

A241675


a(n) = {0 < k < n/2: k is a Fibonacci number with x^2 == k (mod n) for no integer x}.


1



0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 1, 3, 2, 2, 3, 3, 2, 4, 3, 3, 4, 2, 1, 4, 4, 3, 4, 4, 3, 5, 2, 5, 4, 2, 5, 4, 4, 4, 3, 5, 2, 5, 4, 5, 6, 2, 2, 6, 4, 5, 4, 5, 5, 5, 5, 5, 6, 4, 3, 5, 3, 3, 6, 6, 6, 5, 5, 3, 5, 6, 3, 7, 4, 4, 5, 6, 7, 5, 2, 7
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,8


COMMENTS

a(n) > 0 for all n > 4 if and only if the conjecture in A241568 holds.
In fact, if n > 0 is a multiple of 4, then x^2 == F(3) = 2 (mod n) for no integer x. If n > 4 is composite with an odd prime divisor p, then by the conjecture in A241568 there should exist a Fibonacci number k < p <= n/2 such that x^2 == k (mod p) for no integer x and hence x^2 == k (mod n) for no integer x.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
Z.W. Sun, New observations on primitive roots modulo primes, arXiv preprint arXiv:1405.0290, 2014


EXAMPLE

a(5) = 1 since x^2 == F(3) = 2 (mod 5) for no integer x, but 1^2 == F(1) = F(2) = 1 (mod 5), where F(n) denotes the nth Fibonacci number given by A000045.
a(7) = 1 since x^2 == F(4) = 3 (mod 7) for no integer x.
a(22) = 2 since there is no integer x such that x^2 == F(3) = 2 (mod 22) or x^2 == F(6) = 8 (mod 22).
a(23) = 1 since x^2 == F(5) = 5 (mod 23) for no integer x.


MATHEMATICA

f[k_]:=Fibonacci[k]
Do[m=0; Do[If[f[k]>=n/2, Goto[bb]]; Do[If[Mod[i^2, n]==f[k], Goto[aa]], {i, 0, n/2}]; m=m+1; Label[aa]; Continue, {k, 2, (n+1)/2}]; Label[bb]; Print[n, " ", m]; Continue, {n, 1, 80}]


CROSSREFS

Cf. A000045, A000290, A241568, A241604, A241605.
Sequence in context: A257079 A260372 A037180 * A091222 A316506 A294884
Adjacent sequences: A241672 A241673 A241674 * A241676 A241677 A241678


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Apr 27 2014


STATUS

approved



