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A242248
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a(n) = |{0 < g < prime(n): g, 2^g - 1 and (g-1)! are all primitive roots modulo prime(n)}|.
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4
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1, 0, 1, 1, 1, 1, 3, 1, 1, 2, 1, 2, 2, 1, 6, 2, 10, 3, 2, 3, 5, 2, 10, 12, 3, 6, 7, 15, 3, 9, 3, 8, 18, 5, 18, 3, 7, 7, 24, 20, 26, 4, 13, 10, 15, 5, 4, 3, 35, 5, 19, 19, 3, 19, 36, 37, 38, 5, 10, 15, 16, 34, 7, 16, 6, 36, 4, 4, 44, 14
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OFFSET
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1,7
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COMMENTS
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Conjecture: a(n) > 0 for all n > 2. In other words, for any prime p > 3, there exists a positive primitive root g < p modulo p such that 2^g - 1 and (g-1)! are also primitive roots modulo p.
We have verified this for all primes p with 3 < p < 10^6.
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LINKS
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EXAMPLE
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a(4) = 1 since 5, 2^5 - 1 = 31 and (5-1)! = 24 are all primitive roots modulo prime(4) = 7.
a(6) = 1 since 11, 2^(11) - 1 = 2047 and (11-1)! = 3628800 are all primitive roots modulo prime(6) = 13. Note that both 2047 and 3628800 are congruent to 6 modulo 13.
a(14) = 1 since 34, 2^(34) - 1 and (34-1)! are all primititive roots modulo prime(14) = 43. Note that 2^(34) - 1 == 20 (mod 43) and 33! == -14 (mod 43).
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MATHEMATICA
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f[n_]:=2^n-1
g[n_]:=(n-1)!
rMod[m_, n_]:=Mod[m, n, -n/2]
dv[n_]:=Divisors[n]
Do[m=0; Do[If[Mod[f[k], Prime[n]]==0, Goto[aa]]; Do[If[Mod[k^(Part[dv[Prime[n]-1], i]), Prime[n]]==1||Mod[rMod[f[k], Prime[n]]^(Part[dv[Prime[n]-1], i]), Prime[n]]==1||Mod[rMod[g[k], Prime[n]]^(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, Prime[n]-1}]; Print[n, " ", m]; Continue, {n, 1, 70}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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