A101515 Symmetric square array, read by antidiagonals, such that the inverse binomial transform of row n forms the sequence: {C(n,k)*A101514(k), 0<=k<=n}, where A101514 equals the main diagonal shift right.

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 7, 4, 1, 1, 5, 13, 13, 5, 1, 1, 6, 21, 35, 21, 6, 1, 1, 7, 31, 77, 77, 31, 7, 1, 1, 8, 43, 146, 236, 146, 43, 8, 1, 1, 9, 57, 249, 596, 596, 249, 57, 9, 1, 1, 10, 73, 393, 1290, 2037, 1290, 393, 73, 10, 1, 1, 11, 91, 585, 2486, 5772, 5772, 2486, 585

Offset

0,5

Comments

The main diagonal equals A101514 shift one place left. The antidiagonal sums form A101516.

Links

Table of n, a(n) for n=0..74.

Example

Rows begin:
[_1,1,1,1,1,1,1,1,1,...],
[1,_2,3,4,5,6,7,8,9,...],
[1,3,_7,13,21,31,43,57,73,...],
[1,4,13,_35,77,146,249,393,585,...],
[1,5,21,77,_236,596,1290,2486,4387,...],
[1,6,31,146,596,_2037,5772,13987,29987,...],
[1,7,43,249,1290,5772,_21695,67943,181811,...],
[1,8,57,393,2486,13987,67943,_277966,951051,...],
[1,9,73,585,4387,29987,181811,951051,_4198635,...],...
The inverse binomial transform of the rows of this array are generated
from the products of the main diagonal with rows of Pascal's triangle:
BINOMIAL[1*1] = [_1,1,1,1,1,1,1,1,1,...],
BINOMIAL[1*1,1*1] = [1,_2,3,4,5,6,7,8,9,...],
BINOMIAL[1*1,1*2,2*1] = [1,3,_7,13,21,31,43,57,73,...],
BINOMIAL[1*1,1*3,2*3,7*1] = [1,4,13,_35,77,146,249,393,...],
BINOMIAL[1*1,1*4,2*6,7*4,35*1] = [1,5,21,77,_236,596,1290,...],
BINOMIAL[1*1,1*5,2*10,7*10,35*5,236*1] = [1,6,31,146,596,_2037,...],...

Prog

(PARI) T(n, k)=if(n<0 || k<0, 0, if(n==0 || k==0, 1, if(n>k, T(k, n), 1+sum(j=1, k, binomial(k, j)*binomial(n, j)*T(j-1, j-1)); )))

Crossrefs

Cf. A101514, A101516.

Sequence in context: A086617 A094526 A088699 * A327913 A028657 A053534

Adjacent sequences: A101512 A101513 A101514 * A101516 A101517 A101518

Keyword

nonn,tabl

Author

Paul D. Hanna, Dec 06 2004

Status

approved