OFFSET
0,2
COMMENTS
Compare the g.f. to the LambertW identity:
1 = Sum_{n>=0} 5*(n+5)^(n-1) * exp(-(n+5)*x) * x^n/n!.
From Vaclav Kotesovec, May 22 2014: (Start)
Generally, for p>=1, a(n) = 1/n!*Sum_{k=0..n} p*(-1)^(n-k) * binomial(n,k) * k^n * (k+p)^(n-1) = Sum_{j=0..n-1} binomial(n-1,j) * p^(n-j) * StirlingS2(n+j,n).
a(n) ~ p * 2^(2*n-3/2+p/2) * n^(n-3/2) / (sqrt(Pi*(1-c)) * exp(n) * (2-c)^(n-1) * c^(n+p/2)), where c = -LambertW(-2*exp(-2)) = 0.4063757399599599...
(End)
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..340
FORMULA
a(n) = 1/n! * Sum_{k=0..n} 5*(-1)^(n-k)*binomial(n,k) * k^n * (k+5)^(n-1).
a(n) = 1/n! * [x^n] Sum_{k>=0} 5*k^k*(k+5)^(k-1)*x^k / (1 + k*(k+5)*x)^(k+1).
a(n) = [x^n] 1 + 5*x*(1+5*x)^(n-1) / Product_{k=1..n} (1-k*x).
a(n) = [x^n] 1 + 5*x*(1-5*x)^(n-1) / Product_{k=1..n} (1-(k+5)*x).
a(n) ~ 5 * 2^(2*n+1) * n^(n-3/2) / (sqrt(Pi*(1-c)) * exp(n) * (2-c)^(n-1) * c^(n+5/2)), where c = -LambertW(-2*exp(-2)) = 0.4063757399599599... . - Vaclav Kotesovec, May 22 2014
EXAMPLE
O.g.f.: A(x) = 1 + 5*x + 40*x^2 + 550*x^3 + 11000*x^4 + 285380*x^5 + 9064560*x^6 +...
where
A(x) = 1 + 5*1^1*6^0*x*exp(-1*6*x) + 5*2^2*7^1*exp(-2*7*x)*x^2/2! + 5*3^3*8^2*exp(-3*8*x)*x^3/3! + 5*4^4*9^3*exp(-4*9*x)*x^4/4! + 5*5^5*10^4*exp(-5*10*x)*x^5/5! +...
simplifies to a power series in x with integer coefficients.
MATHEMATICA
Flatten[{1, Table[Sum[Binomial[n-1, j]*5^(n-j)*StirlingS2[n+j, n], {j, 0, n-1}], {n, 1, 20}]}] (* Vaclav Kotesovec, May 22 2014 *)
PROG
(PARI) {a(n)=polcoeff(sum(m=0, n, 5*m^m*(m+5)^(m-1)*x^m*exp(-m*(m+5)*x+x*O(x^n))/m!), n)}
(PARI) {a(n)=(1/n!)*polcoeff(sum(k=0, n, 5*k^k*(k+5)^(k-1)*x^k/(1+k*(k+5)*x +x*O(x^n))^(k+1)), n)}
(PARI) {a(n)=1/n!*sum(k=0, n, 5*(-1)^(n-k)*binomial(n, k)*k^n*(k+5)^(n-1))}
(PARI) {a(n)=polcoeff(1+5*x*(1+5*x)^(n-1)/prod(k=0, n, 1-k*x +x*O(x^n)), n)}
(PARI) {a(n)=polcoeff(1+5*x*(1-5*x)^n/prod(k=0, n, 1-(k+5)*x +x*O(x^n)), n)}
for(n=0, 30, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Oct 14 2012
STATUS
approved