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URL: https://oeis.org/A316592

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A316592
a(n) equals the coefficient of x^n in Sum_{m>=0} (x^m + 2 + 1/x^m)^m for n >= 1.
8
1, 4, 15, 57, 210, 798, 3003, 11468, 43759, 168080, 646646, 2496647, 9657700, 37444162, 145422720, 565730729, 2203961430, 8597528644, 33578000610, 131282534380, 513791608421, 2012616897500, 7890371113950, 30957701501466, 121548660036301, 477551187602112, 1877405874750672, 7384942679432199, 29065024282889672, 114449595182606502, 450883717216034179, 1777090076536979756, 7007092303604342400
OFFSET
1,2
COMMENTS
The coefficient of 1/x^n in Sum_{m>=0} (x^m + 2 + 1/x^m)^m equals a(n) for n > 0, while the constant term in the sum increases without limit.
a(n) = Sum_{k=0..n-1} A316590(n,k) * 2^k for n >= 1.
a(n) = A304638(4*n) for n >= 1, where A304638(n) = [x^n] Sum_{m>=0} (x^m + 1/x^m)^m.
LINKS
FORMULA
a(n) ~ 4^n / sqrt(Pi*n). - Vaclav Kotesovec, Jul 10 2018
EXAMPLE
G.f.: A(x) = x + 4*x^2 + 15*x^3 + 57*x^4 + 210*x^5 + 798*x^6 + 3003*x^7 + 11468*x^8 + 43759*x^9 + 168080*x^10 + 646646*x^11 + 2496647*x^12 + ...
such that Sum_{m>=0} (x^m + 2 + 1/x^m)^m = A(x) + A(1/x) + (infinity)*x^0.
PROG
(PARI) {a(n) = polcoeff( sum(m=1, n, (x^-m + 2 + x^m)^m +x*O(x^n)), n, x)}
for(n=1, 40, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jul 08 2018
STATUS
approved