

A242266


a(n) = {0 < g < prime(n): g is a primitive root mod prime(n) with g = sum_{j=1..k} prime(j) for some k > 0}.


2



0, 1, 1, 1, 1, 1, 2, 2, 3, 2, 1, 3, 2, 2, 3, 3, 2, 3, 3, 1, 3, 2, 3, 3, 5, 2, 2, 6, 2, 4, 1, 3, 2, 3, 5, 2, 2, 2, 6, 6, 6, 7, 2, 6, 4, 4, 4, 5, 6, 5, 6, 3, 1, 3, 7, 9, 9, 2, 5, 2, 2, 6, 4, 5, 6, 6, 4, 3, 8, 3, 6, 6, 7, 5, 6, 9, 8, 6, 4, 4
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OFFSET

1,7


COMMENTS

Conjecture: (i) a(n) > 0 for all n > 1. In other words, for any odd prime p, there is a positive integer k such that the sum of the first k primes is not only a primitive root modulo p but also smaller than p.
(ii) For any n > 1, there is a number k among 1, ..., n such that sum_{j=1..k}(1)^(kj)*prime(j) is a primitive root modulo prime(n).
We have verified parts (i) and (ii) for n up to 700000 and 250000 respectively. Note that prime(700000) > 10^7.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
ZhiWei Sun, Notes on primitive roots modulo primes, arXiv:1405.0290 [math.NT], 2014.


EXAMPLE

a(4) = 1 since prime(1) + prime(2) = 2 + 3 = 5 is a primitive root modulo prime(4) = 7 with 5 < 7.
a(7) = 2 since prime(1) = 2 and prime(1) + prime(2) + prime(3) = 2 + 3 + 5 = 10 are not only primitive roots modulo prime(7) = 17 but also smaller than 17.
a(53) = 1 since sum_{j=1..10} prime(j) = 129 is a primitive root modulo prime(53) = 241 with 129 < 241.


MATHEMATICA

f[0]=0
f[n_]:=Prime[n]+f[n1]
dv[n_]:=Divisors[n]
Do[m=0; Do[If[f[k]>=Prime[n], Goto[cc]]; Do[If[Mod[f[k]^(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, n}]; Label[cc]; Print[n, " ", m]; Continue, {n, 1, 80}]


CROSSREFS

Cf. A000040, A008347, A007504, A236966, A239957, A241476, A241504, A241516, A242248, A242250, A242277.
Sequence in context: A346010 A116199 A162915 * A239617 A304737 A092565
Adjacent sequences: A242263 A242264 A242265 * A242267 A242268 A242269


KEYWORD

nonn


AUTHOR

ZhiWei Sun, May 09 2014


STATUS

approved



