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A206826 Number of solutions (n,k) of s(k)=s(n) (mod n), where 1<=k and s(k)=k(k+1)(k+2)/6.

%I A206826 #5 Mar 30 2012 18:58:12
%S A206826 0,2,1,2,1,2,1,2,3,2,2,2,3,4,1,2,1,2,6,3,3,2,4,2,3,2,6,2,5,2,1,3,3,8,
%T A206826 3,2,3,4,6,2,3,2,6,3,3,2,2,2,3,5,6,2,1,8,6,5,3,2,8,2,3,5,1,8,5,2,6,4,
%U A206826 12,2,2,2,3,3,6,8,4,2,6,2,3,2,8,8,3,3,6,2,5,8,6,4,3,8,2,2,3,4,6
%N A206826 Number of solutions (n,k) of s(k)=s(n) (mod n), where 1<=k<n and s(k)=k(k+1)(k+2)/6.
%e A206826 5 divides exactly two of the numbers s(n)-s(k) for k=1,2,3,4, so that a(5)=2.
%t A206826 s[k_] := k (k + 1) (k + 2)/6;
%t A206826 f[n_, k_] := If[Mod[s[n] - s[k], n] == 0, 1, 0];
%t A206826 t[n_] := Flatten[Table[f[n, k], {k, 1, n - 1}]]
%t A206826 a[n_] := Count[Flatten[t[n]], 1]
%t A206826 Table[a[n], {n, 2, 120}]  (* A206826 *)
%Y A206826 Cf. A206590.
%K A206826 nonn
%O A206826 1,2
%A A206826 _Clark Kimberling_, Feb 12 2012