

A241492


a(n) = {0 < g < prime(n): g is a primitive root modulo prime(n) and g is a product of two consecutive integers}.


5



0, 1, 1, 0, 2, 2, 2, 1, 1, 1, 1, 2, 3, 3, 2, 3, 5, 3, 3, 2, 2, 2, 6, 3, 2, 5, 3, 4, 5, 5, 4, 7, 7, 7, 5, 4, 3, 5, 5, 8, 6, 2, 5, 4, 5, 3, 2, 5, 7, 6, 5, 4, 5, 8, 10, 8, 10, 4, 6, 6, 7, 8, 3, 4, 4, 9, 6, 4, 7, 8, 7, 5, 7, 7, 6, 9, 12, 6, 11, 8
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OFFSET

1,5


COMMENTS

Conjecture: a(n) > 0 for all n > 4. In other words, any prime p > 7 has a primitive root g < p of the form k*(k+1).
We have verified this for all n = 5, ..., 2*10^5.
See also A239957 and A239963 for similar conjectures. Clearly, for any prime p > 3, one of the three numbers 1*2, 2*3, 3*4 is a quadratic residue modulo p.


LINKS

ZhiWei 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(9) = 1 since 4*5 = 20 is a primitive root modulo prime(9) = 23.
a(10) = 1 since 1*2 = 2 is a primitive root modulo prime(10) = 29.
a(11) = 1 since 3*4 = 12 is a primitive root modulo prime(11) = 31.


MATHEMATICA

f[k_]:=f[k]=k(k+1)
dv[n_]:=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, 1, (Sqrt[4*Prime[n]3]1)/2}]; Print[n, " ", m]; Continue, {n, 1, 80}]


CROSSREFS

Cf. A000040, A002378, A239957, A239963, A241476.
Sequence in context: A316894 A081503 A176508 * A227739 A047971 A029432
Adjacent sequences: A241489 A241490 A241491 * A241493 A241494 A241495


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Apr 23 2014


STATUS

approved



