login
a(n) = (sum_{k=0}^{n-1}(3k^2+3k+1)*C(n-1,k)^2*C(n+k,k)^2)/n^3, where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).
2

%I #29 Dec 24 2021 08:12:14

%S 1,8,87,1334,25045,529080,12076435,291307490,7325385345,190294925864,

%T 5074233846583,138240914882394,3834434331534781,107990908896551192,

%U 3081524055740420811,88938694296657330170,2592715751635344852505,76252823735941187830920,2260342454730542009915455,67476975730679069406101870

%N a(n) = (sum_{k=0}^{n-1}(3k^2+3k+1)*C(n-1,k)^2*C(n+k,k)^2)/n^3, where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

%C In the latest version of arXiv:1408:5381, the author proved that a(n) is always an integer. Notice that a(65) is relatively prime to 65. - _Zhi-Wei Sun_, Sep 14 2014

%C Conjecture: The sequence a(n+1)/a(n) (n = 1,2,3,...) is strictly increasing to the limit 17+12*sqrt(2), and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n = 1,2,3,...) is strictly decreasing to the limit 1.

%C Note that sum_{k=0}^{n-1}(2k+1)*A(k) = n^5*a(n) for all n > 0, where A(n) = sum_{k=0..n}C(n,k)^2*C(n+k,k)^2*(6k^3+9k^2+5k+1) for n = 0,1,2,....

%H Zhi-Wei Sun, <a href="/A246512/b246512.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):

%F n^3*(n + 1)*(2n + 5)*(3n^2 + 12n + 11)*(6n^2 + 24n + 25)*a(n) - (n + 1)*(2n + 5)*(630n^7 + 6552n^6 + 28137n^5 + 64134n^4 + 82777n^3 + 59512n^2 + 21646n + 3076)*a(n+1) + (n + 2)*(2n + 1)*(630n^7 + 6678n^6 + 29271n^5 + 68751n^4 + 93469n^3 + 73445n^2 + 30640n + 5072)*a(n+2) - (n + 2)*(n + 3)^3*(2n + 1)*(3n^2 + 6n + 2)*(6n^2 + 12n + 7)*a(n+3) = 0.

%e a(2) = 8 since sum_{k=0,1} (3k^2+3k+1)C(1,k)^2*C(2+k,k)^2 = 1 + 7*3^2 = 64 = 2^3*8.

%t a[n_]:=Sum[(3k^2+3k+1)*(Binomial[n-1,k]Binomial[n+k,k])^2,{k,0,n-1}]/(n^3)

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

%o (PARI) a(n) = sum(k=0, n-1, (3*k^2+3*k+1)*binomial(n-1,k)^2*binomial(n+k,k)^2) /n^3; \\ _Michel Marcus_, Dec 24 2021

%Y Cf. A020639, A246065, A246138, A246459, A246460, A246461, A246462, A246511.

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Aug 28 2014