A061573 a(n) = (n!)^2*Sum_{k=1..n} 1/k!.

1, 6, 60, 984, 24720, 890640, 43646400, 2793409920, 226266566400, 22626660268800, 2737825932441600, 394246934750592000, 66627731979077068800, 13059035467986283776000, 2938282980298221523968000, 752200442956365632925696000, 217385928014390023602954240000

Offset

1,2

Links

Harry J. Smith, Table of n, a(n) for n = 1..100

Formula

Recurrence: a(1) = 1, a(2) = 6, a(n) = n*(n+1)*a(n-1) - n*(n-1)^2*a(n-2) for n >=3. The sequence b(n) = n!^2 also satisfies this recurrence with the initial conditions b(1) = 1 and b(2) = 4. Hence we have the finite continued fraction expansion a(n)/b(n) = 1/(1-2/(6-12/(12-...-n*(n-1)^2/(n*(n+1))))). Lim_{n -> infinity} a(n)/b(n) = e - 1 = 1/(1-2/(6-12/(12-...-n*(n-1)^2/(n*(n+1))-...))) = 1/(1-1/(3-2/(4-...-n/(n+2)-...))). Cf. A000522 and A061572. - Peter Bala, Jul 10 2008

a(n) = n!*A002627(n). - R. J. Mathar, Mar 18 2017

Sum_{n>=1} a(n) * x^n / (n!)^2 = (exp(x) - 1) / (1 - x). - Ilya Gutkovskiy, Jul 15 2021

Mathematica

Table[(n!)^2 Sum[1/k!, {k, n}], {n, 20}] (* Harvey P. Dale, Dec 02 2021 *)

Prog

(PARI) a(n) = { n!^2*sum(k=1, n, 1/k!) } \\ Harry J. Smith, Jul 24 2009

Crossrefs

Cf. A000522, A002627, A061572.

Sequence in context: A366335 A308332 A126574 * A248655 A092356 A353196

Adjacent sequences: A061570 A061571 A061572 * A061574 A061575 A061576

Keyword

nonn,changed

Author

N. J. A. Sloane, May 19 2001

Status

approved