OFFSET
0,3
COMMENTS
Coefficient of x^2 in Product_{k=0..n}(x + k^2). - Ralf Stephan, Aug 22 2004
p divides a(p-1) for prime p > 3. p divides a((p-1)/2) for prime p > 3. For prime p, p^2 divides a(n) for n > 2*p+1. - Alexander Adamchuk, Jul 11 2006; last comment corrected by Michel Marcus, May 20 2020
The ratio a(n)/A001044(n) is the partial sum of the reciprocals of squares. E.g., a(4)/A001044(4) = 820/576 = 1/1 + 1/4 + 1/9 + 1/16. - Pierre CAMI, Oct 30 2006
a(n) is the (n-1)-st elementary symmetric function of the squares of the first n numbers. - Anton Zakharov, Nov 06 2016
Primes p such that p^2 | a(p-1) are the Wolstenholme primes A088164. - Amiram Eldar and Thomas Ordowski, Aug 08 2019
REFERENCES
J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..253 (terms 0..50 from T. D. Noe)
Takao Komatsu, Convolution identities of poly-Cauchy numbers with level 2, arXiv:2003.12926 [math.NT], 2020.
Mircea Merca, A Special Case of the Generalized Girard-Waring Formula, J. Integer Sequences, Vol. 15 (2012), Article 12.5.7.
FORMULA
a_n = (n!)^2 * Sum_{k=1..n} 1/k^2. - Joe Keane (jgk(AT)jgk.org)
a(n) ~ (1/3)*Pi^3*n*e^(-2*n)*n^(2*n). - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
Sum_{n>=0} a(n)*x^n/n!^2 = polylog(2, x)/(1-x). - Vladeta Jovovic, Jan 23 2003
a(n) = Sum_{i=1..n} 1/i^2 / Product_{i=1..n} 1/i^2. - Alexander Adamchuk, Jul 11 2006
a(0) = 0, a(n) = a(n-1)*n^2 + A001044(n-1). E.g., a(1) = 0*1 + 1 = 1 since A001044(0) = 1; a(2) = 1*2^2 + 1 = 5 since A001044(1) = 1; a(3) = 5*3^2 + 4 = 49 since A001044(2) = 4; and so on. - Pierre CAMI, Oct 30 2006
Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n^2 + 2*n + 1)*a(n) - n^4*a(n-1). The sequence b(n) = n!^2 satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 1. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(1 - 1^4/(5 - 2^4/(13 - 3^4/(25 - ... -(n-1)^4/((2*n^2 - 2*n + 1)))))), leading to the infinite continued fraction expansion zeta(2) = 1/(1-1^4/(5 - 2^4/(13 - 3^4/(25 - ... - n^4/((2*n^2 + 2*n + 1) - ...))))). Compare with A142995. Compare also with A024167 and A066989. - Peter Bala, Jul 18 2008
a(n)/(n!)^2 -> zeta(2) = A013661 as n -> infinity, rewriting the Keane formula. - Najam Haq (njmalhq(AT)yahoo.com), Jan 13 2010
a(n) = s(n+1,2)^2 - 2*s(n+1,1)*s(n+1,3), where s(n,k) are Stirling numbers of the first kind, A048994. - Mircea Merca, Apr 03 2012
MATHEMATICA
Table[Sum[1/i^2, {i, 1, n}]/Product[1/i^2, {i, 1, n}], {n, 1, 40}] (* Alexander Adamchuk, Jul 11 2006 *)
Table[n!^2*HarmonicNumber[n, 2], {n, 0, 15}] (* Jean-François Alcover, May 09 2012, after Joe Keane *)
PROG
(PARI) a(n)=n!^2*sum(k=1, n, 1/k^2) \\ Charles R Greathouse IV, Nov 06 2016
CROSSREFS
KEYWORD
nonn,easy,nice
AUTHOR
EXTENSIONS
Minor edits by Vaclav Kotesovec, Jan 28 2015
STATUS
approved