A236966 Number of primes p < prime(n)/2 such that 2^p - 1 is a primitive root modulo prime(n).

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

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

Zhi-Wei 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 [math.NT], 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: A070309 A287556 A353748 * A280048 A119910 A130784

Adjacent sequences: A236963 A236964 A236965 * A236967 A236968 A236969

Keyword

nonn

Author

Zhi-Wei Sun, Apr 22 2014

Status

approved