A259871 a(0)=1/2, a(1)=1; a(n) = n*a(n-1) + (n-2)*a(n-2) - 2*Sum_{j=1..n-1} a(j)*a(n-j) + 2*Sum_{j=0..n-1} a(j)*a(n-1-j).

1, 2, 5, 14, 45, 170, 777, 4350, 29513, 236530, 2179133, 22576206, 258821269, 3245286490, 44115311969, 645664173566, 10117122765905, 168922438409826, 2993228077070645, 56090022818326542, 1108099905463382973, 23015655499699484810, 501356717394207256441

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

Cf. A259869, A259870, A259872, A260949.

Sequence in context: A347420 A174300 A059216 * A285201 A007823 A006391

Adjacent sequences: A259868 A259869 A259870 * A259872 A259873 A259874

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