

A229910


a(n) = {0 < g < prime(n): both g and g + g^{1} are primitive roots modulo prime(n)}, where g^{1} is the inverse of g modulo prime(n).


1



0, 0, 0, 0, 2, 0, 4, 2, 4, 4, 2, 4, 8, 6, 10, 8, 14, 4, 4, 12, 8, 6, 20, 24, 16, 16, 12, 26, 8, 16, 14, 12, 24, 14, 32, 10, 20, 18, 40, 48, 44, 4, 30, 16, 32, 18, 14, 18, 56, 8, 60, 40, 12, 40, 64, 64, 72, 20, 40, 32, 36, 80, 22, 44, 24, 72, 22, 36, 86, 32, 84, 88, 40, 24, 28, 94, 104, 28, 76, 28
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OFFSET

1,5


COMMENTS

Conjecture: a(n) > 0 for all n > 6. In other words, for any prime p > 13, there is a primitive root g modulo p such that g + g^{1} is also a primitive root modulo p, where g^{1} is the inverse of g modulo p.
Note that a(n) is even for any n > 1. In fact, if g is a primitive root modulo a prime p > 3, then the inverse of g mod p is different from g since g^2 cannot be congruent to 1 modulo p.
Conjecture: Let F be a finite field with F = q > 13. Then there is a primitive root g of F (i.e., a generator of the cyclic group F\{0}) such that g + g^{1} is also a primitive root of F. If q > 61, then there exists a primitive root g of F such that g  g^{1} is also a primitive root of F.
The author has proved this for any finite field F with F > 2^{66}.


LINKS

M. F. Hasler, Table of n, a(n) for n = 1..1000


EXAMPLE

a(5) = 2 since 2 and 6 are primitive roots modulo prime(5) = 11 with 2*6 == 1 (mod 11) and 2 + 6 = 8 also a primitive root modulo 11.
This example recalls that there is no symmetry g > g (in Z/pZ) (nor a symmetry w.r.t. odd/even g), therefore one cannot (unfortunately) compute a(n) by taking twice the count of the g<prime(n)/2 which satisfy the condition. E.g., for p=19=prime(8), only g = 14 (= 5) and g = 15 (= 4) are in the set.  M. F. Hasler, Oct 06 2013


MATHEMATICA

gp[g_, p_]:=gp[g, p]=Mod[g, p]>0&&Length[Union[Table[Mod[g^k, p], {k, 1, p1}]]]==p1
a[n_]:=Sum[If[gp[g, Prime[n]]&&gp[g+PowerMod[g, 1, Prime[n]], Prime[n]], 1, 0], {g, 1, Prime[n]1}]
Table[a[n], {n, 1, 80}]


PROG

(PARI) A229910(n)=my(p=prime(n)); sum(g=2, p2, znorder(Mod(g, p))==p1 & Mod(g, p)^1+g & znorder(Mod(g, p)^1+g)==p1) \\ M. F. Hasler, Oct 06 2013
(PARI) A229910(n)={my(p=prime(n), u=0, s=0, i); n=p1; for(g=2, p2, bittest(u, g)&next; znorder(Mod(g, p))<n&next; u+=1<<lift(i=Mod(g, p)^1); i+gnext; znorder(i+g)<ns++); s*2} \\ about 20% faster.  M. F. Hasler, Oct 06 2013
(Perl) use ntheory ":all"; sub a229910 { my $p=nth_prime(shift); scalar(grep {is_primitive_root($_, $p) && is_primitive_root($_+invmod($_, $p), $p)} 2..$p2); } # Dana Jacobsen, Sep 19 2016


CROSSREFS

Cf. A001918, A229899.
Cf. A010554.
Sequence in context: A094572 A323905 A079534 * A097042 A332001 A196606
Adjacent sequences: A229907 A229908 A229909 * A229911 A229912 A229913


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Oct 03 2013


EXTENSIONS

Values a(1..400) double checked and extended to n=1000 by M. F. Hasler, Oct 06 2013


STATUS

approved



