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%I #20 Dec 02 2021 13:55:59
%S 1,6,60,984,24720,890640,43646400,2793409920,226266566400,
%T 22626660268800,2737825932441600,394246934750592000,
%U 66627731979077068800,13059035467986283776000,2938282980298221523968000,752200442956365632925696000,217385928014390023602954240000
%N a(n) = (n!)^2*Sum_{k=1..n} 1/k!.
%H Harry J. Smith, <a href="/A061573/b061573.txt">Table of n, a(n) for n = 1..100</a>
%F 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
%F a(n) = n!*A002627(n). - _R. J. Mathar_, Mar 18 2017
%F Sum_{n>=1} a(n) * x^n / (n!)^2 = (exp(x) - 1) / (1 - x). - _Ilya Gutkovskiy_, Jul 15 2021
%t Table[(n!)^2 Sum[1/k!,{k,n}],{n,20}] (* _Harvey P. Dale_, Dec 02 2021 *)
%o (PARI) { for (n=1, 100, write("b061573.txt", n, " ", n!^2*sum(k=1, n, 1/k!)) ) } \\ _Harry J. Smith_, Jul 24 2009
%Y Cf. A000522, A002627, A061572.
%K nonn
%O 1,2
%A _N. J. A. Sloane_, May 19 2001
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