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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).
2
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
2,10
EXAMPLE
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.
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 *)
CROSSREFS
Sequence in context: A113081 A172184 A109865 * A096874 A090046 A202551
KEYWORD
nonn
AUTHOR
Clark Kimberling, Feb 11 2012
STATUS
approved