A246567 - OEIS

%I #29 Dec 24 2021 08:19:14

%S -1,1,9,61,587,7575,117485,2057365,39314175,802816213,17275712297,

%T 387886408443,9020881956707,216101556811603,5309497149531957,

%U 133334756362738885,3412887111988377575,88838285028658754625,2347236720247792005665,62849602943515066525633

%N 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).

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

%C 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.

%C (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.

%C 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).

%C 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

%H Zhi-Wei Sun, <a href="/A246567/b246567.txt">Table of n, a(n) for n = 1..100</a>

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1408.5381">Two new kinds of numbers and related divisibility results</a>, arXiv:1408.5381 [math.NT], 2014-2018.

%H Zuo-Ru Zhang, <a href="https://arxiv.org/abs/2112.12427">Proof of two conjectures of Z.-W. Sun on combinatorial sequences</a>, arXiv:2112.12427 [math.CO], 2021.

%F 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.

%F a(n) ~ (17+12*sqrt(2))^n / (2^(17/4) * Pi^(3/2) * n^(9/2)). - _Vaclav Kotesovec_, Sep 07 2014

%e 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.

%p 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);

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

%t Table[a[n],{n,1,20}]

%o (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

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

%K sign

%O 1,3

%A _Zhi-Wei Sun_, Aug 30 2014