A361888 a(n) = S(5,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, 11, 46, 415, 3265, 30955, 299500, 3173626, 33576266, 386672861, 4340714886, 52846226091, 620906440961, 7857161332715, 95704821415240, 1246162831674580, 15624127945644100, 207990691516965886, 2669841775757784796, 36176886727828945286, 473508685502539872586

Offset: 0

Author

Peter Bala, Mar 29 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(5,n)/S(1,n)}.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..840
• 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) ), A361889 ( S(5,2*n-1)/S(1,2*n-1) ), 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(n,k) - binomial(n,k-1) )^5/binomial(n,floor(n/2)), k = 0..floor(n/2)), n = 0..20);

Mathematica

Table[Sum[(Binomial[n, k]-Binomial[n, k-1])^5/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(5, n)/s(1, n); \\ Seiichi Manyama, Mar 24 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) )^5.

a(n) ~ 2^(4*n + 9) / (125 * Pi^2 * n^4). - Vaclav Kotesovec, Mar 24 2025