A361887 a(n) = S(5,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

1, 1, 2, 33, 276, 4150, 65300, 1083425, 20965000, 399876876, 8461219032, 178642861782, 4010820554664, 90684123972156, 2130950905378152, 50560833176021025, 1231721051614138800, 30294218438009039800, 759645100717216142000, 19213764100954274616908, 493269287121905287769776

Offset

0,3

Comments

For r a positive integer define S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r. The present sequence is {S(5,n)}. 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).

a(n) is the total number of 5-tuples of semi-Dyck paths from (0,0) to (n,n-2*j) for j=0..floor(n/2). - Alois P. Heinz, Apr 02 2023

Links

Table of n, a(n) for n=0..20.

H. W. Gould, Problem E2384, Amer. Math. Monthly, 81 (1974), 170-171.

Formula

a(n) = Sum_{k = 0..floor(n/2)} ( (n - 2*k + 1)/(n - k + 1) * binomial(n,k) )^5.

From Alois P. Heinz, Apr 02 2023: (Start)

a(n) = Sum_{j=0..floor(n/2)} A008315(n,j)^5.

a(n) = Sum_{j=0..n} A120730(n,j)^5.

a(n) = A357824(n,5). (End)

a(n) ~ 2^(5*n + 19/2) / (125 * Pi^(5/2) * n^(9/2)). - Vaclav Kotesovec, Aug 27 2023

Maple

seq(add( ( binomial(n, k) - binomial(n, k-1) )^5, k = 0..floor(n/2)), n = 0..20);

Mathematica

Table[Sum[(Binomial[n, k] - Binomial[n, k-1])^5, {k, 0, Floor[n/2]}], {n, 0, 20}] (* Vaclav Kotesovec, Aug 27 2023 *)

Crossrefs

Cf. A003161 ( S(3,n) ), A003162 ( S(3,n)/S(1,n) ), A183069 ( S(3,2*n+1)/ S(1,2*n+1) ), 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) ), A361892 ( S(7,2*n-1)/S(1,2*n-1) ).

Column k=5 of A357824.

Cf. A008315, A120730.

Sequence in context: A181547 A030448 A093992 * A294272 A231595 A100023

Adjacent sequences: A361884 A361885 A361886 * A361888 A361889 A361890

Keyword

nonn,easy

Author

Peter Bala, Mar 28 2023

Status

approved