Coprime to sum of permutation's terms

[hv at crypt.org]

zak seidov <zakseidov at yahoo.com> wrote:
:--- Antti Karttunen <antti.karttunen at gmail.com> wrote:
[...]
:> "Integers n, such that A117532(1..n) contains all
:> integers from 1 to n."
:> if it's not too short.
:
:Yes, the seq' VERY short:
:A2=(1,2)
:(from first 2,000 terms of A117532)

Well yes, 2 * sum(1..n) is divisible by n. :)

Hugo V.