A277860 a(n) = Sum_{k=0..n-1} binomial(4k, 2k+1)*binomial(2k, k)*48^(n-1-k).

0, 8, 720, 50400, 3220000, 196885920, 11756961216, 692835631488, 40536961717824, 2363784447147552, 137716866109432896, 8030173585594013568, 469162781054378536320, 27486632292027996114560, 1615617140290621588826880, 95302760085090826490672640

Offset

1,2

Comments

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

(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.

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.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..100

Zhi-Wei Sun, Super congruences and Euler numbers, Sci. China Math. 54(2011), no.12, 2509--2535.

Zhi-Wei Sun, New series for some special values of L-functions, Nanjing Univ. J. Math. Biquarterly 32(2015), no.2, 189-218.

Zhi-Wei Sun, Supercongruences involving Lucas sequences, arXiv:1610.03384 [math.NT], 2016.

Formula

a(n) ~ 2^(6*n - 9/2) / (Pi*n). - Vaclav Kotesovec, Nov 06 2021

Example

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

Mathematica

a[n_]:=Sum[Binomial[4k, 2k+1]Binomial[2k, k]48^(n-1-k), {k, 0, n-1}]
Table[a[n], {n, 1, 16}]

Crossrefs

Cf. A000984, A001448, A245089, A277640.

Sequence in context: A037076 A116245 A361222 * A221198 A071308 A046188

Adjacent sequences: A277857 A277858 A277859 * A277861 A277862 A277863

Keyword

nonn

Author

Zhi-Wei Sun, Nov 02 2016

Status

approved