

A241568


a(n) = {0 < k < prime(n)/2: k is not only a quadratic nonresidue modulo prime(n) but also a Fibonacci number}.


5



0, 0, 1, 1, 1, 2, 2, 3, 1, 3, 2, 4, 2, 4, 2, 5, 3, 3, 5, 3, 4, 2, 5, 2, 4, 4, 3, 4, 3, 5, 3, 2, 5, 4, 7, 2, 6, 5, 4, 4, 5, 4, 3, 4, 7, 4, 4, 4, 5, 6, 4, 3, 5, 4, 3, 3, 3, 3, 3, 5, 6, 7, 8, 2, 5, 7, 6, 3, 5, 7, 5, 3, 4, 4, 6, 3, 6, 7, 4, 3
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OFFSET

1,6


COMMENTS

Conjecture: (i) a(n) > 0 for all n > 2. In other words, for any prime p > 3, there is a Fibonacci number among 1, ..., (p1)/2 which is a quadratic nonresidue modulo p.
(ii) For any n > 2, there is a prime q < prime(n) such that the qth Fibonacci number is a quadratic nonresidue modulo prime(n).
(iii) For any odd prime p, there is a Lucas number (i.e., a term of A000032) smaller than p which is a quadratic nonresidue modulo p.
We have checked part (i) for all primes p < 3*10^9, part (ii) for n up to 10^8, and part (iii) for the first 10^7 primes.
See also A241604 for a sequence related to part (i) of the conjecture.


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(3) = 1 since F(3) = 2 is a quadratic nonresidue modulo prime(3) = 5, where F(n) denotes the nth Fibonacci number.
a(4) = 1 since F(4) = 3 is a quadratic nonresidue modulo prime(4) = 7.
a(5) = 1 since F(3) = 2 is a quadratic nonresidue modulo prime(5) = 11.
a(9) = 1 since F(5) = 5 is a quadratic nonresidue modulo prime(9) = 23.


MATHEMATICA

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


CROSSREFS

Cf. A000040, A000032, A000045, A236966, A239957, A239963, A241492, A241504, A241516, A241604.
Sequence in context: A304092 A171691 A088904 * A047972 A004595 A071681
Adjacent sequences: A241565 A241566 A241567 * A241569 A241570 A241571


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Apr 25 2014


STATUS

approved



