OFFSET
1,2
LINKS
Harry J. Smith, Table of n, a(n) for n = 1..100
FORMULA
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
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, May 19 2001
STATUS
approved