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COMMENTS
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Compare the g.f. to the LambertW identity:
1 = Sum_{n>=0} (n+1)^(n-1) * exp(-(n+1)*x) * x^n/n!.
More generally, if we define a(n) for fixed integers m, t, and s>=0, by:
(0) Sum_{n>=0} m * n^(s*n) * (n*t+m)^(n-1) * exp(-n^s*(n*t+m)*x) * x^n/n! = Sum_{n>=0} a(n)*x^n
then the coefficients a(n) are integral and may be expressed by:
(1) a(n) = 1/n! * Sum_{k=0..n} m*(-1)^(n-k)*binomial(n,k) * k^(s*n) * (k*t+m)^(n-1).
(2) a(n) = 1/n! * [x^n] Sum_{k>=0} m*k^(s*k)*(k*t+m)^(k-1)*x^k / (1 + k^s*(k*t+m)*x)^(k+1).
(3) a(n) = 1/t^((s-1)*n) * [x^(s*n)] 1 + m*x*(1+m*x)^(n-1) / Product_{k=1..n} (1-k*t*x).
(4) a(n) = 1/t^((s-1)*n) * [x^(s*n)] 1 + m*x*(1-m*x)^(s*n) / Product_{k=1..n} (1-(k*t+m)*x).
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PROG
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(PARI) {a(n)=polcoeff(sum(m=0, n, m^m*(m+1)^(m-1)*x^m*exp(-m*(m+1)*x+x*O(x^n))/m!), n)}
(PARI) {a(n)=(1/n!)*polcoeff(sum(k=0, n, k^k*(k+1)^(k-1)*x^k/(1+k*(k+1)*x +x*O(x^n))^(k+1)), n)}
(PARI) {a(n)=1/n!*sum(k=0, n, (-1)^(n-k)*binomial(n, k)*k^n*(k+1)^(n-1))}
(PARI) {a(n)=polcoeff(1+x*(1+x)^(n-1)/prod(k=0, n, 1-k*x +x*O(x^n)), n)}
(PARI) {a(n)=polcoeff(1+x*(1-x)^n/prod(k=0, n, 1-(k+1)*x +x*O(x^n)), n)}
for(n=0, 30, print1(a(n), ", "))
(PARI) {Stirling2(n, k)=n!*polcoeff(((exp(x+x*O(x^n))-1)^k)/k!, n)}
{a(n)=if(n==0, 1, sum(k=0, n-1, binomial(n-1, k) * Stirling2(2*n-k-1, n)))} \\ Paul D. Hanna, Nov 13 2012
/* PARI Programs for the General Case (START) ...................... */
(PARI) {a(n, m=1, t=1, s=1)=polcoeff(sum(k=0, n, m*k^(s*k)*(t*k+m)^(k-1)*exp(-k^s*(t*k+m)*x+x*O(x^n))*x^k/k!), n)}
(PARI) {a(n, m=1, t=1, s=1)=(1/n!)*polcoeff(sum(k=0, n, m*k^(s*k)*(t*k+m)^(k-1)*x^k/(1+k^s*(t*k+m)*x +x*O(x^n))^(k+1)), n)}
(PARI) {a(n, m=1, t=1, s=1)=1/n!*sum(k=0, n, m*(-1)^(n-k)*binomial(n, k)*k^(s*n)*(t*k+m)^(n-1))}
(PARI) {a(n, m=1, t=1, s=1)=(1/t^((s-1)*n))*polcoeff(1+m*x*(1+m*x)^(n-1)/prod(k=0, n, 1-t*k*x +x*O(x^(s*n))), s*n)}
(PARI) {a(n, m=1, t=1, s=1)=(1/t^((s-1)*n))*polcoeff(1+m*x*(1-m*x)^(s*n)/prod(k=0, n, 1-(t*k+m)*x +x*O(x^(s*n))), s*n)}
/* (END) ........................................................... */
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