%I #18 Nov 06 2021 10:05:17

%S 0,8,720,50400,3220000,196885920,11756961216,692835631488,

%T 40536961717824,2363784447147552,137716866109432896,

%U 8030173585594013568,469162781054378536320,27486632292027996114560,1615617140290621588826880,95302760085090826490672640

%N a(n) = Sum_{k=0..n-1} binomial(4k, 2k+1)*binomial(2k, k)*48^(n-1-k).

%C Conjecture: For any prime p > 3 and positive integer n, we have

%C (Sum_{k=0..p*n-1} binomial(4k, 2k+1)*binomial(2k, k)/48^k - (p/3)*Sum_{r=0..n-1} binomial(4r, 2r+1)*binomial(2r, r)/48^r)*48^n/((p*n)^2*binomial(4n, 2n)*binomial(2n, n)) == (5/3)*B_{p-2}(1/3) (mod p), where (p/3) is the Legendre symbol and B_{p-2}(x) is the Bernoulli polynomial of degree p-2.

%C This conjecture with n = 1 gives the congruence a(p) == (5/12)*p^2*B_{p-2}(1/3) (mod p^3) for any prime p > 3.

%H Zhi-Wei Sun, <a href="/A277860/b277860.txt">Table of n, a(n) for n = 1..100</a>

%H Zhi-Wei Sun, <a href="http://maths.nju.edu.cn/~zwsun/130e.pdf">Super congruences and Euler numbers</a>, Sci. China Math. 54(2011), no.12, 2509--2535.

%H Zhi-Wei Sun, <a href="http://maths.nju.edu.cn/~zwsun/165s.pdf">New series for some special values of L-functions</a>, Nanjing Univ. J. Math. Biquarterly 32(2015), no.2, 189-218.

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1610.03384">Supercongruences involving Lucas sequences</a>, arXiv:1610.03384 [math.NT], 2016.

%F a(n) ~ 2^(6*n - 9/2) / (Pi*n). - _Vaclav Kotesovec_, Nov 06 2021

%e a(1) = 0 since binomial(4*0, 2*0+1)*binomial(2*0, 0)*48^(1-1-0) = 0.

%e a(2) = 8 since Sum_{k=0..1} binomial(4k, 2k+1)*binomial(2k, k)*48^(2-1-k) = binomial(4, 2+1)*binomial(2, 1)*48^0 = 8.

%t a[n_]:=Sum[Binomial[4k,2k+1]Binomial[2k,k]48^(n-1-k),{k,0,n-1}]

%t Table[a[n],{n,1,16}]

%Y Cf. A000984, A001448, A245089, A277640.

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Nov 02 2016