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A242250
Least positive integer g < prime(n) such that g, 2^g - 1 and (g-1)! are all primitive roots modulo prime(n), or 0 if such a number g does not exist.
4
1, 0, 3, 5, 8, 11, 5, 13, 21, 10, 12, 22, 24, 34, 13, 31, 18, 6, 41, 11, 14, 53, 8, 6, 26, 3, 12, 5, 47, 10, 45, 10, 5, 32, 12, 6, 24, 3, 15, 3, 6, 41, 19, 10, 8, 30, 3, 67, 5, 35, 20, 13, 99, 19, 7, 7, 3, 118, 5, 15, 22, 3, 73, 59, 91, 8, 137, 46, 20, 55
OFFSET
1,3
COMMENTS
According to the conjecture in A242248, a(n) should be positive for all n > 2.
LINKS
Zhi-Wei Sun, Notes on primitive roots modulo primes, arXiv:1405.0290 [math.NT], 2014.
EXAMPLE
a(5) = 8 since 8, 2^8 - 1 = 255 and (8-1)! = 5040 are all primitive roots modulo prime(5) = 11 with 255 == 5040 == 2 (mod 11), but none of 1, 2^2 - 1, 3, 4, 5, (6-1)! and (7-1)!
is a primitive root modulo 11.
MATHEMATICA
f[n_]:=f[n]=2^n-1
g[n_]:=g[n]=(n-1)!
rMod[m_, n_]:=rMod[m, n]=Mod[m, n, -n/2]
dv[n_]:=dv[n]=Divisors[n]
Do[Do[If[Mod[f[k], Prime[n]]==0, Goto[aa]]; Do[If[Mod[k^(Part[dv[Prime[n]-1], i])-1, Prime[n]]==0||Mod[rMod[f[k], Prime[n]]^(Part[dv[Prime[n]-1], i])-1, Prime[n]]==0||Mod[rMod[g[k], Prime[n]]^(Part[dv[Prime[n]-1], i])-1, Prime[n]]==0, Goto[aa]], {i, 1, Length[dv[Prime[n]-1]]-1}]; Print[n, " ", k]; Goto[bb]; Label[aa]; Continue, {k, 1, Prime[n]-1}]; Print[n, " ", 0]; Label[bb]; Continue, {n, 1, 70}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, May 09 2014
STATUS
approved