0,5

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

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,...],...

(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)); )))}

Cf. A101514, A101516.

Sequence in context: A086617 A094526 A088699 this_sequence A028657 A053534 A104881

Adjacent sequences: A101512 A101513 A101514 this_sequence A101516 A101517 A101518

nonn,tabl

Paul D. Hanna (pauldhanna(AT)juno.com), Dec 06 2004