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A193436 exp( Sum_{n>=1} x^n/n^3 ) = Sum_{n>=0} a(n)*x^n/n!^3. 5

%I #13 Jul 18 2020 09:01:20

%S 1,1,5,71,2276,144724,16688884,3249507820,1005334796864,

%T 468967172341824,315409074574480704,294510517409159769024,

%U 369877735410388416241920,608401340784471133062837504,1281569707473914769353921666304,3391681347749396029674738480747264

%N exp( Sum_{n>=1} x^n/n^3 ) = Sum_{n>=0} a(n)*x^n/n!^3.

%C Sum_{n>=0} a(n)/n!^3 = exp(zeta(3)) = 3.326953110002499790...

%F a(0) = 1; a(n) = (n-1)! * (n!)^2 * Sum_{k=0..n-1} a(k) / ((k!)^3 * (n-k)^2). - _Ilya Gutkovskiy_, Jul 18 2020

%e A(x) = 1 + x + 5*x^2/2!^3 + 71*x^3/3!^3 + 2276*x^4/4!^3 +...

%e where

%e log(A(x)) = x + x^2/8 + x^3/27 + x^4/64 + x^5/125 + x^6/216 +...

%o (PARI) {a(n)=n!^3*polcoeff(exp(sum(m=1,n,x^m/m^3)+x*O(x^n)),n)}

%Y Cf. A074707, A217145, A193435.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Jul 25 2011

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