A206590 Number of solutions (n,k) of k^3=n^3 (mod n), where 1<=k

0, 0, 1, 0, 0, 0, 3, 2, 0, 0, 1, 0, 0, 0, 3, 0, 2, 0, 1, 0, 0, 0, 3, 4, 0, 8, 1, 0, 0, 0, 7, 0, 0, 0, 5, 0, 0, 0, 3, 0, 0, 0, 1, 2, 0, 0, 3, 6, 4, 0, 1, 0, 8, 0, 3, 0, 0, 0, 1, 0, 0, 2, 15, 0, 0, 0, 1, 0, 0, 0, 11, 0, 0, 4, 1, 0, 0, 0, 3, 8, 0, 0, 1, 0, 0, 0, 3, 0, 2, 0, 1, 0, 0, 0, 7, 0, 6, 2

Offset: 2

Author

Clark Kimberling, Feb 09 2012

Keywords

• nonn

Examples

			8 divides exactly 3 of the numbers 8^3-k^3 for k = 1, 2 , ..., 7, so that a(8) = 3.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 2..16384

Crossrefs

Cf. A206825.

Programs

Mathematica

f[n_, k_] := If[Mod[n^3 - k^3, n] == 0, 1, 0];
t[n_] := Flatten[Table[f[n, k], {k, 1, n - 1}]]
a[n_] := Count[Flatten[t[n]], 1]
Table[a[n], {n, 2, 120}] (* A206590 *)

PARI

A206590(n) = { my(n3 = n^3); sum(k=1,n-1,!((n3-(k^3))%n)); }; \\ Antti Karttunen, Nov 17 2017