%I #11 Jul 02 2016 08:31:59
%S 1,2,15,560,91390,61124064,163995687856,1756185841659392,
%T 75079359427627897200,12831653340946454374300160,
%U 8777916355714456994772455584000,24054320541767107204031746600673906688
%N Diagonal of square array A136462: a(n) = C((n+1)*2^(n-1), n) for n>=0.
%C a(n) is divisible by (n+1) for n>=0: a(n)/(n+1) = A136464(n).
%F a(n) = [x^n] Sum_{i>=0} ((n+1)/2)^i * log(1 + 2^i*x)^i/i!.
%F a(n) is found in row n, column 0, of matrix power A136467^(n+1) for n>=0.
%F a(n) ~ exp(n+1) * 2^(n*(n-1)) / sqrt(2*Pi*n). - _Vaclav Kotesovec_, Jul 02 2016
%t Table[Binomial[(n+1)2^(n-1),n],{n,0,15}] (* _Harvey P. Dale_, Apr 20 2011 *)
%o (PARI) a(n)=binomial((n+1)*2^(n-1),n)
%o (PARI) /* a(n) = Coefficient of x^n in series: */
%o a(n)=polcoeff(sum(i=0,n,((n+1)/2)^i*log(1+2^i*x +x*O(x^n))^i/i!),n)
%Y Cf. A136462; A136464; rows: A136465, A014070, A136466, A101346; A136467.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Dec 31 2007