OFFSET
1,2
COMMENTS
Sum of generalized harmonic numbers squared multiplied by (n!)^2. agenh(n) = Sum_{k=1..n} HarmonicNumber(k, 2), where HarmonicNumber(n, j) = Sum_{k = 1..n} 1/k^j. - Alexander Adamchuk, Oct 27 2004
LINKS
Harry J. Smith, Table of n, a(n) for n = 1..100
Eric Weisstein's World of Mathematics, Harmonic Number
FORMULA
From Alexander Adamchuk, Oct 27 2004: (Start)
a(n) = (n!)^2 * Sum_{k=0..n-1} (k+1)/(n-k)^2.
a(n) = (n!)^2 * Sum_{k=1..n} HarmonicNumber(k, 2), where HarmonicNumber(k, 2) = A007406(k) / A007407(k). (End)
Sum_{n>=1} a(n) * x^n / (n!)^2 = polylog(2,x) / (1 - x)^2. - Ilya Gutkovskiy, Jul 15 2020
EXAMPLE
a(3) = 6^2 *(1 + (1 + 1/2^2) + (1 + 1/2^2 + 1/3^2)) = 130.
MAPLE
A060944:= n-> (n!)^2*add((1+j)/(n-j)^2, j=0..n-1); seq(A060944(n), n=1..15); # G. C. Greubel, Apr 09 2021
MATHEMATICA
Table[(n!)^2*Sum[(k+1)/(n-k)^2, {k, 0, n-1}], {n, 1, 10}]
PROG
(PARI) a(n)={n!^2 * sum(k=1, n, sum(j=1, k, 1/j^2))} \\ Harry J. Smith, Jul 15 2009
(Magma) [(Factorial(n))^2*(&+[(1+j)/(n-j)^2: j in [0..n-1]]): n in [1..15]]; // G. C. Greubel, Apr 09 2021
(Sage) [(factorial(n))^2*sum((1+j)/(n-j)^2 for j in (0..n-1)) for n in (1..15)] # G. C. Greubel, Apr 09 2021
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Leroy Quet, May 07 2001
STATUS
approved