A121668 Products of consecutive Apery numbers, cf. A006221.

5, 365, 105485, 47686445, 27027984005, 17576522979125, 12539718106476125, 9563891779602510125, 7671490770912738387125, 6401115462988077760992365, 5513180441777884868230908125, 4873728705609344219627834043125

Offset: 1

Author

Angelo B. Mingarelli (amingare(AT)math.carleton.ca), Sep 10 2006

Keywords

• easy
• nonn

Comments

The solutions x_{n-1}:=A_nA_{n-1}, y_n of the four-term recurrence relation defined by x_0=5, x_1= 365, x_2= 105485 and y_0= 0, y_1=8424, y_2= 2438709 are such that y_n/x_n -> 16*zeta(3)^2. Generalizations to products of three or more Apery numbers are to be found in the cited paper.

Examples

			16*y_9/x_9 = 23.11905277493814774261896124285261449340 while
16*zeta(3)^2=23.11905277493814774261896126091180523494.
		

Links

• Angelo B. Mingarelli, Recurrence relations and the algebraic irrationality of zeta(3), arXiv:math.NT/0608577 v1. 23 August, 2006.

Crossrefs

Cf. A005259, A006221.

Programs

Mathematica

Table[Sum[(Binomial[n,k]*Binomial[n + k, k])^2, {k, 0, n}] * Sum[(Binomial[n-1, k] Binomial[n - 1 + k, k])^2, {k, 0, n-1}], {n, 1, 20}] (* Vaclav Kotesovec, Jul 11 2021 *)

Formula

Recurrence:

(n + 3)^3(n + 2)^6(2n + 1)(17n^2 + 17n + 5)z(n + 2) - (2n + 1)(17n^2 + \

17n + 5)(1155n^6 + 13860n^5 + 68535n^4 + 178680n^3 + 259059n^2 + \

198156n + 62531)(n + 2)^3z(n + 1) + (2n + 5)(17n^2 + 85n + 107)(1155n^6 \

+ 6930n^5 + 16560n^4 + 20040n^3 + 12954n^2 + 4308n + 584)(n + 1) ^3z(n) \

- n^3(n + 1)^6(2n + 5)(17n^2 + 85n + 107)z(n - 1) = 0

a(n) ~ (1 + sqrt(2))^(8*n) / (2^(9/2) * Pi^3 * n^3). - Vaclav Kotesovec, Jul 11 2021