OFFSET
0,2
COMMENTS
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.
Given g.f. A(x), note that A(x)^(1/3) is not an integer series.
FORMULA
a(n) == 3 (mod 6) for n>0.
EXAMPLE
G.f.: A(x) = 1 + 3*x + 9*x^2 + 261*x^3 + 13419*x^4 + 7867287*x^5 +...
such that
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 +...
PROG
(PARI) {a(n)=polcoeff(exp(sum(k=1, n, (2^k-(-1)^k)^k*x^k/k)+x*O(x^n)), n)}
for(n=0, 20, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Apr 25 2012
STATUS
approved