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A242748 Number of ordered ways to write n = k + m with 0 < k <= m such that k is a primitive root modulo prime(k) and m is a primitive root modulo prime(m). 11
0, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 3, 3, 3, 1, 3, 3, 2, 2, 3, 3, 2, 1, 2, 3, 3, 2, 3, 3, 3, 3, 1, 2, 3, 3, 3, 2, 1, 4, 2, 3, 3, 3, 3, 2, 5, 3, 4, 2, 4, 6, 6, 1, 5, 4, 6, 7, 4, 6, 4, 6, 3, 6, 3, 7, 5, 5, 6, 7, 4, 6, 8, 5, 6, 4, 6, 4, 8, 3, 7 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,4
COMMENTS
Conjecture: a(n) > 0 for all n > 1.
This implies that there are infinitely many positive integers k which is a primitive root modulo prime(k).
LINKS
EXAMPLE
a(6) = 1 since 6 = 3 + 3 with 3 a primitive root modulo prime(3) = 5.
a(7) = 1 since 7 = 1 + 6 with 1 a primitive root modulo prime(1) = 2 and 6 a primitive root modulo prime(6) = 13.
a(15) = 1 since 15 = 2 + 13 with 2 a primitive root modulo prime(2) = 3 and 13 a primitive root modulo prime(13) = 41.
a(38) = 1 since 38 = 10 + 28 with 10 a primitive root modulo prime(10) = 29 and 28 a primitive root modulo prime(28) = 107.
a(53) = 1 since 53 = 3 + 50 with 3 a primitive root modulo prime(3) = 5 and 50 a primitive root modulo prime(50) = 229.
MATHEMATICA
dv[n_]:=Divisors[n]
Do[m=0; Do[Do[If[Mod[k^(Part[dv[Prime[k]-1], i]), Prime[k]]==1, Goto[aa]], {i, 1, Length[dv[Prime[k]-1]]-1}]; Do[If[Mod[(n-k)^(Part[dv[Prime[n-k]-1], j]), Prime[n-k]]==1, Goto[aa]], {j, 1, Length[dv[Prime[n-k]-1]]-1}]; m=m+1; Label[aa]; Continue, {k, 1, n/2}]; Print[n, " ", m]; Continue, {n, 1, 80}]
CROSSREFS
Sequence in context: A170979 A025864 A070242 * A266012 A202111 A187759
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, May 21 2014
STATUS
approved

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Last modified March 28 22:04 EDT 2024. Contains 371254 sequences. (Running on oeis4.)