A156289 -id:A156289 - cp's OEIS frontend

A136630 Triangular array: T(n,k) counts the partitions of the set [n] into k odd sized blocks.

1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 4, 0, 1, 0, 1, 0, 10, 0, 1, 0, 0, 16, 0, 20, 0, 1, 0, 1, 0, 91, 0, 35, 0, 1, 0, 0, 64, 0, 336, 0, 56, 0, 1, 0, 1, 0, 820, 0, 966, 0, 84, 0, 1, 0, 0, 256, 0, 5440, 0, 2352, 0, 120, 0, 1, 0, 1, 0, 7381, 0, 24970, 0, 5082, 0, 165, 0, 1, 0, 0, 1024, 0, 87296, 0

Offset: 0

Paul D. Hanna, Jan 14 2008

For partitions into blocks of even size see A156289.
Essentially the unsigned matrix inverse of triangle A121408.
From Peter Bala, Jul 28 2014: (Start)
Define a polynomial sequence x_(n) by setting x_(0) = 1 and for n = 1,2,... setting x_(n) = x*(x + n - 2)*(x + n - 4)*...*(x + n - 2*(n - 1)). Then this table is the triangle of connection constants for expressing the monomial polynomials x^n in terms of the basis x_(k), that is, x^n = sum {k = 0..n} T(n,k)*x_(k) for n = 0,1,2,.... An example is given below.
Let M denote the lower unit triangular array A119467 and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0  M/ having the k x k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle, omitting the first row and column, equals the infinite matrix product M(0)*M(1)*M(2)*.... (End)
Also the Bell transform of A000035(n+1). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 27 2016

			Triangle begins:
  1;
  0, 1;
  0, 0,   1;
  0, 1,   0,    1;
  0, 0,   4,    0,    1;
  0, 1,   0,   10,    0,     1;
  0, 0,  16,    0,   20,     0,    1;
  0, 1,   0,   91,    0,    35,    0,    1;
  0, 0,  64,    0,  336,     0,   56,    0,   1;
  0, 1,   0,  820,    0,   966,    0,   84,   0,   1;
  0, 0, 256,    0, 5440,     0, 2352,    0, 120,   0, 1;
  0, 1,   0, 7381,    0, 24970,    0, 5082,   0, 165, 0, 1;
T(5,3) = 10. The ten partitions of the set [5] into 3 odd-sized blocks are
(1)(2)(345), (1)(3)(245), (1)(4)(235), (1)(5)(234), (2)(3)(145),
(2)(4)(135), (2)(5)(134), (3)(4)(125), (3)(5)(124), (4)(5)(123).
Connection constants: Row 5 = [0,1,0,10,0,1]. Hence, with the polynomial sequence x_(n) as defined in the Comments section we have x^5 = x_(1) + 10*x_(3) + x_(5) = x + 10*x*(x+1)*(x-1) + x*(x+3)*(x+1)*(x-1)*(x-3).

L. Comtet, Analyse Combinatoire, Presses Univ. de France, 1970, Vol. II, pages 61-62.
L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 225-226.

Ch. A. Charalambides, Central factorial numbers and related expansions, Fib. Quarterly, Vol. 19, No 5, Dec 1981, pp. 451-456.
Feng Qi and Peter Taylor, Series expansions for powers of sinc function and closed-form expressions for specific partial Bell polynomials, Appl. Anal. Disc. Math. (2024) Vol. 18, No. 1, 1-24. See p. 13.

Cf. A121408; A136631 (antidiagonal sums), A003724 (row sums), A136632; A002452 (column 3), A002453 (column 5); A008958 (central factorial triangle), A156289. A185690, A196776.

A136630 := proc (n, k) option remember; if k < 0 or n < k then 0 elif k = n then 1 else procname(n-2, k-2) + k^2*procname(n-2, k) end if end proc: seq(seq(A136630(n, k), k = 1 .. n), n = 1 .. 12); # Peter Bala, Jul 27 2014
# The function BellMatrix is defined in A264428.
BellMatrix(n -> (n+1) mod 2, 9); # Peter Luschny, Jan 27 2016

t[n_, k_] := Coefficient[ x^k/Product[ 1 - (2*j + k - 2*Quotient[k, 2])^2*x^2, {j, 0, k/2}] + x*O[x]^n, x, n]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 22 2013, after Pari *)
BellMatrix[f_Function, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len-1}, {k, 0, len-1}]];
rows = 13;
M = BellMatrix[Mod[#+1, 2]&, rows];
Table[M[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 23 2018, after Peter Luschny *)

{T(n,k)=polcoeff(x^k/prod(j=0,k\2,1-(2*j+k-2*(k\2))^2*x^2 +x*O(x^n)),n)}

G.f. for column k: x^k/Product_{j=0..floor(k/2)} (1 - (2*j + k-2*floor(k/2))^2 * x^2).
G.f. for column 2*k: x^(2*k)/Product_{j=0..k} (1 - (2*j)^2*x^2).
G.f. for column 2*k+1: x^(2*k+1)/Product_{j=0..k} (1 - (2*j+1)^2*x^2).
From Peter Bala, Feb 21 2011 (Start)
T(n,k) = 1/(2^k*k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(2*j-k)^n,
Recurrence relation T(n+2,k) = T(n,k-2) + k^2*T(n,k).
E.g.f.: F(x,z) = exp(x*sinh(z)) = Sum_{n>=0} R(n,x)*z^n/n! = 1 + x*z + x^2*z^2/2! + (x+x^3)*z^3/3! + ....
The row polynomials R(n,x) begin
R(1,x) = x
R(2,x) = x^2
R(3,x) = x+x^3.
The e.g.f. F(x,z) satisfies the partial differential equation d^2/dz^2(F) = x^2*F + x*F' + x^2*F'' where ' denotes differentiation w.r.t. x.
Hence the row polynomials satisfy the recurrence relation R(n+2,x) = x^2*R(n,x) + x*R'(n,x) + x^2*R''(n,x) with R(0,x) = 1.
The recurrence relation for T(n,k) given above follows from this.
(End)
For the corresponding triangle of ordered partitions into odd-sized blocks see A196776. Let P denote Pascal's triangle A070318 and put M = 1/2*(P-P^-1). M is A162590 (see also A131047). Then the first column of exp(t*M) lists the row polynomials for the present triangle. - Peter Bala, Oct 06 2011
Row generating polynomials equal D^n(exp(x*t)) evaluated at x = 0, where D is the operator sqrt(1+x^2)*d/dx. Cf. A196776. - Peter Bala, Dec 06 2011
From Peter Bala, Jul 28 2014: (Start)
E.g.f.: exp(t*sinh(x)) = 1 + t*x + t^2*x^2/2! + (t + t^3)*x^3/3! + ....
Hockey-stick recurrence: T(n+1,k+1) = Sum_{i = 0..floor((n-k)/2)} binomial(n,2*i)*T(n-2*i,k).
Recurrence equation for the row polynomials R(n,t):
R(n+1,t) = t*Sum_{k = 0..floor(n/2)} binomial(n,2*k)*R(n-2*k,t) with R(0,t) = 1. (End)

A003724 Number of partitions of n-set into odd blocks.

Original entry on oeis.org

1, 1, 1, 2, 5, 12, 37, 128, 457, 1872, 8169, 37600, 188685, 990784, 5497741, 32333824, 197920145, 1272660224, 8541537105, 59527313920, 432381471509, 3252626013184, 25340238127989, 204354574172160, 1699894200469849, 14594815769038848, 129076687233903673

Offset: 0

R. H. Hardin

			G.f. = 1 + x + x^2 + 2*x^3 + 5*x^4 + 12*x^5 + 37*x^6 + 128*x^7 + 457*x^8 + ...

L. Comtet, Analyse Combinatoire, Presses Univ. de France, 1970, Vol. II, pages 61-62.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 225, 2nd line of table.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Alois P. Heinz, Table of n, a(n) for n = 0..592 (first 101 terms from T. D. Noe)
J. Riordan, Letter, Jul 06 1978
Kruchinin Vladimir Victorovich, Composition of ordinary generating functions, arXiv:1009.2565 [math.CO], 2010.

See A136630 for the table of partitions of an n-set into k odd blocks.
For partitions into even blocks see A005046 and A156289.
Cf. A000009, A000110. A002017, A009623.

a:= proc(n) option remember; `if`(n=0, 1, add(
      binomial(n-1, j-1)*irem(j, 2)*a(n-j), j=1..n))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Mar 17 2015

a[n_] := Sum[((-1)^i*(k - 2*i)^n*Binomial[k, i])/(2^k*k!), {k, 1, n}, {i, 0, k}]; a[0] = 1; Table[a[n], {n, 0, 24}] (* Jean-François Alcover, Dec 21 2011, after Vladimir Kruchinin *)
With[{nn=30},CoefficientList[Series[Exp[Sinh[x]],{x,0,nn}],x]Range[0,nn]!] (* Harvey P. Dale, Apr 06 2012 *)
Table[Sum[BellY[n, k, Mod[Range[n], 2]], {k, 0, n}], {n, 0, 24}] (* Vladimir Reshetnikov, Nov 09 2016 *)

a(n):=sum(1/2^k*sum((-1)^i*binomial(k,i)*(k-2*i)^n,i,0,k)/k!,k,1,n); /* Vladimir Kruchinin, Aug 22 2010 */

E.g.f.: exp ( sinh x ).
a(n) = sum(1/2^k*sum((-1)^i*C(k,i)*(k-2*i)^n, i=0..k)/k!, k=1..n). - Vladimir Kruchinin, Aug 22 2010
a(n) = D^n(exp(x)) evaluated at x = 0, where D is the operator sqrt(1+x^2)*d/dx. Cf. A002017 and A009623. - Peter Bala, Dec 06 2011
a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-1,2*k) * a(n-2*k-1). - Ilya Gutkovskiy, Jul 11 2021
O.g.f A(X) satisfies A(x) = 1 + x*( A(x/(1-x))/(1-x) + A(x/(1+x))/(1+x) )/2. - Paul D. Hanna, Aug 19 2024

A005046 Number of partitions of a 2n-set into even blocks.

Original entry on oeis.org

1, 1, 4, 31, 379, 6556, 150349, 4373461, 156297964, 6698486371, 337789490599, 19738202807236, 1319703681935929, 99896787342523081, 8484301665702298804, 802221679220975886631, 83877585692383961052499, 9640193854278691671399436, 1211499609050804749310115589

Offset: 0

N. J. A. Sloane

Conjecture: Taking the sequence modulo an integer k gives an eventually periodic sequence. For example, the sequence taken modulo 10 is [1, 1, 4, 1, 9, 6, 9, 1, 4, 1, 9, 6, 9, 1, 4, 1, 9, 6, 9, ...], with an apparent period [1, 4, 1, 9, 6, 9] beginning at a(1), of length 6. Cf. A006154. - Peter Bala, Apr 12 2023

Louis Comtet, Analyse Combinatoire Tome II, pages 61-62.
Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 225, 3rd line of table.
CRC Standard Mathematical Tables and Formulae, 30th ed. 1996, p. 42.
L. B. W. Jolley, Summation of Series. 2nd ed., Dover, NY, 1961, p. 150.
L. Lovasz, Combinatorial Problems and Exercises, North-Holland, 1993, pp. 15.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Alois P. Heinz, Table of n, a(n) for n = 0..250 (first 51 terms from T. D. Noe)
C. Ahmed, P. Martin, and V. Mazorchuk, On the number of principal ideals in d-tonal partition monoids, arXiv preprint arXiv:1503.06718 [math.CO], 2015-2019.
Steven R. Finch, Moments of sums, April 23, 2004 [Cached copy, with permission of the author]
Vladimir Victorovich Kruchinin, Composition of ordinary generating functions, arXiv:1009.2565 [math.CO], 2010.
J. Riordan, Letter, Jul 06 1978
J. Shallit, Letter to N. J. A. Sloane, Jan 13 1976.
Sebastian Volz, Design and Implementation of Efficient Algorithms for Operations on Partitions of Sets, Bachelor Thesis, Saarland Univ. (Germany, 2023). See p. 45.

See A156289 for the table of partitions of a 2n-set into k even blocks.
For partitions into odd blocks see A003724 and A136630.
Cf. A000110, A003724.

a:= proc(n) option remember;
      `if`(n=0, 1, add(binomial(2*n-1, 2*k-1) *a(n-k), k=1..n))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Apr 12 2011
# second Maple program:
a := n -> add(binomial(2*n,k)*(-1)^k*BellB(k,1/2)*BellB(2*n-k,1/2), k=0..2*n):
seq(a(n), n=0..18); # after Emanuele Munarini,_Peter Luschny_, Sep 10 2017
B := BellMatrix(n -> modp(n, 2), 31): # defined in A264428.
seq(add(k, k in B[2*n + 1]),n=0..15); # Peter Luschny, Aug 13 2019

NestList[ Factor[ D[#, {x, 2}]] &, Exp[ Cosh[x] - 1], 16] /. x -> 0
a[0] = 1; a[n_] := Sum[Sum[(i-k)^(2*n)*Binomial[2*k, i]*(-1)^i, {i, 0, k-1}]/(2^(k-1)*k!), {k, 1, 2*n}]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Apr 07 2015, after Vladimir Kruchinin *)
Table[Sum[BellY[2 n, k, 1 - Mod[Range[2 n], 2]], {k, 0, 2 n}], {n, 0, 20}] (* Vladimir Reshetnikov, Nov 09 2016 *)
With[{nn=40},Abs[Take[CoefficientList[Series[Exp[Cos[x]-1],{x,0,nn}],x] Range[0,nn]!,{1,-1,2}]]] (* Harvey P. Dale, Feb 06 2017 *)

a(n):= sum(1/k!*sum(binomial(k,m)/(2^(m-1))*sum(binomial(m,j) *(2*j-m)^(2*n), j,0,m/2)*(-1)^(k-m), m,0,k), k,1,2*n); /* Vladimir Kruchinin, Aug 05 2010 */

a(n):=sum(sum((i-k)^(2*n)*binomial(2*k,i)*(-1)^(i),i,0,k-1)/(2^(k-1)*k!),k,1,2*n); /* Vladimir Kruchinin, Oct 04 2012 */

from sympy.core.cache import cacheit
from sympy import binomial
@cacheit
def a(n): return 1 if n==0 else sum(binomial(2*n - 1, 2*k - 1)*a(n - k) for k in range(1, n + 1))
print([a(n) for n in range(21)]) # Indranil Ghosh, Sep 11 2017, after Maple program by Alois P. Heinz

E.g.f.: exp(cosh(x) - 1) (or exp(cos(x)-1) ).
a(n) = Sum_{k=1..n} binomial(2*n-1, 2*k-1)*a(n-k). - Vladeta Jovovic, Apr 10 2003
a(n) = sum(1/k!*sum(binomial(k,m)/(2^(m-1))*sum(binomial(m,j)*(2*j-m)^(2*n),j,0,m/2)*(-1)^(k-m),m,0,k),k,1,2*n), n>0. - Vladimir Kruchinin, Aug 05 2010
a(n) = Sum_{k=1..2*n} Sum_{i=0..k-1} ((i-k)^(2*n)*binomial(2*k,i)*(-1)^i)/(2^(k-1)*k!), n>0, a(0)=1. - Vladimir Kruchinin, Oct 04 2012
E.g.f.: E(0)-1, where E(k) = 2 + (cosh(x)-1)/(2*k + 1 - (cosh(x)-1)/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 23 2013
a(n) = Sum_{k=0..2*n} binomial(2*n,k)*(-1)^k*S_k(1/2)*S_{2*n-k}( 1/2), where S_n(x) is the n-th Bell polynomial (or exponential polynomial). - Emanuele Munarini, Sep 10 2017

A241171 Triangle read by rows: Joffe's central differences of zero, T(n,k), 1 <= k <= n.

Original entry on oeis.org

1, 1, 6, 1, 30, 90, 1, 126, 1260, 2520, 1, 510, 13230, 75600, 113400, 1, 2046, 126720, 1580040, 6237000, 7484400, 1, 8190, 1171170, 28828800, 227026800, 681080400, 681080400, 1, 32766, 10663380, 494053560, 6972966000, 39502663200, 95351256000, 81729648000, 1, 131070, 96461910, 8203431600, 196556560200, 1882311631200, 8266953895200, 16672848192000, 12504636144000

Offset: 1

N. J. A. Sloane, Apr 22 2014

T(n,k) gives the number of ordered set partitions of the set {1,2,...,2*n} into k even sized blocks. An example is given below. Cf. A019538 and A156289. - Peter Bala, Aug 20 2014

			Triangle begins:
1,
1, 6,
1, 30, 90,
1, 126, 1260, 2520,
1, 510, 13230, 75600, 113400,
1, 2046, 126720, 1580040, 6237000, 7484400,
1, 8190, 1171170, 28828800, 227026800, 681080400, 681080400,
1, 32766, 10663380, 494053560, 6972966000, 39502663200, 95351256000, 81729648000,
...
From _Peter Bala_, Aug 20 2014: (Start)
Row 2: [1,6]
k  Ordered set partitions of {1,2,3,4} into k blocks    Number
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
1   {1,2,3,4}                                             1
2   {1,2}{3,4}, {3,4}{1,2}, {1,3}{2,4}, {2,4}{1,3},       6
    {1,4}{2,3}, {2,3}{1,4}
(End)

H. T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 283.
S. A. Joffe, Calculation of the first thirty-two Eulerian numbers from central differences of zero, Quart. J. Pure Appl. Math. 47 (1914), 103-126.
S. A. Joffe, Calculation of eighteen more, fifty in all, Eulerian numbers from central differences of zero, Quart. J. Pure Appl. Math. 48 (1917-1920), 193-271.

Muniru A Asiru, Rows n = 1..50 of triangle, flattened

Case m=2 of the polynomials defined in A278073.
Cf. A000680 (diagonal), A094088 (row sums), A000364 (alternating row sums), A281478 (central terms), A327022 (refinement).
Diagonals give A002446, A213455, A241172, A002456.
Cf. A019538, A036969, A156289.

Flat(List([1..10],n->List([1..n],k->1/(2^(k-1))*Sum([1..k],j->(-1)^(k-j)*Binomial(2*k,k-j)*j^(2*n))))); # Muniru A Asiru, Feb 27 2019

T := proc(n,k) option remember;
if k > n then 0
elif k=0 then k^n
elif k=1 then 1
else k*(2*k-1)*T(n-1,k-1)+k^2*T(n-1,k); fi;
end: # Minor edit to make it also work in the (0,0)-offset case. Peter Luschny, Sep 03 2022
for n from 1 to 12 do lprint([seq(T(n,k), k=1..n)]); od:

T[n_, k_] /; 1 <= k <= n := T[n, k] = k(2k-1) T[n-1, k-1] + k^2 T[n-1, k]; T[, 1] = 1; T[, ] = 0; Table[T[n, k], {n, 1, 9}, {k, 1, n}] (* _Jean-François Alcover, Jul 03 2019 *)

@cached_function
def A241171(n, k):
    if n == 0 and k == 0: return 1
    if k < 0 or k > n: return 0
    return (2*k^2 - k)*A241171(n - 1, k - 1) + k^2*A241171(n - 1, k)
for n in (1..6): print([A241171(n, k) for k in (1..n)]) # Peter Luschny, Sep 06 2017

T(n,k) = 0 if k <= 0 or k > n, = 1 if k=1, otherwise T(n,k) = k*(2*k-1)*T(n-1,k-1) + k^2*T(n-1,k).
Related to Euler numbers A000364 by A000364(n) = (-1)^n*Sum_{k=1..n} (-1)^k*T(n,k). For example, A000364(3) = 61 = 90 - 30 + 1.
From Peter Bala, Aug 20 2014: (Start)
T(n,k) = 1/(2^(k-1))*Sum_{j = 1..k} (-1)^(k-j)*binomial(2*k,k-j)*j^(2*n).
T(n,k) = k!*A156289(n,k) = k!*(2*k-1)!!*A036969.
E.g.f.: A(t,z) := 1/( 1 - t*(cosh(z) - 1) ) = 1 + t*z^2/2! + (t + 6*t^2)*z^4/4! + (t + 30*t^2 + 90*t^3)*z^6/6! + ... satisfies the partial differential equation d^2/dz^2(A) = D(A), where D = t^2*(2*t + 1)*d^2/dt^2 + t*(5*t + 1)*d/dt + t.
Hence the row polynomials R(n,t) satisfy the differential equation R(n+1,t) = t^2*(2*t + 1)*R''(n,t) + t*(5*t + 1)*R'(n,t) + t*R(n,t) with R(0,t) = 1, where ' indicates differentiation w.r.t. t. This is equivalent to the above recurrence equation.
Recurrence for row polynomials: R(n,t) = t*( Sum_{k = 1..n} binomial(2*n,2*k)*R(n-k,t) ) with R(0,t) := 1.
Row sums equal A094088(n) for n >= 1.
A100872(n) = (1/2)*R(n,2). (End)

A036969 Triangle read by rows: T(n,k) = T(n-1,k-1) + k^2*T(n-1,k), 1 < k <= n, T(n,1) = 1.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 21, 14, 1, 1, 85, 147, 30, 1, 1, 341, 1408, 627, 55, 1, 1, 1365, 13013, 11440, 2002, 91, 1, 1, 5461, 118482, 196053, 61490, 5278, 140, 1, 1, 21845, 1071799, 3255330, 1733303, 251498, 12138, 204, 1, 1, 87381, 9668036, 53157079, 46587905

Offset: 1

N. J. A. Sloane

Or, triangle of central factorial numbers T(2n,2k) (in Riordan's notation).
Can be used to calculate the Bernoulli numbers via the formula B_2n = (1/2)*Sum_{k = 1..n} (-1)^(k+1)*(k-1)!*k!*T(n,k)/(2*k+1). E.g., n = 1: B_2 = (1/2)*1/3 = 1/6. n = 2: B_4 = (1/2)*(1/3 - 2/5) = -1/30. n = 3: B_6 = (1/2)*(1/3 - 2*5/5 + 2*6/7) = 1/42. - Philippe Deléham, Nov 13 2003
From Peter Bala, Sep 27 2012: (Start)
Generalized Stirling numbers of the second kind. T(n,k) is equal to the number of partitions of the set {1,1',2,2',...,n,n'} into k disjoint nonempty subsets V1,...,Vk such that, for each 1 <= j <= k, if i is the least integer such that either i or i' belongs to Vj then {i,i'} is a subset of Vj. An example is given below.
Thus T(n,k) may be thought of as a two-colored Stirling number of the second kind. See Matsumoto and Novak, who also give another combinatorial interpretation of these numbers. (End)

			Triangle begins:
  1;
  1,    1;
  1,    5,      1;
  1,   21,     14,      1;
  1,   85,    147,     30,     1;
  1,  341,   1408,    627,    55,    1;
  1, 1365,  13013,  11440,  2002,   91,   1;
  1, 5461, 118482, 196053, 61490, 5278, 140, 1;
  ...
T(3,2) = 5: The five set partitions into two sets are {1,1',2,2'}{3,3'}, {1,1',3,3'}{2,2'}, {1,1'}{2,2',3,3'}, {1,1',3}{2,2',3'} and {1,1',3'}{2,2',3}.

L. Carlitz, A conjecture concerning Genocchi numbers. Norske Vid. Selsk. Skr. (Trondheim) 1971, no. 9, 4 pp. [The triangle appears on page 2.]
J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.8.

Vincenzo Librandi, Rows n = 1..100 of triangle, flattened
Thomas Browning, Counting Parabolic Double Cosets in Symmetric Groups, arXiv:2010.13256 [math.CO], 2020.
P. L. Butzer, M. Schmidt, E. L. Stark and L. Vogt. Central factorial numbers; their main properties and some applications, Num. Funct. Anal. Optim., 10 (1989) 419-488.
José L. Cereceda, Sums of powers of integers and the sequence A304330, arXiv:2405.05268 [math.GM], 2024. See p. 2.
M. W. Coffey and M. C. Lettington, On Fibonacci Polynomial Expressions for Sums of mth Powers, their implications for Faulhaber's Formula and some Theorems of Fermat, arXiv:1510.05402 [math.NT], 2015.
D. Dumont, Interprétations combinatoires des nombres de Genocchi, Duke Math. J., 41 (1974), 305-318.
D. Dumont, Interprétations combinatoires des nombres de Genocchi, Duke Math. J., 41 (1974), 305-318. (Annotated scanned copy)
Qi Fang, Ya-Nan Feng, and Shi-Mei Ma, Alternating runs of permutations and the central factorial numbers, arXiv:2202.13978 [math.CO], 2022.
F. G. Garvan, Higher-order spt functions, Adv. Math. 228 (2011), no. 1, 241-265. - From _N. J. A. Sloane_, Jan 02 2013
Petro Kolosov, Polynomial identities involving central factorial numbers, GitHub, 2024. See p. 6.
P. A. MacMahon, Divisors of numbers and their continuations in the theory of partitions, Proc. London Math. Soc., (2) 19 (1919), 75-113; Coll. Papers II, pp. 303-341.
S. Matsumoto and J. Novak, Jucys-Murphy Elements and Unitary Matrix Integrals arXiv.0905.1992 [math.CO], 2009-2012.
B. K. Miceli, Two q-Analogues of Poly-Stirling Numbers, J. Integer Seq., 14 (2011), 11.9.6.
John Riordan, Letter, Apr 28 1976.
John Riordan, Letter, Jul 06 1978
Richard P. Stanley, Hook Lengths and Contents.

Columns are A002450, A002451.
Diagonals are A000330 and A060493.
Transpose of A008957.
(0,0)-based version: A269945.
Cf. A008955, A008956, A156289, A135920 (row sums), A204579 (inverse), A000290.

a036969 n k = a036969_tabl !! (n-1) (k-1)
a036969_row n = a036969_tabl !! (n-1)
a036969_tabl = iterate f [1] where
   f row = zipWith (+)
     ([0] ++ row) (zipWith (*) (tail a000290_list) (row ++ [0]))
-- Reinhard Zumkeller, Feb 18 2013

A036969 := proc(n,k) local j; 2*add(j^(2*n)*(-1)^(k-j)/((k-j)!*(k+j)!),j=1..k); end;

t[n_, k_] := 2*Sum[j^(2*n)*(-1)^(k-j)/((k-j)!*(k+j)!), {j, 1, k}]; Flatten[ Table[t[n, k], {n, 1, 10}, {k, 1, n}]] (* Jean-François Alcover, Oct 11 2011 *)
t1[n_, k_] := (1/(2 k)!) * Sum[Binomial[2 k, j]*(-1)^j*(k - j)^(2 n), {j, 0, 2 k}]; Column[Table[t1[n, k], {n, 1, 10}, {k, 1, n}]] (* Kolosov Petro ,Jul 26 2023 *)

PARI
```
T(n,k)=if(1M. F. Hasler, Feb 03 2012
```

T(n,k)=2*sum(j=1,k,(-1)^(k-j)*j^(2*n)/(k-j)!/(k+j)!)  \\ M. F. Hasler, Feb 03 2012

def A036969(n,k) : return (2/factorial(2*k))*add((-1)^j*binomial(2*k,j)*(k-j)^(2*n) for j in (0..k))
for n in (1..7) : print([A036969(n,k) for k in (1..n)]) # Peter Luschny, Feb 03 2012

T(n,k) = A156289(n,k)/A001147(k). - Peter Bala, Feb 21 2011
From Peter Bala, Oct 14 2011: (Start)
O.g.f.: Sum_{n >= 1} x^n*t^n/Product_{k = 1..n} (1 - k^2*t^2) = x*t + (x + x^2)*t^2 + (x + 5*x^2 + x^3)*t^3 + ....
Define polynomials x^[2*n] = Product_{k = 0..n-1} (x^2 - k^2). This triangle gives the coefficients in the expansion of the monomials x^(2*n) as a linear combination of x^[2*m], 1 <= m <= n. For example, row 4 gives x^8 = x^[2] + 21*x^[4] + 14*x^[6] + x^[8].
A008955 is a signed version of the inverse.
The n-th row sum = A135920(n). (End)
T(n,k) = (2/(2*k)!)*Sum_{j=0..k-1} (-1)^(j+k+1) * binomial(2*k,j+k+1) * (j+1)^(2*n). This formula is valid for n >= 0 and 0 <= k <= n. - Peter Luschny, Feb 03 2012
From Peter Bala, Sep 27 2012: (Start)
Let E(x) = cosh(sqrt(2*x)) = Sum_{n >= 0} x^n/((2*n)!/2^n). A generating function for the triangle is E(t*(E(x)-1)) = 1 + t*x + t*(1 + t)*x^2/6 + t*(1 + 5*t + t^2)*x^3/90 + ..., where the sequence of denominators [1, 1, 6, 90, ...] is given by (2*n)!/2^n. Cf. A008277 which has generating function exp(t*(exp(x)-1)). An e.g.f. is E(t*(E(x^2/2)-1)) = 1 + t*x^2/2! + t*(1 + t)*x^4/4! + t*(1 + 5*t + t^2)*x^6/6! + ....
Put c(n) := (2*n)!/2^n. The column k generating function is (1/c(k))*(E(x)-1)^k = Sum_{n >= k} T(n,k)*x^n/c(n). The inverse array is A204579.
The production array begins:
  1,  1;
  0,  4,  1;
  0,  0,  9,  1;
  0,  0,  0, 16,  1;
  ... (End)
x^n = Sum_{k=1..n} T(n,k)*Product_{i=0..k-1} (x-i^2), see Stanley link. - Michel Marcus, Nov 19 2014; corrected by Kolosov Petro, Jul 26 2023
From Kolosov Petro, Jul 26 2023: (Start)
T(n,k) = (1/(2*k)!) * Sum_{j=0..2k} binomial(2k, j)*(-1)^j*(k - j)^(2n).
T(n,k) = (1/(k*(2k-1)!)) * Sum_{j=0..k} (-1)^(k-j)*binomial(2k, k-j)*j^(2n). (End)

More terms from Vladeta Jovovic, Apr 16 2000

A007106 Number of labeled odd degree trees with 2n nodes.

Original entry on oeis.org

1, 4, 96, 5888, 686080, 130179072, 36590059520, 14290429935616, 7405376630685696, 4917457306800619520, 4071967909087792857088, 4113850542422629363482624, 4980673081258443273955966976, 7119048451600750435732824260608, 11861520124846917915630931846103040

Offset: 1

N. J. A. Sloane

nonn

			From _Peter Bala_, Apr 24 2012: (Start)
Let G(x) = 1 + x^2/2! + 13*x^4/4! + 541*x^6/6! + ... be the e.g.f. for A143601. Then sinh(x*G(x)) = x + 4*x^3/3! + 96*x^5/5! + 5888*x^7/7! + ....
Conjectural e.g.f. as an x-adic limit:
sinh(x) = x + ...; sinh(x*cosh(x)) = x + 4*x^3/3! + ...;
sinh(x*cosh(x*cosh(x))) = x + 4*x^3/3! + 96*x^5/5! + ...;
sinh(x*cosh(x*cosh(x*cosh(x)))) = x + 4*x^3/3! + 96*x^5/5! + 5888*x^7/7! + ....
(End)

R. W. Robinson, personal communication.
R. W. Robinson, Numerical implementation of graph counting algorithms, AGRC Grant, Math. Dept., Univ. Newcastle, Australia, 1976.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Seiichi Manyama, Table of n, a(n) for n = 1..211 (terms 1..39 from R. W. Robinson)
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group, arXiv:2112.11595 [math.CO], 2021.
B. R. Jones, On tree hook length formulas, Feynman rules and B-series, Master's thesis, Simon Fraser University, 2014.
Mathematics Stack Exchange, Marko R. Riedel, Odd degree trees
Mathematics Stack Exchange, Marko R. Riedel, Odd degree trees II
Marko Riedel, Count by Prüfer codes and Stirling numbers
Index entries for sequences related to trees

Cf. A036778, A058014, A060279, A143601, A156289.

A007106(n) = A(2n) where n>=2, A(n) = (add(binomial(n,q)*(n-2*q)^(n-2)/(n-2)!, q=0..n) - add(binomial(n-1,q)*(n-2*q)^(n-3)/(n-3)!, q=0..n-1) + add(binomial(n-1,q)*(n-2-2*q)^(n-3)/(n-3)!, q=0..n-1))*n!/2^(n+1)/(n-1)

{1}~Join~Array[(1/2)*Sum[Binomial[2 #, k]*(# - k)^(2 # - 2), {k, 0, # - 1}] &, 12, 2] (* Michael De Vlieger, Oct 13 2021 *)

a(n) = if(n<=1, n==1, sum(k=0, n-1, binomial(2*n,k) * (n-k)^(2*n-2))/2) \\ Andrew Howroyd, Nov 22 2021

a(n) = A060279(n)/(2*n). - Vladeta Jovovic, Feb 08 2005
Bisection of A058014. The expansion 1/sqrt(1+x^2)*arcsinh(x) = x - 4*x^3/3! + 64*x^5/5! - ... (see A002454) has series reversion x + 4*x^3/3! + 96*x^5/5! + 5888*x^7/7! + .... The coefficients appear to be the terms of this sequence. As an x-adic limit this e.g.f. equals lim_{n -> infinity} sinh(f(n,x)), where f(0,x) = x and f(n,x) = x*cosh(f(n-1,x)) for n >= 1. See the example section below. - Peter Bala, Apr 24 2012
a(n) = Sum_{k=1..n} binomial(n,k) * k! * (n-2)! [z^{n-2}] [u^k] exp(u(exp(z)+exp(-z)-2)/2)). - Marko Riedel, Jun 16 2016
From Alexander Burstein, Oct 13 2021: (Start)
a(n) = (1/2) * Sum_{k=0..n-1} binomial(2*n,k) * (n-k)^(2*n-2) for n >= 2.
a(n) = (2*n-1)!*[x^(2*n-1)] sinh(REVERT(x/cosh(x))), see A036778. (End)
a(n) = Sum_{k=0..n-1} A156289(n-1, k)*(2*n)!/(2*n-k)!. - Peter Luschny, May 07 2022

Corrected and extended by Vladeta Jovovic, Feb 08 2005

A260876 Number of m-shape set partitions, square array read by ascending antidiagonals, A(m,n) for m, n >= 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 4, 5, 5, 1, 1, 11, 31, 15, 7, 1, 1, 36, 365, 379, 52, 11, 1, 1, 127, 6271, 25323, 6556, 203, 15, 1, 1, 463, 129130, 3086331, 3068521, 150349, 877, 22, 1, 1, 1717, 2877421, 512251515, 3309362716, 583027547, 4373461, 4140, 30

Offset: 0

Peter Luschny, Aug 02 2015

A set partition of m-shape is a partition of a set with cardinality m*n for some n >= 0 such that the sizes of the blocks are m times the parts of the integer partitions of n.
If m = 0, all possible sizes are zero. Thus the number of set partitions of 0-shape is the number of integer partitions of n (partition numbers A000041).
If m = 1, the set is {1, 2, ..., n} and the set of all possible sizes are the integer partitions of n. Thus the number of set partitions of 1-shape is the number of set partitions (Bell numbers A000110).
If m = 2, the set is {1, 2, ..., 2n} and the number of set partitions of 2-shape is the number of set partitions into even blocks A005046.
From Petros Hadjicostas, Aug 06 2019: (Start)
Irwin (1916) proved the following combinatorial result: Assume r_1, r_2, ..., r_n are positive integers and we have r_1*r_2*...*r_n objects. We divide them into r_1 classes of r_2*r_3*...*r_n objects each, then each class into r_2 subclasses of r_3*...*r_n objects each, and so on. We call each such classification, without reference to order, a "classification" par excellence. He proved that the total number of classifications is (r_1*r_2*...*r_n)!/( r1! * (r_2!)^(r_1) * (r_3!)^(r_1*r_2) * ... (r_n!)^(r_1*r_2*...*r_{n-1}) ).
Apparently, this problem appeared in Carmichael's "Theory of Numbers".
This result can definitely be used to prove some special cases of my conjecture below. (End)

			[ n ] [0  1   2       3        4           5              6]
[ m ] ------------------------------------------------------
[ 0 ] [1, 1,  2,      3,       5,          7,            11]  A000041
[ 1 ] [1, 1,  2,      5,      15,         52,           203]  A000110
[ 2 ] [1, 1,  4,     31,     379,       6556,        150349]  A005046
[ 3 ] [1, 1, 11,    365,   25323,    3068521,     583027547]  A291973
[ 4 ] [1, 1, 36,   6271, 3086331, 3309362716, 6626013560301]  A291975
        A260878,A309725, ...
For example the number of set partitions of {1,2,...,9} with sizes in [9], [6,3] and [3,3,3] is 1, 84 and 280 respectively. Thus A(3,3) = 365.
Formatted as a triangle:
[1]
[1, 1]
[1, 1,   2]
[1, 1,   2,    3]
[1, 1,   4,    5,     5]
[1, 1,  11,   31,    15,    7]
[1, 1,  36,  365,   379,   52,  11]
[1, 1, 127, 6271, 25323, 6556, 203, 15]
.
From _Peter Luschny_, Aug 14 2019: (Start)
For example consider the case n = 4. There are five integer partitions of 4:
  P = [[4], [3, 1], [2, 2], [2, 1, 1], [1, 1, 1, 1]]. The shapes are m times the parts of the integer partitions: S(m) = [[4m], [3m, m], [2m, 2m], [2m, m, m], [m, m, m, m]].
* In the case m = 1 we look at set partitions of {1, 2, 3, 4} with sizes in  [[4], [3, 1], [2, 2], [2, 1, 1], [1, 1, 1, 1]] which gives rise to [1, 4, 3, 6, 1] with sum 15.
* In the case m = 2 we look at set partitions of {1, 2, .., 8} with sizes in [[8], [6, 2], [4, 4], [4, 2, 2], [2, 2, 2, 2]] which gives rise to [1, 28, 35, 210, 105] with sum 379.
* In the case m = 0 we look at set partitions of {} with sizes in [[0], [0, 0], [0, 0], [0, 0, 0], [0, 0, 0, 0]] which gives rise to [1, 1, 1, 1, 1] with sum 5 (because the only partition of the empty set is the set that contains the empty set, thus from the definition T(0,4) = Sum_{S(0)} card({0}) = A000041(4) = 5).
If n runs through 0, 1, 2,... then the result is an irregular triangle in which the n-th row lists multinomials for partitions of [m*n] which have only parts which are multiples of m. These are the triangles A080575 (m = 1), A257490 (m = 2), A327003 (m = 3), A327004 (m = 4). In the case m = 0 the triangle is A000012 subdivided into rows of length A000041. See the cross references how this case integrates into the full picture.
(End)

Alois P. Heinz, Antidiagonals n = 0..54, flattened
Frank Irwin, Solution to Problem 223 proposed by T. E. Mason in October 1914, Amer. Math. Monthly 23(9) (1916), 352-353.

-----------------------------------------------------------------
[m] |  multi-   |  sum of   |  main     |  by       |  comple-  |
    |  nomials  |  rows     |  diagonal |  size     |  mentary  |
-----------------------------------------------------------------
[0] |  A000012  |  A000041  |  A000012  |  A008284  |  A081362  |
[1] |  A036040  |  A000110  |  A000012  |  A048993  |  A000587  |
[2] |  A257490  |  A005046  |  A001147  |  A156289  |  A260884  |
[3] |  A327003  |  A291973  |  A025035  |  A291451  |  A291974  |
[4] |  A327004  |  A291975  |  A025036  |  A291452  |  A291976  |
Cf. A326996 (main diagonal), A260883 (ordered), A260875 (complementary).
Columns include A000012, A260878, A309725.

A:= proc(m, n) option remember; `if`(m=0, combinat[numbpart](n),
      `if`(n=0, 1, add(binomial(m*n-1, m*k-1)*A(m, n-k), k=1..n)))
    end:
seq(seq(A(d-n, n), n=0..d), d=0..10);  # Alois P. Heinz, Aug 14 2019

A[m_, n_] := A[m, n] = If[m==0, PartitionsP[n], If[n==0, 1, Sum[Binomial[m n - 1, m k - 1] A[m, n - k], {k, 1, n}]]];
Table[Table[A[d - n, n], {n, 0, d}], {d, 0, 10}] // Flatten (* Jean-François Alcover, Dec 06 2019, after Alois P. Heinz *)

def A260876(m, n):
    shapes = ([x*m for x in p] for p in Partitions(n))
    return sum(SetPartitions(sum(s), s).cardinality() for s in shapes)
for m in (0..4): print([A260876(m,n) for n in (0..6)])

From Petros Hadjicostas, Aug 02 2019: (Start)
A(m, 2) = 1 + (1/2) * binomial(2*m, m) for m >= 1.
A(m, 3) = 1 + binomial(3*m, m) + (3*m)!/(6 * (m!)^3) for m >= 1.
A(m, 4) = (1/4!) * multinomial(4*m, [m, m, m, m]) + (1/2) * multinomial(4*m, [2*m, m, m]) + multinomial(4*m, [m, 3*m]) + (1/2) * multinomial(4*m, [2*m, 2*m]) + 1 for m >= 1.
Conjecture: For n >= 0, let P be the set of all possible lists (a_1,...,a_n) of nonnegative integers such that a_1*1 + a_2*2 + ... + a_n*n = n. Consider terms of the form multinomial(n*m, m*[1,..., 1,2,..., 2,..., n,..., n])/(a_1! * a_2! * ... * a_n!), where in the list [1,...,1,2,...,2,...,n,...,n] the number 1 occurs a_1 times, 2 occurs a_2 times, ..., and n occurs a_n times. (Here a_n = 0 or 1.) Summing these terms over P we get A(m, n) provided m >= 1. (End)
Conjecture for a recurrence: A(m, n) = Sum_{k = 0..n-1} binomial(m*n - 1, m*k) * A(m, k) with A(m, 0) = 1 for m >= 1 and n >= 0. (Unfortunately, the recurrence does not hold for m = 0.) - Petros Hadjicostas, Aug 12 2019

A291451 Triangle read by rows, expansion of e.g.f. exp(x(exp(z)/3 + 2exp(-z/2)* cos(z*sqrt(3)/2)/3 - 1)), nonzero coefficients of z.

Original entry on oeis.org

1, 0, 1, 0, 1, 10, 0, 1, 84, 280, 0, 1, 682, 9240, 15400, 0, 1, 5460, 260260, 1401400, 1401400, 0, 1, 43690, 7128576, 99379280, 285885600, 190590400, 0, 1, 349524, 193360720, 6600492080, 42549306800, 76045569600, 36212176000

Offset: 0

Peter Luschny, Sep 07 2017

			Triangle starts:
[1]
[0, 1]
[0, 1,    10]
[0, 1,    84,     280]
[0, 1,   682,    9240,    15400]
[0, 1,  5460,  260260,  1401400,   1401400]
[0, 1, 43690, 7128576, 99379280, 285885600, 190590400]

Cf. A048993 (m=1), A156289 (m=2), this seq. (m=3), A291452 (m=4).
Diagonal: A000012 (m=1), A001147 (m=2), A025035 (m=3), A025036 (m=4).
Row sums: A000110 (m=1), A005046 (m=2), A291973 (m=3), A291975 (m=4).
Alternating row sums: A000587 (m=1), A260884 (m=2), A291974 (m=3), A291976 (m=4).

CL := (f, x) -> PolynomialTools:-CoefficientList(f, x):
A291451_row := proc(n) exp(x*(exp(z)/3+2*exp(-z/2)*cos(z*sqrt(3)/2)/3-1)):
series(%, z, 66): CL((3*n)!*coeff(series(%,z,3*(n+1)),z,3*n),x) end:
for n from 0 to 7 do A291451_row(n) od;
# Alternative:
A291451row := proc(n) local P; P := proc(m, n) option remember;
if n = 0 then 1 else add(binomial(m*n, m*k)*P(m, n-k)*x, k=1..n) fi end:
CL(P(3, n), x); seq(%[k+1]/k!, k=0..n) end: # Peter Luschny, Sep 03 2018

P[m_, n_] := P[m, n] = If[n == 0, 1, Sum[Binomial[m*n, m*k]*P[m, n - k]*x, {k, 1, n}]];
row[n_] := Module[{cl = CoefficientList[P[3, n], x]}, Table[cl[[k + 1]]/k!, {k, 0, n}]];
Table[row[n], {n, 0, 7}] // Flatten (* Jean-François Alcover, Jul 23 2019, after Peter Luschny *)

A291452 Triangle read by rows, expansion of e.g.f. exp(x*(cos(z) + cosh(z) - 2)/2), nonzero coefficients of z.

Original entry on oeis.org

1, 0, 1, 0, 1, 35, 0, 1, 495, 5775, 0, 1, 8255, 450450, 2627625, 0, 1, 130815, 35586525, 727476750, 2546168625, 0, 1, 2098175, 2941884000, 181262956875, 1932541986375, 4509264634875

Offset: 0

Peter Luschny, Sep 07 2017

			Triangle starts:
[1]
[0, 1]
[0, 1,      35]
[0, 1,     495,       5775]
[0, 1,    8255,     450450,      2627625]
[0, 1,  130815,   35586525,    727476750,    2546168625]
[0, 1, 2098175, 2941884000, 181262956875, 1932541986375, 4509264634875]

Cf. A048993 (m=1), A156289 (m=2), A291451 (m=3), this seq. (m=4).
Diagonal: A000012 (m=1), A001147 (m=2), A025035 (m=3), A025036 (m=4).
Row sums: A000110 (m=1), A005046 (m=2), A291973 (m=3), A291975 (m=4).
Alternating row sums: A000587 (m=1), A260884 (m=2), A291974 (m=3), A291976 (m=4).

CL := (f,x) -> PolynomialTools:-CoefficientList(f,x):
A291452_row := proc(n) exp(x*(cos(z)+cosh(z)-2)/2):
series(%, z, 88): CL((4*n)!*coeff(series(%,z,4*(n+1)),z,4*n),x) end:
for n from 0 to 7 do A291452_row(n) od;
# Alternative:
A291452row := proc(n) local P; P := proc(m, n) option remember;
if n = 0 then 1 else add(binomial(m*n, m*k)*P(m, n-k)*x, k=1..n) fi end:
CL(P(4, n), x); seq(%[k+1]/k!, k=0..n) end: # Peter Luschny, Sep 03 2018

P[m_, n_] := P[m, n] = If[n == 0, 1, Sum[Binomial[m*n, m*k]*P[m, n - k]*x, {k, 1, n}]];
row[n_] := Module[{cl = CoefficientList[P[4, n], x]}, Table[cl[[k + 1]]/k!, {k, 0, n}]];
Table[row[n], {n, 0, 6}] // Flatten (* Jean-François Alcover, Jul 23 2019, after Peter Luschny *)

cp's OEIS Frontend

A327416 a(n) = A156289(2*n, n).

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A136630 Triangular array: T(n,k) counts the partitions of the set [n] into k odd sized blocks.

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A003724 Number of partitions of n-set into odd blocks.

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A005046 Number of partitions of a 2n-set into even blocks.

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A241171 Triangle read by rows: Joffe's central differences of zero, T(n,k), 1 <= k <= n.

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A036969 Triangle read by rows: T(n,k) = T(n-1,k-1) + k^2*T(n-1,k), 1 < k <= n, T(n,1) = 1.

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PARI

PARI

A291451 Triangle read by rows, expansion of e.g.f. exp(x(exp(z)/3 + 2exp(-z/2)* cos(z*sqrt(3)/2)/3 - 1)), nonzero coefficients of z.