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%I #23 Jan 06 2022 14:25:56
%S 1,2,1,5,1,1,15,2,1,1,52,5,1,1,1,203,13,2,1,1,1,877,36,6,1,1,1,1,4140,
%T 109,17,2,1,1,1,1,21147,359,44,7,1,1,1,1,1,115975,1266,112,23,2,1,1,1,
%U 1,1,678570,4731,304,65,8,1,1,1,1,1,1,4213597,18657,918,165,30,2,1,1,1,1,1,1
%N Triangle T(n,k), n>=1, 1<=k<=n, read by rows, where sequence a_k of column k has a_k(0)=1, followed by (k-1)-fold 0 and a_k(n) shifts k places down under binomial transform.
%C The matrix inverse starts:
%C 1;
%C -2, 1;
%C -3, -1, 1;
%C -8, -1, -1, 1;
%C -31, -3, 0, -1, 1;
%C -132, -7, -1, 0, -1, 1;
%C -616, -19, -4, 0, 0, -1, 1; - _R. J. Mathar_, Mar 22 2013
%H Alois P. Heinz, <a href="/A143983/b143983.txt">Rows n = 1..141, flattened</a>
%H N. J. A. Sloane, <a href="/transforms.txt">Transforms</a>
%F T(n,k) = Sum_{j=0..n-k} C(n-k,j)*T(j,k) if n>=k, else T(n,k) = 1 if n=1, else T(n,k) = 0.
%e T(5,2) = 5, because [1,3,3,1] * [1,0,1,1] = 5.
%e Triangle begins:
%e : 1;
%e : 2, 1;
%e : 5, 1, 1;
%e : 15, 2, 1, 1;
%e : 52, 5, 1, 1, 1;
%e : 203, 13, 2, 1, 1, 1;
%e : 877, 36, 6, 1, 1, 1, 1;
%e : 4140, 109, 17, 2, 1, 1, 1, 1;
%e : 21147, 359, 44, 7, 1, 1, 1, 1, 1;
%e : 115975, 1266, 112, 23, 2, 1, 1, 1, 1, 1;
%p T:= proc(n, k) option remember; `if`(n<k, `if`(n=0, 1, 0),
%p add(binomial(n-k, j) *T(j,k), j=0..n-k))
%p end:
%p seq(seq(T(n, k), k=1..n), n=1..14);
%t t[n_, k_] := t[n, k] = If[n < k, If[n == 0, 1, 0], Sum[Binomial[n-k, j]*t[j, k], {j, 0, n-k}]]; Table[Table[t[n, k], {k, 1, n}], {n, 1, 13}] // Flatten (* _Jean-François Alcover_, Dec 18 2013, translated from Maple *)
%Y Columns 1-6 give: A000110, A000994, A000996, A010748, A010749, A010750.
%Y Cf. A007318.
%K eigen,nonn,tabl
%O 1,2
%A _Alois P. Heinz_, Sep 06 2008
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