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A331506
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Least primitive root g < prime(n) of the n-th prime with g a product of two Fibonacci numbers, or 0 if such a number g does not exist.
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1
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1, 2, 2, 3, 2, 2, 3, 2, 5, 2, 3, 2, 6, 3, 5, 2, 2, 2, 2, 13, 5, 3, 2, 3, 5, 2, 5, 2, 6, 3, 3, 2, 3, 2, 2, 6, 5, 2, 5, 2, 2, 2, 21, 5, 2, 3, 2, 3, 2, 6, 3, 13, 13, 6, 3, 5, 2, 6, 5, 3, 3, 2, 5, 34, 10, 2, 3, 10, 2, 2, 3, 13, 6, 2, 2, 5, 2, 5, 3, 21, 2, 2, 13, 5, 15, 2, 3, 13, 2, 3, 2, 13, 3, 2, 10, 5, 2, 3, 2, 2
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OFFSET
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1,2
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COMMENTS
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Conjecture 1: a(n) > 0 for all n > 0. In other words, for each prime p there are two Fibonacci numbers F(k) and F(m) with F(k)*F(m) < p such that F(k)*F(m) is a primitive root modulo p.
This implies that for each odd prime p there exists a Fibonacci number F(k) < p which is a quadratic nonresidue modulo p.
It seems that Conjecture 1 can be strengthened as follows: For any prime p, there is a primitive root g < p modulo p such that g/F(2) = g or g/F(3) = g/2 or g/F(4) = g/3 is a Fibonacci number. We have verified this strong version for all primes p < 5*10^9.
We also have the following conjecture similar to Conjecture 1.
Conjecture 2. For any prime p, there are two Lucas numbers L(k) and L(m) with k >= m >= 0 and L(k)*L(m) < p such that L(k)*L(m) is a primitive root modulo p.
This has been verified for all primes p < 10^9.
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LINKS
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EXAMPLE
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a(85) = 15 with 15 = 3*5 = F(4)*F(5) a primitive root modulo prime(85) = 439.
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MATHEMATICA
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p[n_]:=p[n]=Prime[n];
Dv[n_]:=Dv[n]=Divisors[p[n]-1];
ls={};
Do[If[Fibonacci[k]Fibonacci[m]<p[100], ls=Append[ls, Fibonacci[k]Fibonacci[m]]], {k, 2, 14}, {m, 2, k}]
LL:=LL=Sort[DeleteDuplicates[ls]];
ff[r_]:=ff[r]=LL[[r]];
tab={}; Do[r=1; Label[aa]; If[ff[r]>=p[n], tab=Append[tab, 0]; Goto[bb]]; Do[If[PowerMod[ff[r], Dv[n][[i]], p[n]]==1, r=r+1; Goto[aa]], {i, 1, Length[Dv[n]]-1}]; tab=Append[tab, ff[r]]; Label[bb], {n, 1, 100}]; Print[tab]
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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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