

A091499


Triangle, read by rows, such that T(n,k) equals the kth term of the convolution of the two prior rows indexed by (nk) and (k1).


3



1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 2, 3, 3, 1, 1, 1, 2, 4, 4, 3, 1, 1, 1, 2, 4, 6, 5, 4, 1, 1, 1, 2, 4, 7, 8, 7, 4, 1, 1, 1, 2, 4, 8, 10, 12, 8, 5, 1, 1, 1, 2, 4, 9, 13, 15, 16, 10, 5, 1, 1, 1, 2, 4, 9, 15, 20, 22, 21, 12, 6, 1, 1, 1, 2, 4, 9, 17, 25, 31, 31, 27, 14, 6, 1, 1, 1, 2, 4, 9
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OFFSET

0,9


COMMENTS

Row sums are A091501. Convergent of rows is A091500: {1,1,2,4,9,20,47,113,275,676,1685,4271,10843,27801,71611,185795,...}.


LINKS

Table of n, a(n) for n=0..95.


FORMULA

T(n, k) = Sum T(nk, j)*T(k1, kj1) {j=0..min(nk, k)}, with T(n, 0)=1, T(n, n)=1 for n>=0.


EXAMPLE

T(7,3) = 4 = third term of convolution[{1,1,2,2,1},{1,1,1}] = 1*1 + 1*1 + 2*1 = T(4,0)*T(2,2) + T(4,1)*T(2,1) + T(4,2)*T(2,0).
T(14,5) = 19 = 5th term of convolution[{1,1,2,4,8,10,12,8,5,1},{1,1,2,2,1}] = 1*1 + 1*2 + 2*2 + 4*1 + 8*1 = T(9,0)*T(4,4) + T(9,1)*T(4,3) + T(9,2)*T(4,2) + T(9,3)*T(4,1) + T(9,4)*T(4,0).
Rows begin:
{1},
{1,1},
{1,1,1},
{1,1,2,1},
{1,1,2,2,1},
{1,1,2,3,3,1},
{1,1,2,4,4,3,1},
{1,1,2,4,6,5,4,1},
{1,1,2,4,7,8,7,4,1},
{1,1,2,4,8,10,12,8,5,1},
{1,1,2,4,9,13,15,16,10,5,1},
{1,1,2,4,9,15,20,22,21,12,6,1},
{1,1,2,4,9,17,25,31,31,27,14,6,1},...


PROG

(PARI) {T(n, k)=if(k>nn<0k<0, 0, if(k<=1k==n, 1, sum(j=0, min(nk, k), T(nk, j)*T(k1, kj1))))}


CROSSREFS

Cf. A091500, A091501.
Sequence in context: A055215 A239550 A058398 * A284249 A137350 A166240
Adjacent sequences: A091496 A091497 A091498 * A091500 A091501 A091502


KEYWORD

nonn,tabl


AUTHOR

Paul D. Hanna, Jan 16 2004


STATUS

approved



