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%I #8 Jun 13 2017 22:15:44
%S 1,1,1,1,3,1,1,5,5,1,1,7,14,7,1,1,9,28,28,9,1,1,11,47,76,47,11,1,1,13,
%T 71,163,163,71,13,1,1,15,100,301,435,301,100,15,1,1,17,134,502,971,
%U 971,502,134,17,1,1,19,173,778,1909,2577,1909,778,173,19,1,1,21,217,1141
%N Symmetric square array, read by antidiagonals, where the inverse binomial transform of row n equals: [C(n,0)*1, C(n,1)*2,..., C(n,k)*A051163(k), ..., C(n,n)*A051163(n)] and where A051162 equals the antidiagonal sums.
%C Antidiagonal sums form A051163. Main diagonal is A100937. Different from A086620.
%F T(n, k) = Sum_{j=0..n} C(k, j)*C(n, j)*A051162(j), with T(0, 0) = 1 and where Sum_{i=0..n} T(n-i, i) = A051162(n).
%e Rows begin:
%e [1,1,1,1,1,1,1,1,1,...],
%e [1,3,5,7,9,11,13,15,17,...],
%e [1,5,14,28,47,71,100,134,...],
%e [1,7,28,76,163,301,502,778,...],
%e [1,9,47,163,435,971,1909,3417,...],
%e [1,11,71,301,971,2577,5917,12167,...],
%e [1,13,100,502,1909,5917,15678,36744,...],
%e [1,15,134,778,3417,12167,36744,97272,...],...
%e Antidiagonal sums form A051163: [1,2,5,12,30,76,194,496,1269,3250,8337,...].
%e The inverse binomial transform of the rows form the respective rows of the triangle B:
%e [1*1],
%e [1*1,1*2],
%e [1*1,2*2,1*5],
%e [1*1,3*2,3*5,1*12],
%e [1*1,4*2,6*5,4*12,1*30],...
%e where B(n,k) = binomial(n,k)*A051163(k).
%o (PARI) T(n,k)=if(n==0 || k==0,1, sum(j=0,n,binomial(k,j)*binomial(n,j)*sum(i=0,j,T(j-i,i)));)
%Y Cf. A051163, A100937.
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Nov 23 2004
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