

A181783


Array described in comments to A053482, here read by increasing diagonals. See comments below.


3



1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 4, 1, 1, 1, 16, 21, 7, 1, 1, 1, 65, 142, 63, 11, 1, 1, 1, 326, 1201, 709, 151, 16, 1, 1, 1, 1957, 12336, 9709, 2521, 311, 22, 1, 1, 1, 13700, 149989, 157971, 50045, 7186, 575, 29, 1, 1, 1, 10960
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OFFSET

0,9


COMMENTS

We denote by a(n,p) the number in row number n>=0 and column number p>=0. The recurrence which defines the array is a(n,p)=n(p1)a(n1,p)+a(n,p1). The initials values are given by a(n,0)=1=a(0,p) for all n>=0 and p>=0.


LINKS

Table of n, a(n) for n=0..57.


FORMULA

If we consider the e.g.f Psi(p) of column number p we have: Psi(p)(z)=Psi(p1)(z)/((1(p1)z)) with Psi(1)(z)=exp(z). Then Psi(p)(z)=exp(z)/Prod_{k=0..p1}(1kz). We conclude that a(n,p)=n!sum_{m=0..n} sum_{k=1}^{p1}(1)^{p1k}k^{m+p2}/((nm)!(k1)!(p1k)!). It seems after the recurrence (and its proof) in A053482 that:
a(n,p) = sum_{k=1}^{p1}s1(p,pk)n(n1)...(np+1)a(nk,p)+1 where s1(m,n) are the classical Stirling numbers of first kind.
a(n,1) = 1 for every n.
a(1,p) = 1+p(p1)/2 for every p.


EXAMPLE

Array read row after row:
1, 1, 1, 1, 1, 1, 1, ...
1, 1, 2, 4, 7, 11, 16, ...
1, 1, 5, 21, 63, 151, 311, ...
1, 1, 16, 142, 709, 2521, ...
1, 1, 65, 1201, 9709, ...
a(4,3)=1201.


MAPLE

A181783 := proc(n, k)
option remember;
if n =0 or k = 0 then
1;
else
n*(k1)*procname(n1, k)+procname(n, k1) ;
end if;
end proc:
seq(seq(A181783(dk, k), k=0..d), d=0..12) ; # R. J. Mathar, Mar 02 2016


CROSSREFS

Cf. A000522, A053482, A185106.
Sequence in context: A008326 A181196 A227578 * A121395 A275377 A219585
Adjacent sequences: A181780 A181781 A181782 * A181784 A181785 A181786


KEYWORD

nonn,tabl,easy


AUTHOR

Richard Choulet, Dec 23 2012


EXTENSIONS

Edited by N. J. A. Sloane, Dec 24 2012


STATUS

approved



