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%I #14 Mar 02 2015 16:08:50
%S 1,3,-1,5,-6,1,7,-15,11,-1,9,-28,38,-20,1,11,-45,90,-90,37,-1,13,-66,
%T 175,-260,207,-70,1,15,-91,301,-595,707,-469,135,-1,17,-120,476,-1176,
%U 1862,-1848,1052,-264,1,19,-153,708,-2100,4158,-5502,4692,-2340,521,-1
%N Triangle T(n,k) read by rows: consider the sequence a(m) = a(m-1) + sum_{0<j<=m/n} a(m-j*n) with a(0)=1. Row n of T(n,k) is formed by the coefficients of the recurrence relation of sequence b(i) = a(n*i).
%C The sequence a(m) is also the expansion of (1-x^n)/(1-x-2x^n+x^{n+1}).
%C Instead of b(i) = a(n*i) one can take b(i) = a(n*i+p) for p=1..n-1.
%H Alois P. Heinz, <a href="/A113445/b113445.txt">Rows n = 0..140, flattened</a>
%F From _Alois P. Heinz_, Jul 16 2009: (Start)
%F T(n,k) = (-1)^k/(k+1)! * (1+k+(n-k)*2^(k+1)) * Product_{j=1..k}(n+j-k).
%F G.f. of column k: (-1)^k * x^k * (1+(2^(k+1)-1)*x)/(1-x)^(k+2). (End)
%e For n=5 (A113444) the recurrence relation is b(i) = 11b(i-1)-45b(i-2) +90b(i-3)-90b(i-4)+37b(i-5)-b(i-6), so the fifth row reads 11, -45, 90, -90, 37, -1.
%p T:= (n,k)-> (-1)^k /(k+1)! *(1+k +(n-k) *2^(k+1)) *mul (n+j-k, j=1..k):
%p seq(seq(T(n,k), k=0..n), n=0..11); # _Alois P. Heinz_, Jul 16 2009
%Y Columns k=0..2 give: A005408, -A000384, A007585(n-1) for n>=1. - _Alois P. Heinz_, Jul 16 2009
%K sign,tabl
%O 0,2
%A _Floor van Lamoen_, Nov 04 2005
%E More terms from _Alois P. Heinz_, Jul 16 2009
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