

A242345


Number of primes p < prime(n) with p and 2^p  p both primitive roots modulo prime(n).


5



0, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 2, 2, 1, 4, 4, 7, 1, 2, 1, 1, 1, 6, 4, 1, 4, 2, 6, 3, 7, 1, 3, 7, 4, 6, 1, 5, 6, 9, 12, 7, 5, 6, 4, 11, 2, 3, 6, 12, 6, 18, 13, 3, 14, 13, 14, 15, 4, 9, 6, 3, 13, 8, 12, 5, 12, 6, 6, 20, 8, 14, 19, 8, 5, 5, 22, 20, 6, 18, 6
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OFFSET

1,7


COMMENTS

Conjecture: a(n) > 0 for all n > 1. In other words, for any odd prime p, there is a prime q < p such that both q and 2^q  q are primitive roots modulo p.
According to page 377 in Guy's book, ErdÅ‘s asked whether for any sufficiently large prime p there exists a prime q < p which is a primitive root modulo p.


REFERENCES

R. K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, New York, 2004.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..3500
ZhiWei Sun, Notes on primitive roots modulo primes, arXiv:1405.0290 [math.NT], 2014.


EXAMPLE

a(4) = 1 since 3 is a prime smaller than prime(4) = 7, and both 3 and 2^3  3 = 5 are primitive roots modulo 7.
a(10) = 1 since 2 is a prime smaller than prime(10) = 29, and 2 and 2^2  2 are primitive roots modulo 29.
a(36) = 1 since 71 is a prime smaller than prime(36) = 151, and both 71 and 2^(71)  71 ( == 14 (mod 151)) are primitive roots modulo 151.


MATHEMATICA

f[k_]:=2^(Prime[k])Prime[k]
dv[n_]:=Divisors[n]
Do[m=0; Do[If[Mod[f[k], Prime[n]]==0, Goto[aa], Do[If[Mod[(Prime[k])^(Part[dv[Prime[n]1], i]), Prime[n]]==1Mod[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, n1}];
Print[n, " ", m]; Continue, {n, 1, 80}]


CROSSREFS

Cf. A000040, A000325, A234972, A236966, A242248, A242250, A242292.
Sequence in context: A023591 A165661 A107711 * A179067 A061893 A078530
Adjacent sequences: A242342 A242343 A242344 * A242346 A242347 A242348


KEYWORD

nonn


AUTHOR

ZhiWei Sun, May 11 2014


STATUS

approved



