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A061572 a(n) = (n!)^2 * Sum_{k=1..n} 1/(k^2*(k-1)!). 3

%I

%S 1,5,47,758,18974,683184,33476736,2142516144,173543847984,

%T 17354385161280,2099880608143680,302382807612606720,

%U 51102694487009537280,10016128119460096327680,2253628826878608852019200,576928979680925173791283200,166732475127787396148470732800

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

%H Harry J. Smith, <a href="/A061572/b061572.txt">Table of n, a(n) for n = 1..100</a>

%F Recurrence: a(1) = 1, a(2) = 5, a(n) = (n^2+n-1)*a(n-1) - (n-1)^3*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/(1-1^3/(5-2^3/(11-...-(n-1)^3/(n^2+n-1)))). Lim n -> infinity a(n)/b(n) = Ei(1) - gamma = 1/(1-1^3/(5-2^3/(11-...-(n-1)^3/(n^2+n-1)-...))). Cf. A061573. - _Peter Bala_, Jul 10 2008

%o (PARI) { for (n=1, 100, write("b061572.txt", n, " ", n!^2*sum(k=1, n, 1/(k^2*(k-1)!))) ) } \\ _Harry J. Smith_, Jul 24 2009

%Y Cf. A061573.

%K nonn

%O 1,2

%A _N. J. A. Sloane_, May 19 2001

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Last modified May 9 20:59 EDT 2021. Contains 343746 sequences. (Running on oeis4.)