OFFSET
1,2
COMMENTS
The sequence officially starts with a(0)=1/2, but since the OEIS only uses integers, we show it with offset 1.
LINKS
Eric M. Schmidt, Table of n, a(n) for n = 1..300
Richard J. Martin, and Michael J. Kearney, Integral representation of certain combinatorial recurrences, Combinatorica: 35:3 (2015), 309-315.
FORMULA
Martin and Kearney (2015) give a g.f.
a(n) ~ (n-1)! / exp(1) * (1 + 4/n + 16/n^2 + 76/n^3 + 416/n^4 + 2576/n^5 + 17840/n^6 + 137268/n^7 + 1170104/n^8 + 11050940/n^9 + 115885968/n^10), for coefficients see A260949. - Vaclav Kotesovec, Jul 29 2015
MATHEMATICA
nmax = 25; Rest[CoefficientList[Assuming[Element[x, Reals], Series[-1/(2*ExpIntegralEi[1 + 1/x]/Exp[1 + 1/x] - 1)/2, {x, 0, nmax}]], x]] (* Vaclav Kotesovec, Aug 05 2015 *)
PROG
(Sage)
@CachedFunction
def a(n) : return 1 if n==1 else 2 if n==2 else (n+2)*a(n-1) + (n-2)*a(n-2) - 2*sum(a(j)*a(n-j) for j in [1..n-1]) + 2*sum(a(j)*a(n-1-j) for j in [1..n-2])
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Jul 09 2015
EXTENSIONS
More terms from Eric M. Schmidt, Jul 10 2015
The offset 1 is correct. - N. J. A. Sloane, Jun 16 2021
STATUS
approved