A206713 Number of solutions (n,k) of s(k)=s(n) (mod n), where 1<=kA000045(k+1) (Fibonacci numbers).

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, 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, 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

Offset: 2

Author

Clark Kimberling, Feb 11 2012

Keywords

• nonn

Examples

			s(11)=144 = A000045(12); the numbers s(j) for j<11 are 143,142,141,139,136,131,110,89,55,
of which three are multiples of 3, so that a(11)=3.
		

Programs

Mathematica

s[k_] := Fibonacci[k + 1];
f[n_, k_] := If[Mod[s[n] - s[k], 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}]  (* A206713 *)