

A239957


a(n) = {0 < g < prime(n): g is a primitive root modulo prime(n) of the form k^2 + 1}.


11



1, 1, 1, 1, 1, 1, 2, 2, 3, 3, 1, 3, 2, 2, 3, 4, 4, 4, 2, 1, 2, 1, 3, 3, 6, 3, 3, 7, 4, 5, 2, 8, 3, 5, 6, 1, 2, 5, 8, 10, 7, 3, 2, 6, 8, 2, 3, 5, 8, 4, 7, 4, 2, 5, 8, 9, 10, 5, 8, 6, 10, 6, 4, 9, 6, 9, 5, 3, 13, 5, 8, 9, 5, 6, 8, 13, 13, 6, 6, 5
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OFFSET

1,7


COMMENTS

Conjecture: (i) a(n) > 0 for all n > 0. In other words, every prime p has a primitive root 0 < g < p of the form k^2 + 1, where k is an integer.
(ii) If p > 3 is a prime not equal to 13, then p has a primitive root 0 < g < p which is of the form k^2  1, where k is a positive integer.
See also A239963 for a similar conjecture.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000


EXAMPLE

a(6) = 1 since 1^2 + 1 = 2 is a primitive root modulo prime(6) = 13.
a(11) = 1 since 4^2 + 1 = 17 is a primitive root modulo prime(11) = 31.
a(20) = 1 since 8^2 + 1 = 65 is a primitive root modulo prime(20) = 71.
a(22) = 1 since 6^2 + 1 = 37 is a primitive root modulo prime(22) = 79.
a(36) = 1 since 9^2 + 1 = 82 is a primitive root modulo prime(36) = 151.


MATHEMATICA

f[k_]:=k^2+1
dv[n_]:=Divisors[n]
Do[m=0; Do[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, 0, Sqrt[Prime[n]2]}]; Print[n, " ", m]; Continue, {n, 1, 80}]


PROG

(PARI) ispr(n, p)=my(f=factor(p1)[, 1], m=Mod(n, p)); for(i=1, #f, if(m^(p\f[i])==1, return(0))); m^(p1)==1
a(n)=my(p=prime(n)); sum(k=0, sqrtint(p2), ispr(k^2+1, p)) \\ Charles R Greathouse IV, May 01 2014


CROSSREFS

Cf. A000040, A002522, A236308, A236966, A237112, A237121, A237594, A239963.
Sequence in context: A071434 A227314 A128924 * A230040 A242361 A116464
Adjacent sequences: A239954 A239955 A239956 * A239958 A239959 A239960


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Apr 23 2014


STATUS

approved



