A101515 - OEIS

%I #5 Jun 13 2017 22:17:09

%S 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,

%T 77,77,31,7,1,1,8,43,146,236,146,43,8,1,1,9,57,249,596,596,249,57,9,1,

%U 1,10,73,393,1290,2037,1290,393,73,10,1,1,11,91,585,2486,5772,5772,2486,585

%N 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.

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

%e Rows begin:

%e [_1,1,1,1,1,1,1,1,1,...],

%e [1,_2,3,4,5,6,7,8,9,...],

%e [1,3,_7,13,21,31,43,57,73,...],

%e [1,4,13,_35,77,146,249,393,585,...],

%e [1,5,21,77,_236,596,1290,2486,4387,...],

%e [1,6,31,146,596,_2037,5772,13987,29987,...],

%e [1,7,43,249,1290,5772,_21695,67943,181811,...],

%e [1,8,57,393,2486,13987,67943,_277966,951051,...],

%e [1,9,73,585,4387,29987,181811,951051,_4198635,...],...

%e The inverse binomial transform of the rows of this array are generated

%e from the products of the main diagonal with rows of Pascal's triangle:

%e BINOMIAL[1*1] = [_1,1,1,1,1,1,1,1,1,...],

%e BINOMIAL[1*1,1*1] = [1,_2,3,4,5,6,7,8,9,...],

%e BINOMIAL[1*1,1*2,2*1] = [1,3,_7,13,21,31,43,57,73,...],

%e BINOMIAL[1*1,1*3,2*3,7*1] = [1,4,13,_35,77,146,249,393,...],

%e BINOMIAL[1*1,1*4,2*6,7*4,35*1] = [1,5,21,77,_236,596,1290,...],

%e BINOMIAL[1*1,1*5,2*10,7*10,35*5,236*1] = [1,6,31,146,596,_2037,...],...

%o (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));)))

%Y Cf. A101514, A101516.

%K nonn,tabl

%O 0,5

%A _Paul D. Hanna_, Dec 06 2004