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A160563 Table of the number of (n,k)-Riordan complexes, read by rows. 1
1, 1, 1, 9, 10, 1, 225, 259, 35, 1, 11025, 12916, 1974, 84, 1, 893025, 1057221, 172810, 8778, 165, 1, 108056025, 128816766, 21967231, 1234948, 28743, 286, 1, 18261468225, 21878089479, 3841278805, 230673443, 6092515, 77077, 455, 1, 4108830350625, 4940831601000 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

0,4

COMMENTS

From Table 4, right-hand side, of Gelineau and Zeng. The Jacobi-Stirling numbers of the first and second kinds were introduced in 2006 in the spectral theory and are polynomial refinements of the Legendre-Stirling numbers.

Andrews and Littlejohn have recently given a combinatorial interpretation for the second kind of the latter numbers. Noticing that these numbers are very similar to the classical central factorial numbers, we give combinatorial interpretations for the Jacobi-Stirling numbers of both kinds, which provide a unified treatment of the combinatorial theories for the two previous sequences and also for the Stirling numbers of both kinds.

Essentially a row-reversal of A008956. - R. J. Mathar, May 20 2009

LINKS

Table of n, a(n) for n=0..37.

Yoann Gelineau and Jiang Zeng, Combinatorial Interpretations of the Jacobi-Stirling Numbers, arXiv:0905.2899 [math.CO], May 2009.

W. Zhang, Some identities involving the Euler and the central factorial numbers, The Fibonacci Quarterly, Vol. 36, Number 2, May 1998.

FORMULA

a(n,k) = |v(n,k)| where v(n,k) = v(n-1,k-1) - (2n-1)^2*v(n-1,k); eq (4.2).

Let F(x) = 1/cos(x). Then (2*n)!*(1/cos(x))^(2*n+1) = Sum_{k=0..n} T(n,k)*F^(2*k)(x), where F^(r) denotes the r-th derivative of F(x) (Zhang 1998). An example is given below. - Peter Bala, Feb 06 2012

EXAMPLE

For row 3: F(x) := 1/cos(x). Then 225*F(x) + 259*(d/dx)^2(F(x)) + 35*(d/dx)^4(F(x)) + (d/dx)^6(F(x)) = 720*(1/cos(x))^7, where F^(r) denotes the r-th derivative of F(x).

MAPLE

t := proc(n, k) option remember ; expand(x*mul(x+n/2-i, i=1..n-1)) ; coeftayl(%, x=0, k) ; end:

v := proc(n, k) option remember ; 4^(n-k)*t(2*n+1, 2*k+1) ; end:

A160563 := proc(n, k) abs(v(n, k)) ; end: for n from 0 to 10 do for k from 0 to n do printf("%d, ", A160563(n, k)) ; od: od: # R. J. Mathar, May 20 2009

MATHEMATICA

t[_, 0] = 1; t[n_, n_] := t[n, n] = ((2*n - 1)!!)^2; t[n_, k_] := t[n, k] = (2*n - 1)^2*t[n - 1, k - 1] + t[n - 1, k];

T[n_, k_] := t[n, n - k];

Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-Fran├žois Alcover, Nov 28 2017, after R. J. Mathar's comment *)

CROSSREFS

Cf. A001819, A008275, A008277, A160562, A091885.

Sequence in context: A324663 A109409 A262551 * A158286 A126839 A341818

Adjacent sequences:  A160560 A160561 A160562 * A160564 A160565 A160566

KEYWORD

nonn,tabl

AUTHOR

Jonathan Vos Post, May 19 2009

EXTENSIONS

Extended by R. J. Mathar, May 20 2009

STATUS

approved

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Last modified July 26 02:10 EDT 2021. Contains 346294 sequences. (Running on oeis4.)