A001819 Central factorial numbers: second right-hand column of triangle A008955. (Formerly M4008 N1661)

0, 1, 5, 49, 820, 21076, 773136, 38402064, 2483133696, 202759531776, 20407635072000, 2482492033152000, 359072203696128000, 60912644957448192000, 11977654199703478272000, 2702572249389834608640000

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.

Index entries for sequences related to factorial numbers

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

Cf. A000254, A001044, A002455, A007406, A007407, A066989, A024167, A142995.

Second right-hand column of triangle A008955.

Equals row sums of A162990(n)/(n+1)^2 for n >= 1.

Sequence in context: A221972 A002111 A305114 * A064618 A249588 A348901

Adjacent sequences: A001816 A001817 A001818 * A001820 A001821 A001822

Keyword

nonn,easy,nice

Author

N. J. A. Sloane

Extensions

Minor edits by Vaclav Kotesovec, Jan 28 2015

Status

approved