A361889 a(n) = S(5,2*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, 11, 415, 30955, 3173626, 386672861, 52846226091, 7857161332715, 1246162831674580, 207990691516965886, 36176886727828945286, 6510211391453319830461, 1205449991704260042021490, 228686327051301858363357905, 44299708036441260810228742915, 8738765548899621077157770551275

Offset: 1

Author

Peter Bala, Mar 29 2023

Keywords

• nonn
• easy

Comments

Odd bisection of A361888.

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.

Examples

			Examples of supercongruences:
a(13) - a(1) = 1205449991704260042021490 - 1 = 3*(13^3)*182893338143568508879 == 0 (mod 13^3).
a(2*5) - a(2) = 207990691516965886 - 11 = (5^3)*7*237703647447961 == 0 (mod 5^3)
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..420
• H. W. Gould, Problem E2384, Amer. Math. Monthly, 81 (1974), 170-171.

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) ), A361890 ( S(7,n) ), A361891 ( S(7,n)/S(1,n) ), A361892 ( S(7,2*n-1)/S(1,2*n-1) ).

Programs

Maple

seq(add( ( binomial(2*n-1,k) - binomial(2*n-1,k-1) )^5/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])^5 / Binomial[2*n-1, n-1], {k, 0, n-1}], {n, 1, 20}] (* Vaclav Kotesovec, Mar 24 2025 *)

Python

from math import comb
def A361889(n): return sum((comb((n<<1)-1,j)*(m:=n-j<<1)//(m+j))**5 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) )^5 for n >= 1.

a(n) ~ 2^(8*n + 1) / (125 * Pi^2 * n^4). - Vaclav Kotesovec, Mar 24 2025