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Number of solutions (n,k) of s(k)=s(n) (mod n), where 1<=k<n and s(k)=2^(k-1).
2

%I #5 Mar 30 2012 18:58:12

%S 0,1,1,1,2,2,4,1,2,1,4,1,4,3,11,2,2,1,4,3,2,2,10,1,2,1,8,1,7,6,26,3,4,

%T 2,5,1,2,3,9,2,6,3,4,3,4,2,21,2,2,6,4,1,2,2,17,3,2,1,14,1,12,10,57,5,

%U 6,1,8,3,5,2,11,8,2,3,4,2,6,2,18,1,4,1,13,10,6,3,8,8,7,7,8,9,4

%N Number of solutions (n,k) of s(k)=s(n) (mod n), where 1<=k<n and s(k)=2^(k-1).

%e 2^6=64; the numbers 64-s(j) are 63,62,60,56,48,32, of which two are multiples of 6, so that a(6)=2.

%t s[k_] := 2^(k - 1);

%t f[n_, k_] := If[Mod[s[n] - s[k], n] == 0, 1, 0];

%t t[n_] := Flatten[Table[f[n, k], {k, 1, n - 1}]]

%t a[n_] := Count[Flatten[t[n]], 1]

%t Table[a[n], {n, 2, 120}] (* A206714 *)

%K nonn

%O 2,5

%A _Clark Kimberling_, Feb 11 2012