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

%I A206590 #11 Nov 17 2017 20:34:27
%S A206590 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,
%T A206590 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,
%U A206590 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
%N A206590 Number of solutions (n,k) of k^3=n^3 (mod n), where 1<=k<n.
%H A206590 Antti Karttunen, <a href="/A206590/b206590.txt">Table of n, a(n) for n = 2..16384</a>
%e A206590 8 divides exactly 3 of the numbers 8^3-k^3 for k = 1, 2 , ..., 7, so that a(8) = 3.
%t A206590 f[n_, k_] := If[Mod[n^3 - k^3, n] == 0, 1, 0];
%t A206590 t[n_] := Flatten[Table[f[n, k], {k, 1, n - 1}]]
%t A206590 a[n_] := Count[Flatten[t[n]], 1]
%t A206590 Table[a[n], {n, 2, 120}] (* A206590 *)
%o A206590 (PARI) A206590(n) = { my(n3 = n^3); sum(k=1,n-1,!((n3-(k^3))%n)); }; \\ _Antti Karttunen_, Nov 17 2017
%Y A206590 Cf. A206825.
%K A206590 nonn
%O A206590 2,7
%A A206590 _Clark Kimberling_, Feb 09 2012