cp's OEIS Frontend

A206713 Number of solutions (n,k) of s(k)=s(n) (mod n), where 1<=k and s(k)=A000045(k+1) (Fibonacci numbers).

%I A206713 #6 Mar 30 2012 18:58:12
%S A206713 0,0,1,1,1,0,1,1,0,3,2,1,1,0,1,1,1,3,0,1,3,0,2,1,1,1,2,6,0,3,5,0,2,0,
%T A206713 4,1,4,1,2,4,3,0,1,2,2,2,5,2,0,1,2,1,1,2,2,0,5,3,2,4,4,2,2,1,4,0,3,1,
%U A206713 1,3,8,1,1,0,8,1,2,3,2,3,3,0,6,1,1,3,6,8,1,0,3,3,6,0,5,1,0,1,1
%N A206713 Number of solutions (n,k) of s(k)=s(n) (mod n), where 1<=k<n and s(k)=A000045(k+1) (Fibonacci numbers).
%e A206713 s(11)=144 = A000045(12); the numbers s(j) for j<11 are 143,142,141,139,136,131,110,89,55,
%e A206713 of which three are multiples of 3, so that a(11)=3.
%t A206713 s[k_] := Fibonacci[k + 1];
%t A206713 f[n_, k_] := If[Mod[s[n] - s[k], n] == 0, 1, 0];
%t A206713 t[n_] := Flatten[Table[f[n, k], {k, 1, n - 1}]]
%t A206713 a[n_] := Count[Flatten[t[n]], 1]
%t A206713 Table[a[n], {n, 2, 120}]  (* A206713 *)
%K A206713 nonn
%O A206713 2,10
%A A206713 _Clark Kimberling_, Feb 11 2012