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A305048
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Number of ordered pairs (k, m) of nonnegative integers such that 5^k + 10^m is not only a primitive root modulo prime(n) but also smaller than prime(n).
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2
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0, 1, 1, 0, 2, 3, 2, 2, 2, 4, 1, 3, 5, 1, 4, 3, 3, 4, 2, 2, 3, 2, 4, 4, 2, 5, 4, 4, 2, 2, 2, 4, 4, 6, 6, 4, 3, 3, 7, 6, 6, 2, 4, 3, 5, 3, 2, 3, 8, 3, 4, 4, 1, 3, 5, 5, 6, 5, 6, 4, 3, 5, 1, 1, 3, 4, 4, 2, 7, 2, 4, 4, 2, 8, 3, 7, 7, 3, 5, 4, 6, 1, 3, 4, 4, 7, 5, 4, 6, 2
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OFFSET
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1,5
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COMMENTS
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Conjecture: a(n) > 0 for all n > 4. In other words, any prime p > 7 has a primitive root g < p of the form 5^k + 10^m with k and m nonnegative integers.
We have verified this for any prime p > 7 not exceeding 10^9.
It seems that a(n) = 1 only for n = 2, 3, 11, 14, 53, 63, 64, 82, 99, 101, 111, 129, 344, 369, 391, 795, 1170, 1587, 5629, 5718, 6613, 430516.
See also A305048 for similar conjectures.
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LINKS
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Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28--Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.
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EXAMPLE
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a(14) = 1 with 5^2 + 10^0 = 26 a primitive root modulo prime(14) = 43.
a(101) = 1 with 5^0 + 10^0 = 2 a primitive root modulo prime(101) = 547.
a(111) = 1 with 5^2 + 10 = 35 a primitive root modulo prime(111) = 607.
a(5718) = 1 with 5^0 + 10^3 = 1001 a primitive root modulo prime(5718) = 56401.
a(6613) = 1 with 5^1 + 10^3 = 1005 a primitive root modulo prime(6613) = 66301.
a(430516) = 1 with 5^5 + 10^1 = 3135 a primitive root modulo prime(430516) = 6276271.
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MATHEMATICA
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p[n_]:=p[n]=Prime[n];
Dv[n_]:=Dv[n]=Divisors[n];
gp[g_, p_]:=gp[g, p]=Mod[g, p]>0&&Sum[Boole[PowerMod[g, Dv[p-1][[k]], p]==1], {k, 1, Length[Dv[p-1]]-1}]==0;
tab={}; Do[r=0; Do[If[gp[5^a+10^b, p[n]], r=r+1], {a, 0, Log[5, p[n]-1]}, {b, 0, Log[10, p[n]-5^a]}]; tab=Append[tab, r], {n, 1, 90}]; 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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