A246567 a(n) = (sum_{k=0}^{n-1}C(n-1,k)^2*C(-n-1,k)^2/(4*k^2-1))/n, where C(x,k) refers to binomial(x,k).

-1, 1, 9, 61, 587, 7575, 117485, 2057365, 39314175, 802816213, 17275712297, 387886408443, 9020881956707, 216101556811603, 5309497149531957, 133334756362738885, 3412887111988377575, 88838285028658754625, 2347236720247792005665, 62849602943515066525633

Offset: 1

Author

Zhi-Wei Sun, Aug 30 2014

Keywords

• sign

Comments

The following conjecture implies that a(n) is an integer.

Conjecture: (i) For any positive integers m and n, the sum sum_{k=0}^{n-1}C(n-1,k)^m*C(-n-1,k)^m/(4k^2-1) is always an integer divisible by n.

(ii) The sequence a(n+1)/a(n) (n > 2) is strictly increasing to the limit 17+12*sqrt(2), and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n > 1) is strictly decreasing to the limit 1.

For any positive integer n, we have sum_{k=0}^{n-1}C(n-1,k)*C(-n-1,k)/(4k^2-1) = -n, and n^3*a(n) = sum_{k=0}^{n-1}(2*k+1)*sum_{j=0..k}C(k,j)^2*C(k+j,j)^2/(2*j-1).

In the latest version of arXiv:1408.5381, the author proved part (i) of the conjecture, thus a(n) is indeed integral. - Zhi-Wei Sun, Sep 04 2014

Examples

			a(2) = 1 since 1/2*sum_{k=0,1}C(1,k)^2*C(-3,k)^2/(4*k^2-1) = 1/2*(-1+9/3) = 1.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..100
• Zhi-Wei Sun, Two new kinds of numbers and related divisibility results, arXiv:1408.5381 [math.NT], 2014-2018.
• Zuo-Ru Zhang, Proof of two conjectures of Z.-W. Sun on combinatorial sequences, arXiv:2112.12427 [math.CO], 2021.

Crossrefs

Cf. A243101, A246460, A246511, A246542, A246543.

Programs

Maple

A246567:=n->add((binomial(n-1,k)*binomial(-n-1,k))^2/(4*k^2-1), k=0..n-1)/n: seq(A246567(n), n=1..20);

Mathematica

a[n_]:=Sum[(Binomial[n-1,k]*Binomial[-n-1,k])^2/(4*k^2-1),{k,0,n-1}]/n
Table[a[n],{n,1,20}]

PARI

a(n) = sum(k=0, n-1, binomial(n-1,k)^2*binomial(n+k,k)^2/(4*k^2-1))/n; \\ Michel Marcus, Dec 24 2021

Formula

Recurrence (obtained via the Zeilberger algorithm): n^3*(n+1)*(2*n+5)*a(n) - (n+1)*(2*n+5)*(35*n^3+152*n^2+191*n+62)*a(n+1) + (n+2)*(2*n+1)*(35*n^3+163*n^2+224*n+88)*a(n+2) - (n+2)*(n+3)^3*(2*n+1)*a(n+3) = 0.

a(n) ~ (17+12*sqrt(2))^n / (2^(17/4) * Pi^(3/2) * n^(9/2)). - Vaclav Kotesovec, Sep 07 2014