

A236966


Number of primes p < prime(n)/2 such that 2^p  1 is a primitive root modulo prime(n).


13



0, 0, 1, 1, 1, 1, 3, 2, 1, 3, 2, 1, 2, 2, 5, 6, 3, 4, 3, 5, 4, 5, 7, 9, 3, 5, 2, 10, 7, 7, 7, 7, 9, 5, 10, 4, 5, 7, 12, 11, 14, 6, 7, 5, 10, 9, 8, 5, 12, 15, 14, 8, 12, 11, 16, 12, 16, 9, 12, 10, 10, 14, 15, 10, 12, 14, 9, 10, 21, 9, 22, 21, 11, 9, 18, 24, 20, 17, 17, 16
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OFFSET

1,7


COMMENTS

Conjecture: a(n) > 0 for all n > 2. In other words, for any prime p > 3, there is a prime q < p/2 with the Mersenne number 2^q  1 a primitive root modulo p.
We have verified this for all n = 3, ..., 530000.
See also the comment in A234972.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..1200
Z.W. Sun, New observations on primitive roots modulo primes, arXiv preprint arXiv:1405.0290, 2014


EXAMPLE

a(12) = 1 since 17 is a prime smaller than prime(12)/2 = 37/2 with 2^(17)  1 = 131071 a primitive root modulo prime(12) = 37.


MATHEMATICA

f[k_]:=2^(Prime[k])1
dv[n_]:=Divisors[n]
Do[m=0; Do[If[Mod[f[k], Prime[n]]==0, Goto[aa], Do[If[Mod[f[k]^(Part[dv[Prime[n]1], i]), Prime[n]]==1, Goto[aa]], {i, 1, Length[dv[Prime[n]1]]1}]]; m=m+1; Label[aa]; Continue, {k, 1, PrimePi[(Prime[n]1)/2]}]; Print[n, " ", m]; Continue, {n, 1, 80}]


CROSSREFS

Cf. A000040, A001348, A001918, A234972, A235709, A235712, A236306, A236308.
Sequence in context: A130827 A070309 A287556 * A280048 A119910 A130784
Adjacent sequences: A236963 A236964 A236965 * A236967 A236968 A236969


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Apr 22 2014


STATUS

approved



