Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #2 Mar 30 2012 18:37:20
%S 1,4,20,320,21064,5030400,4056470528,10872157339648,98162974155542592,
%T 3052890463194814939136,334052589949087491382968320,
%U 130858881562759880830581892710400
%N G.f.: A(x) = Sum_{n>=0} 2^n*log(G(2^n*x))^n/n! where G(x) = g.f. of A167000.
%C The g.f. of A167000, G(x), satisfies:
%C Sum_{n>=0} log(G(2^n*x))^n/n! = 1 + Sum_{n>=0} 2^(2^n-n)*x^(2^n).
%F a(n) = [x^n] G(x)^(2^(n+1)) for n>=0 where G(x) = g.f. of A167000.
%e G.f.: A(x) = 1 + 4*x + 20*x^2 + 320*x^3 + 21064*x^4 + 5030400*x^5 +...
%e Let G(x) equal the g.f. of A167000:
%e G(x) = 1 + x - x^2 - 16*x^4 - 1767*x^5 - 493164*x^6 - 422963721*x^7 +...
%e then the g.f. A(x) of this sequence equals the series:
%e A(x) = 1 + 2*log(G(2x)) + 4*log(G(4x))^2/2! + 8*log(G(8x))^3/3! + 16*log(G(16x))^4/4! +...
%e ILLUSTRATE (2^n)-th POWERS OF G.F. G(x) OF A167000.
%e The coefficients in the expansion of G(x)^(2^n), n>=0, begin:
%e G^1: [1,1,-1,0,-16,-1767,-493164,-422963721,-1130568823448,...];
%e G^2: [(1),2,-1,-2,-31,-3566,-989830,-846910236,...];
%e G^4: [1,(4),2,-8,-69,-7252,-1993858,-1697772536,...];
%e G^8: [1,8,(20),0,-198,-15088,-4045944,-3411523840,...];
%e G^16: [1,16,104,(320),4,-33344,-8341216,-6888386304,...];
%e G^32: [1,32,464,3968,(21064),0,-17646208,-14050624512,...];
%e G^64: [1,64,1952,37632,511376,(5030400),0,-29063442432,...];
%e G^128: [1,128,8000,325120,9649952,222432256,(4056470528),0,...]; ...
%e where the coefficients along the diagonal (shown in parenthesis) form the initial terms of this sequence.
%o (PARI) {a(n)=local(A=[1,4],B=[1,2],G=[1,1]);for(i=1,n,G=concat(G,0); B=Vec(sum(m=0,#G,log(subst(Ser(G),x,2^m*x))^m/m!)); G[ #G]=-floor(B[ #G]/2^(#G-1))); A=Vec(sum(m=0,#G,2^m*log(subst(Ser(G),x,2^m*x))^m/m!)); A[n+1]}
%Y Cf. A167000, A167001.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Nov 14 2009