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%I #22 Feb 14 2024 07:34:42
%S 1,6,63,866,13899,246366,4676768,93322596,1934035965,41286407510,
%T 902562584556,20119266633060,455832458083577,10470568749165246,
%U 243361203186769659,5714294570067499930,135377464019074334826,3232534121305720233264,77726654423445817800164
%N G.f.: exp( Sum_{n>=1} (3*n)!/n!^3 * x^n/n ).
%C The sixth root of the o.g.f. A(x)^(1/6) = 1 + x + 8*x^2 + 101*x^3 + 1569*x^4 + 27445*x^5 + ... appears to have integer coefficients. See A229452. More generally, if A(m,x) := exp( Sum_{n >= 1} (m*n)!/n!^m * x^n/n ), m = 1,2,3,..., then it can be shown that the expansion of A(m,x) has integer coefficients. We conjecture that the expansion of A(m,x)^(1/m!) also has integer coefficients. - _Peter Bala_, Feb 16 2020
%H Seiichi Manyama, <a href="/A229451/b229451.txt">Table of n, a(n) for n = 0..702</a>
%F a(n) ~ c * 3^(3*n)/n^2, where c = 2^11 * 3^(7/2) * Pi^5 * A370293^6 = 0.406436497... - _Vaclav Kotesovec_, Dec 25 2013, updated Feb 14 2024
%F a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} A006480(k) * a(n-k). - _Seiichi Manyama_, Feb 09 2024
%e G.f.: A(x) = 1 + 6*x + 63*x^2 + 866*x^3 + 13899*x^4 + 246366*x^5 +...
%e where
%e log(A(x)) = 6*x + 90*x^2/2 + 1680*x^3/3 + 34650*x^4/4 + 756756*x^5/5 +...+ A006480(n)*x^n/n +...
%t CoefficientList[Series[Exp[6*x*HypergeometricPFQ[{1,1,4/3,5/3},{2,2,2},27*x]],{x,0,20}],x] (* _Vaclav Kotesovec_, Dec 25 2013 *)
%o (PARI) {a(n)=polcoeff(exp(sum(k=1,n,(3*k)!/k!^3*x^k/k) +x*O(x^n)),n)}
%o for(n=0,25,print1(a(n),", "))
%Y Cf. A229452, A006480 (De Bruijn's S(3,n)), A333042, A333043, A370288, A370289, A370293.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Sep 23 2013
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