OFFSET
1,9
COMMENTS
Conjecture: a(n) > 0 for all n > 2. In other words, for any prime p > 3, there exists a prime q < p/2 such that the Catalan number C(q) = binomial(2q, q)/(q+1) is a primitive root modulo p.
We have verified this for all n = 3, ..., 2*10^5.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..1000
Z.-W. Sun, New observations on primitive roots modulo primes, arXiv preprint arXiv:1405.0290 [math.NT], 2014.
EXAMPLE
a(13) = 1 since C(7) = 429 is a primitive root modulo prime(13) = 41.
MATHEMATICA
f[k_]:=CatalanNumber[Prime[k]]
dv[n_]:=Divisors[n]
Do[m=0; Do[If[Mod[f[k], Prime[n]]==0, Goto[aa], 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, PrimePi[(Prime[n]-1)/2]}]; Print[n, " ", m]; Continue, {n, 1, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 21 2014
STATUS
approved