%I M4008 N1661 #80 Jul 08 2020 20:38:06
%S 0,1,5,49,820,21076,773136,38402064,2483133696,202759531776,
%T 20407635072000,2482492033152000,359072203696128000,
%U 60912644957448192000,11977654199703478272000,2702572249389834608640000
%N Central factorial numbers: second right-hand column of triangle A008955.
%C Coefficient of x^2 in Product_{k=0..n}(x + k^2). - _Ralf Stephan_, Aug 22 2004
%C 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
%C 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
%C a(n) is the (n-1)-st elementary symmetric function of the squares of the first n numbers. - _Anton Zakharov_, Nov 06 2016
%C Primes p such that p^2 | a(p-1) are the Wolstenholme primes A088164. - _Amiram Eldar_ and _Thomas Ordowski_, Aug 08 2019
%D J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H Seiichi Manyama, <a href="/A001819/b001819.txt">Table of n, a(n) for n = 0..253</a> (terms 0..50 from T. D. Noe)
%H Takao Komatsu, <a href="https://arxiv.org/abs/2003.12926">Convolution identities of poly-Cauchy numbers with level 2</a>, arXiv:2003.12926 [math.NT], 2020.
%H Mircea Merca, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL15/Merca2/merca7.html">A Special Case of the Generalized Girard-Waring Formula</a>, J. Integer Sequences, Vol. 15 (2012), Article 12.5.7.
%H <a href="/index/Fa#factorial">Index entries for sequences related to factorial numbers</a>
%F a_n = (n!)^2 * Sum_{k=1..n} 1/k^2. - Joe Keane (jgk(AT)jgk.org)
%F a(n) ~ (1/3)*Pi^3*n*e^(-2*n)*n^(2*n). - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
%F Sum_{n>=0} a(n)*x^n/n!^2 = polylog(2, x)/(1-x). - _Vladeta Jovovic_, Jan 23 2003
%F a(n) = Sum_{i=1..n} 1/i^2 / Product_{i=1..n} 1/i^2. - _Alexander Adamchuk_, Jul 11 2006
%F 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
%F 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
%F a(n)/(n!)^2 -> zeta(2) = A013661 as n -> infinity, rewriting the Keane formula. - Najam Haq (njmalhq(AT)yahoo.com), Jan 13 2010
%F 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
%t Table[Sum[1/i^2,{i,1,n}]/Product[1/i^2,{i,1,n}],{n,1,40}] (* _Alexander Adamchuk_, Jul 11 2006 *)
%t Table[n!^2*HarmonicNumber[n, 2], {n, 0, 15}] (* _Jean-François Alcover_, May 09 2012, after Joe Keane *)
%o (PARI) a(n)=n!^2*sum(k=1,n,1/k^2) \\ _Charles R Greathouse IV_, Nov 06 2016
%Y Cf. A000254, A001044, A002455, A007406, A007407, A066989, A024167, A142995.
%Y Second right-hand column of triangle A008955.
%Y Equals row sums of A162990(n)/(n+1)^2 for n >= 1.
%K nonn,easy,nice
%O 0,3
%A _N. J. A. Sloane_
%E Minor edits by _Vaclav Kotesovec_, Jan 28 2015