A361891 a(n) = S(7,n)/S(1,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

1, 1, 1, 43, 386, 9451, 246961, 6031627, 212559508, 6571985126, 243940325734, 9140730357409, 352312505157354, 14801600281919487, 600054439936968241, 26927918031565051915, 1149140935414286560040, 53804800109969394477580, 2401141625752684697505820

Offset: 0

Author

Peter Bala, Mar 30 2023

Keywords

• nonn
• easy

Comments

For r a positive integer define S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r. Gould (1974) conjectured that S(3,n) was always divisible by S(1,n). See A183069 for {S(3,n)/S(1,n)}. In fact, calculation suggests that if r is odd then S(r,n) is always divisible by S(1,n). The present sequence is {S(7,n)/S(1,n)}.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..563
• 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) ), 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) ), A361892 ( S(7,2*n-1)/S(1,2*n-1) ).

Programs

Maple

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

Mathematica

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

PARI

s(r, n) = sum(k=0, n\2, (binomial(n, k)-binomial(n, k-1))^r);
a(n) = s(7, n)/s(1, n); \\ Seiichi Manyama, Mar 24 2025

Python

from math import comb
def A361891(n): return sum((comb(n,j)*(m:=n-(j<<1)+1)//(m+j))**7 for j in range((n>>1)+1))//comb(n,n>>1) # Chai Wah Wu, Mar 25 2025

Formula

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

a(n) ~ 3 * 2^(6*n+13) / (2401 * Pi^3 * n^6). - Vaclav Kotesovec, Mar 24 2025