A361892 a(n) = S(7,2*n-1)/S(1,2*n-1), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

1, 43, 9451, 6031627, 6571985126, 9140730357409, 14801600281919487, 26927918031565051915, 53804800109969394477580, 116002825041515533807200418, 266118189111094898593879923346, 642598035707739308769581970619393

Offset: 1

Author

Peter Bala, Mar 30 2023

Keywords

• nonn
• easy

Comments

Odd bisection of A361891.

Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for positive integers n and r and all primes p >= 5.

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..282

Crossrefs

Cf. A003161 ( S(3,n) ), A003162 ( S(3,n)/S(1,n) ), A382394 ( S(3,2*n-1) ), A183069 ( S(3,2*n-1)/ S(1,2*n-1) ), A361887 ( S(5,n) ), A361888 ( S(5,n)/S(1,n) ), A361889 ( S(5,2*n-1)/S(1,2*n-1) ), A361890 ( S(7,n) ), A361891 ( S(7,n)/S(1,n) ).

Programs

Maple

seq(add( ( binomial(2*n-1,k) - binomial(2*n-1,k-1) )^7/binomial(2*n-1,n-1), k = 0..n-1), n = 1..20);

Mathematica

Table[Sum[(Binomial[2*n-1, k]-Binomial[2*n-1, k-1])^7 / Binomial[2*n-1, n-1], {k, 0, n-1}], {n, 1, 20}] (* Vaclav Kotesovec, Mar 24 2025 *)

Python

from math import comb
def A361892(n): return sum((comb((n<<1)-1,j)*(m:=n-j<<1)//(m+j))**7 for j in range(n))//comb((n<<1)-1,n-1) # Chai Wah Wu, Mar 25 2025

Formula

a(n) = 1/binomial(2*n-1,n-1) * Sum_{k = 0..n-1} ( (2*n - 2*k)/(2*n - k) * binomial(2*n-1,k) )^7 for n >= 1.

a(n) ~ 3 * 2^(12*n+1) / (2401 * Pi^3 * n^6). - Vaclav Kotesovec, Mar 24 2025

Extensions

Offset changed to 1 by Georg Fischer, Nov 20 2024