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%I #24 Jul 03 2021 09:03:29
%S 1,1,7,199,17713,4572529,3426693463,7575807034711,49908659904426337,
%T 983868034228748840161,58130023275752925902247847,
%U 10299771730830080877230000021479,5474153833417147528343683843805979793,8727821227226586439546709016484604992020049
%N E.g.f.: ( Sum_{n>=0} 3^(n*(n + 1)/2) * x^n/n! )^(1/3).
%H Alois P. Heinz, <a href="/A178319/b178319.txt">Table of n, a(n) for n = 0..64</a>
%H Richard Stanley, <a href="https://mathoverflow.net/q/385402">Proof of the general conjecture</a>, MathOverflow, March 2021.
%F a(n) = 1 (mod 6) for n >= 0 (conjecture).
%F General conjecture: [x^n/n!] E(q*x, q)^(1/q) = 1 (mod q(q-1)) for n >= 0 and integer q > 1 where E(x, q) = Sum_{n>=0} q^(n*(n - 1)/2)*x^n/n!.
%e E.g.f.: A(x) = 1 + x + 7*x^2/2! + 199*x^3/3! + 17713*x^4/4! +...
%e A(x)^3 = 1 + 3*x + 3^3*x^2/2! + 3^6*x^3/3! + 3^10*x^4/4! +...
%e Let E(x, q) = Sum_{n>=0} q^(n*(n - 1)/2)*x^n/n!, then the coefficients of (x^n/n!) in E(qx, q)^(1/q) begin:
%e 1;
%e 1;
%e q^2 - q + 1;
%e q^5 - 3*q^3 + 5*q^2 - 3*q + 1;
%e q^9 - 4*q^6 + q^5 + 15*q^4 - 24*q^3 + 17*q^2 - 6*q + 1;
%e q^14 - 5*q^10 + 5*q^9 - 10*q^8 + 30*q^7 - 95*q^5 + 149*q^4 - 110*q^3 + 45*q^2 - 10*q + 1; ...
%e Setting q = 3 generates this sequence.
%p a:= n-> n!*coeff(series(add(3^binomial(j+1, 2)
%p *x^j/j!, j=0..n)^(1/3), x, n+1), x, n):
%p seq(a(n), n=0..14); # _Alois P. Heinz_, Mar 15 2021
%o (PARI) {a(n)=n!*polcoeff(sum(m=0,n,3^(m*(m+1)/2)*x^m/m!+x*O(x^n))^(1/3),n)}
%Y Cf. A178315 (sqrt case).
%Y Column k=3 of A346061.
%K nonn
%O 0,3
%A _Paul D. Hanna_, May 24 2010
%E General conjecture restated by _Paul D. Hanna_, May 25 2010
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