A346744 - OEIS

%I #21 Aug 07 2021 17:23:11

%S 0,1,2,3,2,3,2,5,4,7,4,7,2,3,2,9,2,7,6,7,8,3,2,15,6,9,10,11,4,17,4,17,

%T 4,9,8,15,2,5,10,15,2,13,4,11,6,3,2,19,8,15,2,11,2,19,12,19,8,11,4,19,

%U 4,5,14,33,6,13,6,17,2,25,2,31,2,9,12,11,6,29,4

%N The number of congruencies k^(n-k) + (n-k)^k == 0 (mod n) with 0 < k < n.

%C Of course for any n, k being equal to either 1 or n-1 would work.

%H Seiichi Manyama, <a href="/A346744/b346744.txt">Table of n, a(n) for n = 1..10000</a>

%p a:= n-> add(`if`(k&^(n-k)+(n-k)&^k mod n=0, 1, 0), k=1..n-1):

%p seq(a(n), n=1..100);  # _Alois P. Heinz_, Aug 06 2021

%t f[n_] := Block[{c = 0, k = 1}, While[k < n, If[ Mod[ PowerMod[k, n - k, n] + PowerMod[n - k, k, n], n] == 0, c++]; k++]; c]; Array[f@# &, 100]

%o (Python)

%o def a(n): return sum((k**(n-k) + (n-k)**k)%n == 0 for k in range(1, n))

%o print([a(n) for n in range(1, 101)]) # _Michael S. Branicky_, Jul 31 2021

%o (PARI) a(n) = sum(k=1, n-1, Mod(k, n)^(n-k) + Mod(n-k, n)^k == 0); \\ _Michel Marcus_, Aug 06 2021

%Y Cf. A114977, A114978, A114979, A114980, A114981.

%K easy,nonn

%O 1,3

%A _Robert G. Wilson v_, Jul 31 2021