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A212775 Number of partitions of 2^(2^n) into powers of 2.

%I A212775 #17 Feb 28 2017 22:54:57
%S A212775 2,4,36,692004,114788185359199234852802340
%N A212775 Number of partitions of 2^(2^n) into powers of 2.
%C A212775 Lengths (in decimal digits) of the terms a(0), a(1), ... are: 1, 1, 2, 6, 27, 119, 525, 2241, 9330, ... .
%H A212775 Alois P. Heinz, <a href="/A212775/b212775.txt">Table of n, a(n) for n = 0..6</a>
%F A212775 a(n) = [x^2^(2^n-1)] 1/(1-x) * 1/Product_{j=0..2^n-1} (1-x^(2^j)).
%e A212775 a(0) = 2 because the number of partitions of 2^2^0 = 2 into powers of 2 is 2: [2], [1,1].
%e A212775 a(1) = 4: [4], [2,2], [2,1,1], [1,1,1,1].
%p A212775 b:= proc(n, j) option remember; local nn, r;
%p A212775       if n<0 then 0
%p A212775     elif j=0 then 1
%p A212775     elif j=1 then n+1
%p A212775     elif n<j then b(n, j):= b(n-1, j) +b(2*n, j-1)
%p A212775              else nn:= 1 +floor(n);
%p A212775                   r:= n-nn;
%p A212775                   (nn-j) *binomial(nn, j) *add(binomial(j, h)
%p A212775                   /(nn-j+h) *b(j-h+r, j) *(-1)^h, h=0..j-1)
%p A212775       fi
%p A212775     end:
%p A212775 a:= n-> b(1, 2^n):
%p A212775 seq(a(n), n=0..6);
%t A212775 b[n_, j_] := b[n, j] = Module[{nn, r}, Which[n<0, 0, j==0, 1, j==1, n+1, n < j, b[n, j] = b[n-1, j] + b[2*n, j-1], True, nn = 1+Floor[n]; r = n-nn; (nn-j)*Binomial[nn, j]*Sum[Binomial[j, h]/(nn-j+h)*b[j-h+r, j]*(-1)^h, {h, 0, j-1}]]]; a[n_] := b[1, 2^n]; Table[a[n], {n, 0, 6}] (* _Jean-François Alcover_, Feb 28 2017, translated from Maple *)
%Y A212775 Cf. A000123, A002577, A172288, A182135.
%K A212775 nonn
%O A212775 0,1
%A A212775 _Alois P. Heinz_, May 26 2012