A160562 Triangle of scaled central factorial numbers, T(n,k) = A008958(n,n-k).

1, 1, 1, 1, 10, 1, 1, 91, 35, 1, 1, 820, 966, 84, 1, 1, 7381, 24970, 5082, 165, 1, 1, 66430, 631631, 273988, 18447, 286, 1, 1, 597871, 15857205, 14057043, 1768195, 53053, 455, 1, 1, 5380840, 397027996, 704652312, 157280838, 8187608, 129948, 680, 1

Offset: 0

Author

Jonathan Vos Post, May 19 2009

Keywords

• nonn
• tabl

Comments

This is table 4 on page 12 of Gelineau and Zeng, read downwards by columns.

Reversing rows gives A008958.

Apparently the table can also be obtained by deleting each second row and column of A136630.

Examples

			Triangle starts:
  1;
  1,     1;
  1,    10,      1;
  1,    91,     35,      1;
  1,   820,    966,     84,     1;
  1,  7381,  24970,   5082,   165,   1;
  1, 66430, 631631, 273988, 18447, 286, 1;
  ...
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows n = 0..150, flattened)
• Qi Fang, Ya-Nan Feng, and Shi-Mei Ma, Alternating runs of permutations and the central factorial numbers, arXiv:2202.13978 [math.CO], 2022.
• Yoann Gelineau and Jiang Zeng, Combinatorial Interpretations of the Jacobi-Stirling Numbers, arXiv:0905.2899 [math.CO], May 18 2009.

Crossrefs

Cf. A002452 (column k=1), A002453 (column k=2), A000447 (right column k=n-1), A185375 (right column k=n-2).

Cf. A001819, A008275, A008277, A008958, A039599, A136630.

Programs

Maple

A160562 := proc(n,k) npr := 2*n+1 ; kpr := 2*k+1 ; sinh(t*sinh(x)) ; npr!*coeftayl(%,x=0,npr) ; coeftayl(%,t=0,kpr) ; end: seq(seq(A160562(n,k),k=0..n),n=0..15) ; # R. J. Mathar, Sep 09 2009

Mathematica

T[n_, k_] := Sum[(-1)^(k - m)*(2m + 1)^(2n + 1)*Binomial[2k, k + m]/(k + m + 1), {m, 0, k}]/(4^k*(2k)!);
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 22 2017 *)

Formula

T(n,k) = (1/((2*k)!*4^k)) * Sum_{m=0..k} (-1)^(k-m)*A039599(k,m)*(2*m+1)^(2*n). - Werner Schulte, Nov 01 2015

T(n,k) = ((-1)^(n-k)*(2*n+1)!/(2*k+1)!) * [x^(2*n+1)]sin(x)^(2*k+1) = ((2*n+1)!/(2*k+1)!) * [x^(2*n+1)]sinh(x)^(2*k+1). Note that sin(x)^(2*k+1) = (Sum_{i=0..k} (-1)^i*binomial(2*k+1,k-i)*sin((2*i+1)*x))/(2^(2*k)). - Jianing Song, Oct 29 2023

Extensions

More terms from R. J. Mathar, Sep 09 2009