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A181783 Array described in comments to A053482, here read by increasing diagonals. See comments below. 3

%I

%S 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,

%T 326,1201,709,151,16,1,1,1,1957,12336,9709,2521,311,22,1,1,1,13700,

%U 149989,157971,50045,7186,575,29,1,1,1,10960

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

%C 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(p-1)a(n-1,p)+a(n,p-1). The initials values are given by a(n,0)=1=a(0,p) for all n>=0 and p>=0.

%F If we consider the e.g.f Psi(p) of column number p we have: Psi(p)(z)=Psi(p-1)(z)/((1-(p-1)z)) with Psi(1)(z)=exp(z). Then Psi(p)(z)=exp(z)/Prod_{k=0..p-1}(1-kz). We conclude that a(n,p)=n!sum_{m=0..n} sum_{k=1}^{p-1}(-1)^{p-1-k}k^{m+p-2}/((n-m)!(k-1)!(p-1-k)!). It seems after the recurrence (and its proof) in A053482 that:

%F a(n,p) = -sum_{k=1}^{p-1}s1(p,p-k)n(n-1)...(n-p+1)a(n-k,p)+1 where s1(m,n) are the classical Stirling numbers of first kind.

%F a(n,1) = 1 for every n.

%F a(1,p) = 1+p(p-1)/2 for every p.

%e Array read row after row:

%e 1, 1, 1, 1, 1, 1, 1, ...

%e 1, 1, 2, 4, 7, 11, 16, ...

%e 1, 1, 5, 21, 63, 151, 311, ...

%e 1, 1, 16, 142, 709, 2521, ...

%e 1, 1, 65, 1201, 9709, ...

%e a(4,3)=1201.

%p A181783 := proc(n,k)

%p option remember;

%p if n =0 or k = 0 then

%p 1;

%p else

%p n*(k-1)*procname(n-1,k)+procname(n,k-1) ;

%p end if;

%p end proc:

%p seq(seq(A181783(d-k,k),k=0..d),d=0..12) ; # _R. J. Mathar_, Mar 02 2016

%Y Cf. A000522, A053482, A185106.

%K nonn,tabl,easy

%O 0,9

%A _Richard Choulet_, Dec 23 2012

%E Edited by _N. J. A. Sloane_, Dec 24 2012

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Last modified April 23 06:38 EDT 2021. Contains 343201 sequences. (Running on oeis4.)