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A091499 Triangle, read by rows, such that T(n,k) equals the k-th term of the convolution of the two prior rows indexed by (n-k) and (k-1). 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 (list; table; graph; refs; listen; history; text; internal format)
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(n-k, j)*T(k-1, k-j-1) {j=0..min(n-k, 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>n||n<0||k<0, 0, if(k<=1||k==n, 1, sum(j=0, min(n-k, k), T(n-k, j)*T(k-1, k-j-1))))}

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

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Last modified February 27 13:18 EST 2020. Contains 332306 sequences. (Running on oeis4.)