A345633 Sum of terms of odd index in the binomial decomposition of n^(n-1).

0, 1, 4, 36, 272, 4400, 51012, 1188544, 18640960, 567108864, 11225320100, 421504185344, 10079828372880, 450353989316608, 12627774819845668, 654244800082329600, 21046391759976988928, 1240529732459024678912, 45032132922921758270916, 2975557672677668838178816

Offset

1,3

Comments

When writing n^(n-1) (A000169) as a sum of powers of n using the binomial theorem, one can separately sum the even and the odd powers of n. This is the odd part. See the Formula section.

Links

Table of n, a(n) for n=1..20.

Formula

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

a(n+1) = ((1 + n)^n - (1 - n)^n)/2.

a(n) ~ (1 + (-1)^n/e^2) * n^(n-1) / 2. - Amiram Eldar, Dec 29 2025

Mathematica

Table[Plus @@ Table[(n - 1)^(2 k + 1) Binomial[n - 1, 2 k + 1], {k, 0, Floor[(n - 1)/2]}], {n, 1, 21}]
a[n_] := (n^(n-1) - (2-n)^(n-1))/2; Array[a, 20] (* Amiram Eldar, Dec 29 2025 *)

Crossrefs

Cf. A345632 (even part).

Cf. A062024, A302583.

Cf. A000169, A007778, A092364, A081131.

Sequence in context: A043024 A144889 A176097 * A173429 A172134 A098916

Adjacent sequences: A345630 A345631 A345632 * A345634 A345635 A345636

Keyword

nonn,easy

Author

Olivier Gérard, Jun 21 2021

Status

approved