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A241476 Least positive primitive root g < prime(n) modulo prime(n) with g - 1 a square, or 0 if such a number g does not exist. 7
1, 2, 2, 5, 2, 2, 5, 2, 5, 2, 17, 2, 17, 5, 5, 2, 2, 2, 2, 65, 5, 37, 2, 26, 5, 2, 5, 2, 10, 5, 65, 2, 5, 2, 2, 82, 5, 2, 5, 2, 2, 2, 101, 5, 2, 170, 2, 5, 2, 10, 5, 26, 37, 26, 5, 5, 2, 26, 5, 26, 5, 2, 5, 17, 10, 2, 37, 10, 2, 2, 5, 26, 10, 2, 2, 5, 2, 5, 17, 26 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

According to the conjecture in A239957, a(n) should be always positive.

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Z.-W. Sun, New observations on primitive roots modulo primes, arXiv preprint arXiv:1405.0290 [math.NT], 2014.

EXAMPLE

a(4) = 5 since 2^2 + 1 = 5 is a primitive root modulo prime(4) = 7, but neither 0^1 + 1 = 1 nor 1^1 + 1 = 2 is a primitive root modulo prime(4) = 7.

MATHEMATICA

f[k_]:=k^2+1

dv[n_]:=Divisors[n]

Do[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}]; Print[n, " ", k^2+1]; Goto[bb]; Label[aa]; Continue, {k, 0, Sqrt[Prime[n]-2]}]; Print[n, " ", 0]; Label[bb]; Continue, {n, 1, 80}]

PROG

(PARI) ispr(n, p)=my(f=factor(p-1)[, 1], m=Mod(n, p)); for(i=1, #f, if(m^(p\f[i])==1, return(0))); m^(p-1)==1

a(n)=my(p=prime(n)); for(k=0, sqrtint(p-2), if(ispr(k^2+1, p), return(k^2+1))); 0 \\ Charles R Greathouse IV, May 01 2014

CROSSREFS

Cf. A000040, A002522, A239957, A239963, A241472.

Sequence in context: A021448 A229709 A242277 * A309727 A195719 A108053

Adjacent sequences:  A241473 A241474 A241475 * A241477 A241478 A241479

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Apr 23 2014

STATUS

approved

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Last modified October 19 13:01 EDT 2019. Contains 328222 sequences. (Running on oeis4.)