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A183235 Sums of the cubes of multinomial coefficients.

%I A183235 #13 Feb 19 2015 16:17:36
%S A183235 1,1,9,244,15833,1980126,428447592,146966837193,75263273895385,
%T A183235 54867365927680618,54868847079435960134,73030508546599681432983,
%U A183235 126197144644287414997433576,277255161467330877411064074059
%N A183235 Sums of the cubes of multinomial coefficients.
%C A183235 Equals sums of the cubes of terms in rows of the triangle of multinomial coefficients (A036038).
%C A183235 Ignoring initial term, equals the logarithmic derivative of A182963.
%H A183235 Vaclav Kotesovec, <a href="/A183235/b183235.txt">Table of n, a(n) for n = 0..180</a>
%F A183235 G.f.: Sum_{n>=0} a(n)*x^n/n!^3 = Product_{n>=1} 1/(1 - x^n/n!^3).
%F A183235 a(n) ~ c * (n!)^3, where c = Product_{k>=2} 1/(1-1/(k!)^3) = 1.14825648754771664323845829539510031170864046029463094659207423270573478812675... . - _Vaclav Kotesovec_, Feb 19 2015
%e A183235 G.f.: A(x) = 1 + x + 9*x^2/2!^3 + 244*x^3/3!^3 + 15833*x^4/4!^3 +...
%e A183235 A(x) = 1/((1-x)*(1-x^2/2!^3)*(1-x^3/3!^3)*(1-x^4/4!^3)*...).
%e A183235 ...
%e A183235 After the initial term a(0)=1, the next few terms are
%e A183235 a(1) = 1^3 = 1,
%e A183235 a(2) = 1^3 + 2^3 = 9,
%e A183235 a(3) = 1^3 + 3^3 + 6^3 = 244,
%e A183235 a(4) = 1^3 + 4^3 + 6^3 + 12^3 + 24^3 = 15833,
%e A183235 a(5) = 1^3 + 5^3 + 10^3 + 20^3 + 30^3 + 60^3 + 120^3 = 1980126, ...;
%e A183235 and continue with the sums of cubes of the terms in triangle A036038.
%o A183235 (PARI) {a(n)=n!^3*polcoeff(1/prod(k=1, n, 1-x^k/k!^3 +x*O(x^n)), n)}
%Y A183235 Cf. A036038, A005651, A183240, A183236, A183237, A183238; A182963.
%K A183235 nonn
%O A183235 0,3
%A A183235 _Paul D. Hanna_, Jan 04 2011
%E A183235 Examples added and name changed by _Paul D. Hanna_, Jan 05 2011