A236306 Least prime p < prime(n) for which both p and p! are primitive roots modulo prime(n), or 0 if such a prime does not exist.

0, 2, 2, 0, 2, 2, 3, 2, 5, 2, 11, 2, 17, 5, 5, 2, 2, 2, 2, 13, 5, 3, 2, 3, 5, 2, 11, 2, 67, 3, 3, 2, 3, 2, 2, 13, 53, 2, 5, 2, 2, 2, 47, 5, 2, 3, 2, 3, 2, 29, 3, 7, 137, 11, 3, 5, 2, 59, 31, 13, 17, 2, 5, 23, 47, 2, 101, 23, 2, 2, 13, 7, 43, 2, 2, 5, 2, 109, 3, 127

Offset

1,2

Comments

Conjecture: a(n) > 0 for all n > 4. In other words, for any prime p > 7, there exists a prime q < p such that both q and q! are primitive roots modulo p.

P. Erdős asked whether for any sufficiently large prime p there exists a prime q < p which is a primitive root modulo p (see page 377 of Guy's book in the reference).

References

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

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Example

a(7) = 3 since both 3 and 3! = 6 are primititive roots modulo prime(7) = 17, but 2 is not a primitive root modulo 17.

Mathematica

f[k_]:=Prime[k]!
dv[n_]:=Divisors[n]
Do[Do[Do[If[Mod[Prime[k]^(Part[dv[Prime[n]-1], i]), Prime[n]]==1||Mod[f[k]^(Part[dv[Prime[n]-1], i]), Prime[n]]==1, Goto[aa]], {i, 1, Length[dv[Prime[n]-1]]-1}]; Print[n, " ", Prime[k]]; Goto[bb]; Label[aa]; Continue, {k, 1, n-1}]; Print[n, " ", 0]; Label[bb]; Continue, {n, 1, 80}]

Crossrefs

Cf. A000040, A000142, A234972, A235709, A235712.

Sequence in context: A356583 A217943 A177225 * A153239 A229502 A356359

Adjacent sequences: A236303 A236304 A236305 * A236307 A236308 A236309

Keyword

nonn

Author

Zhi-Wei Sun, Apr 21 2014

Status

approved