <= 1 + max{j = 1..i} N_j for i >= 1 (restricted growth functions). For example, Stirling(4,2) = 7 since the sequences of length 4 having 2 distinct entries that satisfy the conditions are (1,1,1,2), (1,1,2,1), (1,2,1,1), (1,1,2,2), (1,2,2,2), (1,2,2,1) and (1,2,1,2).
(End)
Number of combinations of subsets in the plane. - Mats Granvik, Jan 13 2009
S2(n+1,k+1) is the number of size k collections of pairwise disjoint, nonempty subsets of [n]. For example: S2(4,3)=6 because there are six such collections of subsets of [3] that have cardinality two: {(1)(23)},{(12)(3)}, {(13)(2)}, {(1)(2)}, {(1)(3)}, {(2)(3)}. - Geoffrey Critzer, Apr 06 2009
Consider a set of A000217(n) balls of n colors in which, for each integer k = 1 to n, exactly one color appears in the set a total of k times. (Each ball has exactly one color and is indistinguishable from other balls of the same color.) a(n+1, k+1) equals the number of ways to choose 0 or more balls of each color in such a way that exactly (n-k) colors are chosen at least once, and no two colors are chosen the same positive number of times. - Matthew Vandermast, Nov 22 2010
S2(n,k) is the number of monotonic-labeled forests on n vertices with exactly k rooted trees, each of height one or less. See link "Counting forests with Stirling and Bell numbers" below. - Dennis P. Walsh, Nov 16 2011
If D is the operator d/dx, and E the operator xd/dx, Stirling numbers are given by:  E^n = Sum_{k=1..n} S2(n,k) * x^k*D^k. - Hyunwoo Jang, Dec 13 2011
The Stirling polynomials of the second kind (a.k.a. the Bell / Touchard polynomials) are the umbral compositional inverses of the falling factorials (a.k.a. the Pochhammer symbol or Stirling polynomials of the first kind), i.e., binomial(Bell(.,x),n) = x^n/n! (cf. Copeland's 2007 formulas), implying binomial(xD,n) = binomial(Bell(.,:xD:),n) = :xD:^n/n! where D = d/dx and :xD:^n = x^n*D^n. - Tom Copeland, Apr 17 2014
S2(n,k) is the number of ways to nest n matryoshkas (Russian nesting dolls) so that exactly k matryoshkas are not contained in any other matryoshka. - Carlo Sanna, Oct 17 2015
The row polynomials R(n, x) = Sum_{k=1..n} S2(n, k)*x^k appear in the numerator of the e.g.f. of n-th powers, E(n, x) = Sum_{m>=0} m^n*x^m/m!, as E(n, x) = exp(x)*x*R(n, x), for n >= 1. - Wolfdieter Lang, Apr 02 2017
With offsets 0 for n and k this is the Sheffer product matrix A007318*A048993 denoted by (exp(t), (exp(t) - 1)) with e.g.f. exp(t)*exp(x*(exp(t) - 1)). - Wolfdieter Lang, Jun 20 2017
Number of words on k+1 unlabeled letters of length n+1 with no repeated letters. - Thomas Anton, Mar 14 2019
Also coefficients of moments of Poisson distribution about the origin expressed as polynomials in lambda. [Haight] (see also A331155). - N. J. A. Sloane, Jan 14 2020
k!*S2(n,k) is the number of surjections from an n-element set to a k-element set. - Jianing Song, Jun 01 2022

			The triangle S2(n, k) begins:
\ k    1       2       3        4         5         6         7         8        9
n \   10      11      12       13        14        15       ...
----------------------------------------------------------------------------------
1  |   1
2  |   1       1
3  |   1       3       1
4  |   1       7       6        1
5  |   1      15      25       10         1
6  |   1      31      90       65        15         1
7  |   1      63     301      350       140        21         1
8  |   1     127     966     1701      1050       266        28         1
9  |   1     255    3025     7770      6951      2646       462        36        1
10 |   1     511    9330    34105     42525     22827      5880       750       45
       1
11 |   1    1023   28501   145750    246730    179487     63987     11880     1155
      55       1
12 |   1    2047   86526   611501   1379400   1323652    627396    159027    22275
    1705      66       1
13 |   1    4095  261625  2532530   7508501   9321312   5715424   1899612   359502
   39325    2431      78        1
14 |   1    8191  788970 10391745  40075035  63436373  49329280  20912320  5135130
  752752   66066    3367       91         1
15 |   1   16383 2375101 42355950 210766920 420693273 408741333 216627840 67128490
12662650 1479478  106470     4550       105         1
...
----------------------------------------------------------------------------------
x^4 = 1 x_(1) + 7 x_(2) + 6 x_(3) + 1 x_(4), where x_(k) = P(x,k) = k!*C(x,k). - _Daniel Forgues_, Jan 16 2016

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Index entries for "core" sequences

Cf. A008275 (Stirling numbers of first kind), A048993 (another version of this triangle).
Cf. A000217, A001296, A001297, A001298, A007318, A028246, A039810-A039813, A048994, A087107-A087111, A087127, A094262, A127701.
See also A331155.
Cf. A038207, A086885, A132440, A138378, A173018, A238385.
Cf. A055154, A163353, A256894.
Cf. A000110 (row sums), A102661 (partial row sums).

a008277 n k = a008277_tabl !! (n-1) !! (k-1)
a008277_row n = a008277_tabl !! (n-1)
a008277_tabl = map tail $ a048993_tabl  -- Reinhard Zumkeller, Mar 26 2012

n ((] (1 % !)) * +/@((^~ * (] (1 ^ |.)) * (! {:)@]) i.@>:)) k NB. _Stephen Makdisi, Apr 06 2016

[[StirlingSecond(n,k): k in [1..n]]: n in [1..12]]; // G. C. Greubel, May 22 2019

seq(seq(combinat[stirling2](n, k), k=1..n), n=1..10); # Zerinvary Lajos, Jun 02 2007
stirling_2 := (n,k) -> (1/k!) * add((-1)^(k-i)*binomial(k,i)*i^n, i=0..k);

Table[StirlingS2[n, k], {n, 11}, {k, n}] // Flatten (* Robert G. Wilson v, May 23 2006 *)
BellMatrix[f_, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len - 1}, {k, 0, len - 1}]];
rows = 12;
B = BellMatrix[1&, rows];
Table[B[[n, k]], {n, 2, rows}, {k, 2, n}] // Flatten (* Jean-François Alcover, Jun 28 2018, after Peter Luschny *)
a[n_, n_] := 1; a[n_, 1] := 1;
a[n_, k_] := a[n, k] = a[n-1, k-1] + k a[n-1, k]; Flatten@
Table[a[n, k], {n, 1, 11}, {k, 1, n}] (* Oliver Seipel, Jun 12 2024 *)
With[{m = 11},
 Flatten@MapIndexed[Take[#, #2[[1]]] &,
   Transpose@
    Table[Range[1, m]! Coefficient[(E^x-1)^k/k! + O[x]^(m+1), x,
Range[1, m]], {k, 1, m}]]] (* Oliver Seipel, Jun 12 2024 *)

create_list(stirling2(n+1,k+1),n,0,30,k,0,n); /* Emanuele Munarini, Jun 01 2012 */

for(n=1,22,for(k=1,n,print1(stirling(n,k,2),", "));print()); \\ Joerg Arndt, Apr 21 2013

Stirling2(n,k)=sum(i=0,k,(-1)^i*binomial(k,i)*i^n)*(-1)^k/k!  \\ M. F. Hasler, Mar 06 2012

stirling_number2 # Danny Rorabaugh, Oct 11 2015

S2(n, k) = k*S2(n-1, k) + S2(n-1, k-1), n > 1. S2(1, k) = 0, k > 1. S2(1, 1) = 1.
E.g.f.: A(x, y) = e^(y*e^x-y). E.g.f. for m-th column: (e^x-1)^m/m!.
S2(n, k) = (1/k!) * Sum_{i=0..k} (-1)^(k-i)*binomial(k, i)*i^n.
Row sums: Bell number A000110(n) = Sum_{k=1..n} S2(n, k), n>0.
S(n, k) = Sum (i_1*i_2*...*i_(n-k)) summed over all (n-k)-combinations {i_1, i_2, ..., i_k} with repetitions of the numbers {1, 2, ..., k}. Also S(n, k) = Sum (1^(r_1)*2^(r_2)*...* k^(r_k)) summed over integers r_j >= 0, for j=1..k, with Sum{j=1..k} r_j = n-k. [Charalambides]. - Wolfdieter Lang, Aug 15 2019.
A019538(n, k) = k! * S2(n, k).
A028248(n, k) = (k-1)! * S2(n, k).
For asymptotics see Hsu (1948), among other sources.
Sum_{n>=0} S2(n, k)*x^n = x^k/((1-x)(1-2x)(1-3x)...(1-kx)).
Let P(n) = the number of integer partitions of n (A000041), p(i) = the number of parts of the i-th partition of n, d(i) = the number of distinct parts of the i-th partition of n, p(j, i) = the j-th part of the i-th partition of n, m(i, j) = multiplicity of the j-th part of the i-th partition of n, and Sum_{i=1..P(n), p(i)=m} = sum running from i=1 to i=P(n) but taking only partitions with p(i)=m parts into account. Then S2(n, m) = Sum_{i=1..P(n), p(i)=m} n!/(Product_{j=1..p(i)} p(i, j)!) * 1/(Product_{j=1..d(i)} m(i, j)!). For example, S2(6, 3) = 90 because n=6 has the following partitions with m=3 parts: (114), (123), (222). Their complexions are: (114): 6!/(1!*1!*4!) * 1/(2!*1!) = 15, (123): 6!/(1!*2!*3!) * 1/(1!*1!*1!) = 60, (222): 6!/(2!*2!*2!) * 1/(3!) = 15. The sum of the complexions is 15+60+15 = 90 = S2(6, 3). - Thomas Wieder, Jun 02 2005
Sum_{k=1..n} k*S2(n,k) = B(n+1)-B(n), where B(q) are the Bell numbers (A000110). - Emeric Deutsch, Nov 01 2006
Recurrence: S2(n+1,k) = Sum_{i=0..n} binomial(n,i)*S2(i,k-1). With the starting conditions S2(n,k) = 1 for n = 0 or k = 1 and S2(n,k) = 0 for k = 0 we have the closely related recurrence S2(n,k) = Sum_{i=k..n} binomial(n-1,i-1)*S2(i-1,k-1). - Thomas Wieder, Jan 27 2007
Representation of Stirling numbers of the second kind S2(n,k), n=1,2,..., k=1,2,...,n, as special values of hypergeometric function of type (n)F(n-1): S2(n,k)= (-1)^(k-1)*hypergeom([ -k+1,2,2,...,2],[1,1,...,1],1)/(k-1)!, i.e., having n parameters in the numerator: one equal to -k+1 and n-1 parameters all equal to 2; and having n-1 parameters in the denominator all equal to 1 and the value of the argument equal to 1. Example: S2(6,k)= seq(evalf((-1)^(k-1)*hypergeom([ -k+1,2,2,2,2,2],[1,1,1,1,1],1)/(k-1)!),k=1..6)=1,31,90,65,15,1. - Karol A. Penson, Mar 28 2007
From Tom Copeland, Oct 10 2007: (Start)
Bell_n(x) = Sum_{j=0..n} S2(n,j) * x^j = Sum_{j=0..n} E(n,j) * Lag(n,-x, j-n) = Sum_{j=0..n} (E(n,j)/n!) * (n!*Lag(n,-x, j-n)) = Sum_{j=0..n} E(n,j) * binomial(Bell.(x)+j, n) umbrally where Bell_n(x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; and Lag(n,x,m), the associated Laguerre polynomials of order m.
For x = 0, the equation gives Sum_{j=0..n} E(n,j) * binomial(j,n) = 1 for n=0 and 0 for all other n. By substituting the umbral compositional inverse of the Bell polynomials, the lower factorial n!*binomial(y,n), for x in the equation, the Worpitzky identity is obtained; y^n = Sum_{j=0..n} E(n,j) * binomial(y+j,n).
Note that E(n,j)/n! = E(n,j)/(Sum_{k=0..n}E(n,k)). Also (n!*Lag(n, -1, j-n)) is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to the equation for x=1; n!*Bell_n(1) = n!*Sum_{j=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * (n!*Lag(n, -1, j-n)).
(Appended Sep 16 2020) For connections to the Bernoulli numbers, extensions, proofs, and a clear presentation of the number arrays involved in the identities above, see my post Reciprocity and Umbral Witchcraft. (End)
n-th row = leftmost column of nonzero terms of A127701^(n-1). Also, (n+1)-th row of the triangle = A127701 * n-th row; deleting the zeros. Example: A127701 * [1, 3, 1, 0, 0, 0, ...] = [1, 7, 6, 1, 0, 0, 0, ...]. - Gary W. Adamson, Nov 21 2007
The row polynomials are given by D^n(e^(x*t)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A147315 and A094198. See also A185422. - Peter Bala, Nov 25 2011
Let f(x) = e^(e^x). Then for n >= 1, 1/f(x)*(d/dx)^n(f(x)) = 1/f(x)*(d/dx)^(n-1)(e^x*f(x)) = Sum_{k=1..n} S2(n,k)*e^(k*x). Similar formulas hold for A039755, A105794, A111577, A143494 and A154537. - Peter Bala, Mar 01 2012
S2(n,k) = A048993(n,k), 1 <= k <= n. - Reinhard Zumkeller, Mar 26 2012
O.g.f. for the n-th diagonal is D^n(x), where D is the operator x/(1-x)*d/dx. - Peter Bala, Jul 02 2012
n*i!*S2(n-1,i) = Sum_{j=(i+1)..n} (-1)^(j-i+1)*j!/(j-i)*S2(n,j). - Leonid Bedratyuk, Aug 19 2012
G.f.: (1/Q(0)-1)/(x*y), where Q(k) = 1 - (y+k)*x - (k+1)*y*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
From Tom Copeland, Apr 17 2014: (Start)
Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result as A007318(x) = P(x).
With Bell(n,x)=B(n,x) defined above, D = d/dx, and :xD:^n = x^n*D^n, a Dobinski formula gives umbrally f(y)^B(.,x) = e^(-x)*e^(f(y)*x). Then f(y)^B(.,:xD:)g(x) = [f(y)^(xD)]g(x) = e^[-(1-f(y)):xD:]g(x) = g[f(y)x].
In particular, for f(y) = (1+y),
A) (1+y)^B(.,x) = e^(-x)*e^((1+y)*x) = e^(x*y) = e^[log(1+y)B(.,x)],
B) (I+dP)^B(.,x) = e^(x*dP) = P(x) = e^[x*(e^M-I)]= e^[M*B(.,x)] with dP = A132440, M = A238385-I = log(I+dP), and I = identity matrix, and
C) (1+dP)^(xD) = e^(dP:xD:) = P(:xD:) = e^[(e^M-I):xD:] = e^[M*xD] with action e^(dP:xD:)g(x) = g[(I+dP)*x].
D) P(x)^m = P(m*x), which implies (Sum_{k=1..m} a_k)^j = B(j,m*x) where the sum is umbrally evaluated only after exponentiation with (a_k)^q = B(.,x)^q = B(q,x). E.g., (a1+a2+a3)^2=a1^2+a2^2+a3^2+2(a1*a2+a1*a3+a2*a3) = 3*B(2,x)+6*B(1,x)^2 = 9x^2+3x = B(2,3x).
E) P(x)^2 = P(2x) = e^[M*B(.,2x)] = A038207(x), the face vectors of the n-Dim hypercubes.
(End)
As a matrix equivalent of some inversions mentioned above, A008277*A008275 = I, the identity matrix, regarded as lower triangular matrices. - Tom Copeland, Apr 26 2014
O.g.f. for the n-th diagonal of the triangle (n = 0,1,2,...): Sum_{k>=0} k^(k+n)*(x*e^(-x))^k/k!. Cf. the generating functions of the diagonals of A039755. Also cf. A112492. - Peter Bala, Jun 22 2014
Floor(1/(-1 + Sum_{n>=k} 1/S2(n,k))) = A034856(k-1), for k>=2. The fractional portion goes to zero at large k. - Richard R. Forberg, Jan 17 2015
From Daniel Forgues, Jan 16 2016: (Start)
Let x_(n), called a factorial term (Boole, 1970) or a factorial polynomial (Elaydi, 2005: p. 60), denote the falling factorial Product_{k=0..n-1} (x-k). Then, for n >= 1, x_(n) = Sum_{k=1..n} A008275(n,k) * x^k, x^n = Sum_{k=1..n} T(n,k) * x_(k), where A008275(n,k) are Stirling numbers of the first kind.
For n >= 1, the row sums yield the exponential numbers (or Bell numbers): Sum_{k=1..n} T(n,k) = A000110(n), and Sum_{k=1..n} (-1)^(n+k) * T(n,k) = (-1)^n * Sum_{k=1..n} (-1)^k * T(n,k) = (-1)^n * A000587(n), where A000587 are the complementary Bell numbers. (End)
Sum_{k=1..n} k*S2(n,k) = A138378(n). - Alois P. Heinz, Jan 07 2022
O.g.f. for the m-th column: x^m/(Product_{j=1..m} 1-j*x). - Daniel Checa, Aug 25 2022
S2(n,k) ~ (k^n)/k!, for fixed k as n->oo. - Daniel Checa, Nov 08 2022
S2(2n+k, n) ~ (2^(2n+k-1/2) * n^(n+k-1/2)) / (sqrt(Pi*(1-c)) * exp(n) * c^n * (2-c)^(n+k)), where c = -LambertW(-2 * exp(-2)). - Miko Labalan, Dec 21 2024
From Mikhail Kurkov, Mar 05 2025: (Start)
For a general proof of the formulas below via generating functions, see Mathematics Stack Exchange link.
Recursion for the n-th row (independently of other rows): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} (j-2)!*binomial(-k,j)*T(n,k+j-1) for 1 <= k < n with T(n,n) = 1 (see Fedor Petrov link).
Recursion for the k-th column (independently of other columns): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} binomial(n,j)*T(n-j+1,k)*(-1)^j for 1 <= k < n with T(n,n) = 1. (End)

A001296 4-dimensional pyramidal numbers: a(n) = (3n+1)binomial(n+2, 3)/4. Also Stirling2(n+2, n).

Original entry on oeis.org

0, 1, 7, 25, 65, 140, 266, 462, 750, 1155, 1705, 2431, 3367, 4550, 6020, 7820, 9996, 12597, 15675, 19285, 23485, 28336, 33902, 40250, 47450, 55575, 64701, 74907, 86275, 98890, 112840, 128216, 145112, 163625, 183855, 205905, 229881, 255892, 284050, 314470

Offset: 0

N. J. A. Sloane

Permutations avoiding 12-3 that contain the pattern 31-2 exactly once.
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Nov 18 2005
Partial sums of A002411. - Jonathan Vos Post, Mar 16 2006
If Y is a 3-subset of an n-set X then, for n>=6, a(n-5) is the number of 6-subsets of X having at least two elements in common with Y. - Milan Janjic, Nov 23 2007
Starting with 1 = binomial transform of [1, 6, 12, 10, 3, 0, 0, 0, ...]. Equals row sums of triangle A143037. - Gary W. Adamson, Jul 18 2008
Rephrasing the Perry formula of 2003: a(n) is the sum of all products of all two  numbers less than or equal to n, including the squares. Example: for n=3 the sum of these products is 1*1 + 1*2 + 1*3 + 2*2 + 2*3 + 3*3 = 25. - J. M. Bergot, Jul 16 2011
Half of the partial sums of A011379. [Jolley, Summation of Series, Dover (1961), page 12 eq (66).] - R. J. Mathar, Oct 03 2011
Also the number of (w,x,y,z) with all terms in {1,...,n+1} and w < x >= y > z (see A211795). - Clark Kimberling, May 19 2012
Convolution of A000027 with A000326. - Bruno Berselli, Dec 06 2012
This sequence is related to A000292 by a(n) = n*A000292(n) - Sum_{i=0..n-1} A000292(i) for n>0. - Bruno Berselli, Nov 23 2017
a(n-2) is the maximum number of intersections made from the perpendicular bisectors of all pair combinations of n points. - Ian Tam, Dec 22 2020

			G.f. = x + 7*x^2 + 25*x^3 + 65*x^4 + 140*x^5 + 266*x^6 + 462*x^7 + 750*x^8 + 1155*x^9 + ...

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 835.
A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 195.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 227, #16.
S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see p. 166, Table 10.4/I/3).
F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 223.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
P. Aluffi, Degrees of projections of rank loci, arXiv:1408.1702 [math.AG], 2014. ["After compiling the results of many explicit computations, we noticed that many of the numbers d_{n,r,S} appear in the existing literature in contexts far removed from the enumerative geometry of rank conditions; we owe this surprising (to us) observation to perusal of [Slo14]."]
S. Butler and P. Karasik, A note on nested sums, J. Int. Seq. 13 (2010), 10.4.4, page 5.
M. Griffiths, Remodified Bessel Functions via Coincidences and Near Coincidences, Journal of Integer Sequences, Vol. 14 (2011), Article 11.7.1.
L. Hogben, Choice and Chance by Cardpack and Chessboard, Vol. 1, Chanticleer Press, NY, 1950, p. 36.
C. Krishnamachaki, The operator (xD)^n, J. Indian Math. Soc., 15 (1923),3-4. [Annotated scanned copy]
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 29.
T. Mansour, Restricted permutations by patterns of type 2-1, arXiv:math/0202219 [math.CO], 2002.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Eric Weisstein's World of Mathematics, Stirling numbers of the 2nd kind.
Index to sequences related to pyramidal numbers
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

a(n)=f(n, 2) where f is given in A034261.
Cf. A000217, A000326, A001297, A001298, A002411, A008277, A008517, A094262.
a(n)= A093560(n+3, 4), (3, 1)-Pascal column.
Cf. A220212 for a list of sequences produced by the convolution of the natural numbers with the k-gonal numbers.
Cf. similar sequences listed in A241765 and A254142.
Cf. A000914.

/* A000027 convolved with A000326: */ A000326:=func; [&+[(n-i+1)*A000326(i): i in [0..n]]: n in [0..40]]; // Bruno Berselli, Dec 06 2012

[(3*n+1)*Binomial(n+2,3)/4: n in [0..40]]; // Vincenzo Librandi, Jul 30 2014

A001296:=-(1+2*z)/(z-1)**5; # Simon Plouffe in his 1992 dissertation for sequence without the leading zero

Table[n*(1+n)*(2+n)*(1+3*n)/24, {n, 0, 100}]
CoefficientList[Series[x (1 + 2 x)/(1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Jul 30 2014 *)
Table[StirlingS2[n+2, n], {n, 0, 40}] (* Jean-François Alcover, Jun 24 2015 *)
Table[ListCorrelate[Accumulate[Range[n]],Range[n]],{n,0,40}]//Flatten (* or *) LinearRecurrence[{5,-10,10,-5,1},{0,1,7,25,65},40] (* Harvey P. Dale, Aug 14 2017 *)

t(n)=n*(n+1)/2
for(i=1,30,print1(","sum(j=1,i,j*t(j))))

{a(n) = n * (n+1) * (n+2) * (3*n+1) / 24}; /* Michael Somos, Sep 04 2017 */

[stirling_number2(n+2,n) for n in range(0,38)] # Zerinvary Lajos, Mar 14 2009

a(n) = n*(1+n)*(2+n)*(1+3*n)/24. - T. D. Noe, Jan 21 2008
G.f.: x*(1+2*x)/(1-x)^5. - Paul Barry, Jul 23 2003
a(n) = Sum_{j=0..n} j*A000217(j). - Jon Perry, Jul 28 2003
E.g.f. with offset -1: exp(x)*(1*(x^2)/2! + 4*(x^3)/3! + 3*(x^4)/4!). For the coefficients [1, 4, 3] see triangle A112493.
E.g.f. x*exp(x)*(24 + 60*x + 28*x^2 + 3*x^3)/24 (above e.g.f. differentiated).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + 3. - Kieren MacMillan, Sep 29 2008
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Jaume Oliver Lafont, Nov 23 2008
O.g.f. is D^2(x/(1-x)) = D^3(x), where D is the operator x/(1-x)*d/dx. - Peter Bala, Jul 02 2012
a(n) = A153978(n)/2. - J. M. Bergot, Aug 09 2013
a(n) = A002817(n) + A000292(n-1). - J. M. Bergot, Aug 29 2013; [corrected by Cyril Damamme, Feb 26 2018]
a(n) = A000914(n+1) - 2 * A000330(n+1). - Antal Pinter, Dec 31 2015
a(n) = A080852(3,n-1). - R. J. Mathar, Jul 28 2016
a(n) = 1*(1+2+...+n) + 2*(2+3+...+n) + ... + n*n. For example, a(6) = 266 = 1(1+2+3+4+5+6) + 2*(2+3+4+5+6) + 3*(3+4+5+6) + 4*(4+5+6) + 5*(5+6) + 6*(6).- J. M. Bergot, Apr 20 2017
a(n) = A000914(-2-n) for all n in Z. - Michael Somos, Sep 04 2017
a(n) = A000292(n) + A050534(n+1). - Cyril Damamme, Feb 26 2018
From Amiram Eldar, Jul 02 2020: (Start)
Sum_{n>=1} 1/a(n) = (6/5) * (47 - 3*sqrt(3)*Pi - 27*log(3)).
Sum_{n>=1} (-1)^(n+1)/a(n) = (6/5) * (16*log(2) + 6*sqrt(3)*Pi - 43). (End)

A062196 Triangle read by rows, T(n, k) = binomial(n, k)*binomial(n + 2, k).

Original entry on oeis.org

1, 1, 3, 1, 8, 6, 1, 15, 30, 10, 1, 24, 90, 80, 15, 1, 35, 210, 350, 175, 21, 1, 48, 420, 1120, 1050, 336, 28, 1, 63, 756, 2940, 4410, 2646, 588, 36, 1, 80, 1260, 6720, 14700, 14112, 5880, 960, 45, 1, 99, 1980, 13860, 41580, 58212, 38808, 11880, 1485, 55

Offset: 0

Wolfdieter Lang, Jun 19 2001

Also the coefficient triangle of certain polynomials N(2; m,x) := Sum_{k=0..m} T(m,k)*x^k. The e.g.f. of the m-th (unsigned) column sequence without leading zeros of the generalized (a=2) Laguerre triangle L(2; n+m,m) = A062139(n+m,m), n >= 0, is N(2; m,x)/(1-x)^(3+2*m), with the row polynomials N(2; m,x).

			Triangle starts:
  n\k 0...1.....2......3..... 4.....;
  [0] 1;
  [1] 1,  3;
  [2] 1,  8,    6;
  [3] 1, 15,   30,    10;
  [4] 1, 24,   90,    80,    15;
  [5] 1, 35,  210,   350,   175,    21;
  [6] 1, 48,  420,  1120,  1050,   336,    28;
  [7] 1, 63,  756,  2940,  4410,  2646,   588,    36;
  [8] 1, 80, 1260,  6720, 14700, 14112,  5880,   960,   45;
  [9] 1, 99, 1980, 13860, 41580, 58212, 38808, 11880, 1485, 55.

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Family of polynomials (see A062145): A008459 (c=1), A132813 (c=2), this sequence (c=3), A062145 (c=4), A062264 (c=5), A062190 (c=6).
Sums include: A001791 (row), (-1)^n*A089849(n+1) (alternating sign row).
Diagonals: A000217 (k=n), A002417 (k=n-1), A001297 (k=n-2), A105946 (k=n-3), A105947 (k=n-4), A105948 (k=n-5), A107319 (k=n-6).
Columns: A005563 (k=1), A033487 (k=2), A027790 (k=3), A107395 (k=4), A107396 (k=5), A107397 (k=6), A107398 (k=7), A107399 (k=8).
Cf. A000108, A089849.

A062196:= func;
[A062196(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 21 2025

T := (n, k) -> binomial(n, k)*binomial(n + 2, k);
seq(seq(T(n, k), k=0..n), n=0..9); # Peter Luschny, Sep 30 2021

A062196[n_, k_]:= Binomial[n, k]*Binomial[n+2, k];
Table[A062196[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 21 2025 *)

def A062196(n,k): return binomial(n,k)*binomial(n+2,k)
print(flatten([[A062196(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Feb 21 2025

T(m, k) = [x^k] N(2; m, x), where N(2; m, x) = ((1-x)^(3+2*m))*(d^m/dx^m)(x^m/(m!*(1-x)^(m+3))).
N(2; m, x) = Sum_{j=0..m} ((binomial(m, j)*(2*m+2-j)!/((m+2)!*(m-j)!)*(x^(m-j)))*(1-x)^j).
T(n,m) = binomial(n, m)*(binomial(n+1, m) + binomial(n+1, m-1)). - Vladimir Kruchinin, Apr 06 2018
From G. C. Greubel, Feb 21 2025: (Start)
T(2*n, n) = (n+1)^2*A000108(n)*A000108(n+1).
T(2*n-1, n) = (4*n^2 - 1)*A000108(n-1)*A000108(n), n >= 1.
T(2*n+1, n) = (1/2)*binomial(n+2,2)*A000108(n+1)*A000108(n+2). (End)

New name by Peter Luschny, Sep 30 2021

A001303 Stirling numbers of first kind, s(n+3, n), negated.

Original entry on oeis.org

6, 50, 225, 735, 1960, 4536, 9450, 18150, 32670, 55770, 91091, 143325, 218400, 323680, 468180, 662796, 920550, 1256850, 1689765, 2240315, 2932776, 3795000, 4858750, 6160050, 7739550, 9642906, 11921175, 14631225, 17836160, 21605760, 26016936, 31154200

Offset: 1

N. J. A. Sloane

a(n) is equal to the sum of the products of each distinct grouping of 3 members of the set {1, 2, 3, ..., n + 2} (a(1) = 1*2*3, a(2) = 1*2*3 + 1*2*4 + 1*3*4 + 2*3*4, a(3) = 1*2*3 + 1*2*4 + 1*2*5 + 1*3*4 + 1*3*5 + 1*4*5 + 2*3*4 + 2*3*5 + 2*4*5 + 3*4*5). - Jeffreylee R. Snow, Sep 23 2013

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 833.
Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 227, #16.
F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 226.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..1000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Karl Dienger, Beiträge zur Lehre von den arithmetischen und geometrischen Reihen höherer Ordnung, Jahres-Bericht Ludwig-Wilhelm-Gymnasium Rastatt, Rastatt, 1910. [Annotated scanned copy]
G. C. Greubel, A Note on Jain basis functions, arXiv:1612.09385 [math.CA], 2016.
Robert E. Moritz, On the sum of products of n consecutive integers, Univ. Washington Publications in Math., Vol. 1, No. 3 (1926), pp. 44-49. [Annotated scanned copy]
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000217, A001297, A008275.

seq(numbperm (n,2)*numbperm (n,4)/48, n=4..33); # Zerinvary Lajos, Apr 26 2007
seq(15*binomial(n+2,6)-10*binomial(n+1,5)+binomial(n,4),n=4..30); # Miklos Kristof, Nov 04 2007
A001303 := proc(n)
    -combinat[stirling1](n+3,n) ;
end proc: # R. J. Mathar, May 19 2016

Table[-StirlingS1[n + 3, n], {n, 100}] (* T. D. Noe, Jun 27 2012 *)
a[ n_] := n (n + 1) (n + 2)^2 (n + 3)^2 / 48; (* Michael Somos, Sep 04 2017 *)

a(n) = n*(n+1)*(n+2)^2*(n+3)^2/48; \\ Altug Alkan, Aug 29 2017

[stirling_number1(n,n-3) for n in range(4, 34)] # Zerinvary Lajos, May 16 2009

a(n) = binomial(n+3, 4)*binomial(n+3, 2).
G.f.: x*(6 + 8*x + x^2)/(1 - x)^7. - Simon Plouffe in his 1992 dissertation
E.g.f. with offset 3: exp(x)*(6*(x^3)/3! + 26*(x^4)/4! + 35*(x^5)/5! + 15*(x^6)/6!). See row k=3 of A112486 for the coefficients [6, 26, 35, 15].
a(n) = (f(n+2, 3)/6!)*Sum_{m=0..min(3, n)} A112486(3,m)*f(6, 3-m)*f(n-1, m), with the falling factorials notation f(n, m):=n*(n-1)*...*(n-(m-1)).
From Jason Lang, Oct 03 2006: (Start)
a(n) = A000217(n) * n! / ( 4! * (n-4)! ) [for n > 4 and A000217 = the triangular numbers];
a(n) = ((n+4)! / n! ) ^2 / ( (n+2) * (n+1) * 2*4!);
a(n) = (n-0)^2 * (n-1)^2 * (n-2) * (n-3) / (2*4!). (End)
From Miklos Kristof, Nov 04 2007: (Start)
a(n) = 15*binomial(n+5,6) - 10*binomial(n+4,5) + binomial(n+3,4).
E.g.f. with offset 4: exp(x)*((1/4)*x^4 + (1/6)*x^5 + (1/48)*x^6). (End)
a(n) = n*(n+1)(n+2)^2*(n+3)^2/48. - Jeremy Galvagni, Mar 03 2009
From Gary Detlefs, Jun 06 2010: (Start)
a(n) = (n+3)^2/(n^2-1)*a(n-1), n > 1;
a(n) = 6*Product_{k=2..n} (k+3)^2/(k^2 - 1). (End)
a(n) = A001297(-3-n) for all n in Z. - Michael Somos, Sep 04 2017
From Amiram Eldar, Jan 10 2022: (Start)
Sum_{n>=1} 1/a(n) = 16*Pi^2/3 - 472/9.
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*Pi^2/3 + 16/9 - 64*log(2)/3. (End)

More terms from Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Jan 17 2000

Notation of the polynomial formula edited by R. J. Mathar, Sep 15 2009

A094262 Triangle read by rows: T(n,k) is the number of rooted trees with k nodes which are disjoint sets of labels with union {1..n}. If a node has an empty set of labels then it must have at least two children.

Original entry on oeis.org

1, 1, 2, 1, 1, 6, 12, 10, 3, 1, 14, 61, 124, 131, 70, 15, 1, 30, 240, 890, 1830, 2226, 1600, 630, 105, 1, 62, 841, 5060, 16990, 35216, 47062, 40796, 22225, 6930, 945, 1, 126, 2772, 25410, 127953, 401436, 836976, 1196532, 1182195, 795718, 349020, 90090, 10395

Offset: 1

André F. Labossière, Jun 01 2004

The original name for this sequence was "Triangle read by rows giving the coefficients of formulas generating each variety of S2(n,k) (Stirling numbers of 2nd kind). The p-th row (p>=1) contains T(i,p) for i=1 to 2*p-1, where T(i,p) satisfies Sum_{i=1..2*p-1} T(i,p) * C(n-p,i-1)".
The terms of the n-th diagonal sequence of the triangle of Stirling numbers of the second kind A008277, i.e., (Stirling2(N + n - 1,N)), N>=1, are given by a polynomial in N of degree 2*n - 2. This polynomial may be expressed as a linear combination of the falling factorial polynomials binomial(N - n,0), binomial(N - n,1), ... , binomial(N - n,2*n - 2). This table gives the coefficients in these expansions.
The formulas obtained are those for Stirling2(N+1,N) (A000217), Stirling2(N+2,N) (A001296), Stirling2(N+3,N) (A001297), Stirling2(N+4,N) (A001298), Stirling2(N+5,N) (A112494), Stirling2(N+6,N) (A144969) and so on.

			Row 5 contains 1,30,240,890,1830,2226,1600,630,105, so the formula generating Stirling2(n+4,n) numbers (A001298) will be the following: 1 + 30*(n-5) + 240*C(n-5,2) + 890*C(n-5,3) + 1830*C(n-5,4) + 2226*C(n-5,5) + 1600*C(n-5,6) + 630*C(n-5,7) + 105*C(n-5,8). For example, taking n = 9 gives Stirling2(13,9) = 359502.
Triangle starts:
  1;
  1,  2,   1;
  1,  6,  12,  10,    3;
  1, 14,  61, 124,  131,   70,   15;
  1, 30, 240, 890, 1830, 2226, 1600, 630, 105;
  ...
From _Peter Bala_, Jun 14 2016: (Start)
Connection with row polynomials of A134991:
  R(2,z) = (1 + z)^2*z
  R(3,z) = (1 + z)^2*(z + 3*z^2)
  R(4,z) = (1 + z)^4*(z + 10*z^2 + 15*z^3)
  R(5,z) = (1 + z)^5*(z + 25*z^2 + 105*z^3 + 105*z^4). (End)
From _Andrew Howroyd_, Mar 28 2025: (Start)
The T(3,3) = 12 trees up to relabeling have one of the following 3 forms:
     {}         {1}        {1}
    /  \       /   \        |
  {1} {2,3}   {2}  {3}     {2}
                            |
                           {3}
(End)

Andrew Howroyd, Table of n, a(n) for n = 1..2500 (rows 1..50)
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions (with Formulas, Graphs and Mathematical Tables), U.S. Dept. of Commerce, National Bureau of Standards, Applied Math. Series 55, 1964, 1046 pages (9th Printing: November 1970) - Combinatorial Analysis, Table 24.4, Stirling Numbers of the Second Kind (author: Francis L. Miksa), p. 835.
J. Fernando Barbero G., Jesús Salas, Eduardo J. S. Villaseñor, Generalized Stirling permutations and forests: Higher-order Eulerian and Ward numbers, Electronic Journal of Combinatorics 22(3) (2015), #P3.37.
M. Kazarian, KP hierarchy for Hodge integrals, p. 2, arxiv:0809.3263 [math.AG], 18 Sep 2008. [From _Tom Copeland_, Jun 12 2015]
F. R. McMorris and T. Zaslavsky, The number of cladistic characters, Math. Biosciences, 54 (1981), 3-10.
F. R. McMorris and T. Zaslavsky, The number of cladistic characters, Math. Biosciences, 54 (1981), 3-10. [Annotated scanned copy]
Eric Weisstein's World of Mathematics, Stirling numbers of the 2nd kind.

Row sums are A005172.
Columns 1..4 are A000012, A000918, A005173, A005174, A005175.
Cf. A008277, A000217, A001296, A001297, A001298, A008517, A134991, A112494, A144969.

row_poly := n -> (1+z)^(n+1)*add(z^k*add((-1)^(m+k)*binomial(n+k,n+m)*Stirling2(n+m,m), m=0..k), k=0..n): T_row := n -> seq(coeff(row_poly(n),z,j),j=1..2*n+1):
seq(T_row(n),n=0..6); # Peter Luschny, Jun 15 2016

Clear[T, q, u]; T[0] = q[1];T[n_] := Sum[m*(u^2*q[m] + 2*u*q[m+1] + q[m+2])*D[T[n-1], q[m]], {m, 1, 2*n+1}]; row[n_] := List @@ Expand[T[n-1]] /. {u -> 1, q[] -> 1}; Table[row[n], {n, 1, 7}] // Flatten (* _Jean-François Alcover, Jun 12 2015 *)

T(n)={my(g=serreverse(log(((1+1/y)*x+1)/exp(x + O(x*x^n))))); [Vecrev(p/y) | p<-Vec(serlaplace(g))]}
{ my(A=T(5)); for(i=1, #A, print(A[i])) } \\ Andrew Howroyd, Mar 28 2025

Apparently, a raising operator for bivariate polynomials P(n,u,z) having these coefficients is R = (u+z)^2 * z * d/dz with P(0,u,z) = z. E.g., R P(1,u,z) = R^2 P(0,u,z) = R^2 z = u^4 z + 6 u^3 z^2 + 12 u^2 z^3 + 10 u z^4 + 3 z^5 = P(2,u,z). See the Kazarian link. - Tom Copeland, Jun 12 2015
Reverse polynomials seem to be generated by 1 + exp[t*(x+1+z)^2*(1+z)d/dz]z evaluated at z = 0. - Tom Copeland, Jun 13 2015
From Peter Bala, Jun 14 2016: (Start)
T(n,k) = k*T(n,k) + 2*(k - 1)*T(n,k-1) + (k - 2)*T(n,k-2).
n-th diagonal of A008277: Stirling2(N + n - 1,N) = Sum_{k = 1..2*n - 1} T(n,k)*binomial(N - n,k - 1) for N = 1,2,3,....
Row polynomials R(n,z) = Sum_{k >= 1} k^(n+k-1)*( z/(1 + z)*exp(-z/(1 + z)) )^k/k!, n = 1,2,..., follows from the formula given in A008277 for the o.g.f.'s of the diagonals of the Stirling numbers of the second kind.
Consequently, R(n+1,z) = (1 + z)^2*z*d/dz(R(n,z)) for n >= 1 as conjectured above by Copeland.
R(n,z) = (1 + z)^n*P(n,z) where P(n,z) are the row polynomials of A134991.
R(n,z) = (1 + z)^(2*n+1)*B(n,z/(1 + z)), where B(n,z) are the row polynomials of the triangle of second-order Eulerian numbers A008517 (see Barbero et al., Section 6, equation 27). (End)
Based on the comment of Bala the row polynomials have the explicit form R(n, z) = (1+z)^(n+1)*Sum_{k=0..n}(z^k*Sum_{m=0..k}((-1)^(m+k)*binomial(n+k, n+m)* Stirling2(n+m,m))). - Peter Luschny, Jun 15 2016
E.g.f. G(x,y) satisfies G(x,y) = y*(exp(x)*exp(G(x,y)) - G(x,y) - 1). - Andrew Howroyd, Mar 28 2025

Edited and Name changed by Peter Bala, Jun 16 2016

Name changed by Andrew Howroyd, Mar 28 2025

cp's OEIS Frontend

A008277 Triangle of Stirling numbers of the second kind, S2(n,k), n >= 1, 1 <= k <= n.

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A001296 4-dimensional pyramidal numbers: a(n) = (3*n+1)*binomial(n+2, 3)/4. Also Stirling2(n+2, n).

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A062196 Triangle read by rows, T(n, k) = binomial(n, k)*binomial(n + 2, k).

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A001303 Stirling numbers of first kind, s(n+3, n), negated.

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A094262 Triangle read by rows: T(n,k) is the number of rooted trees with k nodes which are disjoint sets of labels with union {1..n}. If a node has an empty set of labels then it must have at least two children.

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A001296 4-dimensional pyramidal numbers: a(n) = (3n+1)binomial(n+2, 3)/4. Also Stirling2(n+2, n).

A037961 a(n) = n^2(n+1)(n+3)!/48.

A351105 a(n) = Sum_{k=1..n} Sum_{j=1..k} Sum_{i=1..j} (ijk)^2.