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%I #9 Apr 25 2012 00:49:55
%S 1,3,9,261,13419,7867287,10444212819,84955235950827,
%T 2235017786095822257,273416315791427558035965,
%U 125533366255776787874473759857,242979442003484538229530424638338553,1852958949086213206247388599213928431454549
%N G.f.: exp( Sum_{n>=1} (2^n - (-1)^n)^n * x^n/n ).
%C CONJECTURE: the highest power of 3 dividing a(n) equals 3^A089792(n) for n>=0; that is, n!*a(n)/3^n is an integer not divisible by 3 for n>=0.
%C Given g.f. A(x), note that A(x)^(1/3) is not an integer series.
%F a(n) == 3 (mod 6) for n>0.
%e G.f.: A(x) = 1 + 3*x + 9*x^2 + 261*x^3 + 13419*x^4 + 7867287*x^5 +...
%e such that
%e log(A(x)) = 3*x + 3^2*x^2/2 + 9^3*x^3/3 + 15^4*x^4/4 + 33^5*x^5/5 + 63^6*x^6/6 + 129^7*x^7/7 + 255^8*x^8/8 +...+ (2^n - (-1)^n)^n*x^n/n +...
%o (PARI) {a(n)=polcoeff(exp(sum(k=1, n, (2^k-(-1)^k)^k*x^k/k)+x*O(x^n)), n)}
%o for(n=0, 20, print1(a(n), ", "))
%Y Cf. A211897, A155200, A089792.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Apr 25 2012
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