A236308 Number of primes q < prime(n)/2 such that the Catalan number C(q) is a primitive root modulo prime(n).

0, 0, 1, 1, 1, 1, 1, 1, 4, 2, 1, 4, 1, 3, 3, 5, 5, 5, 2, 4, 5, 4, 10, 4, 7, 7, 4, 7, 4, 9, 5, 6, 10, 9, 7, 5, 5, 12, 12, 13, 12, 4, 10, 7, 13, 4, 7, 10, 18, 9, 14, 13, 9, 9, 15, 17, 16, 8, 9, 12, 10, 19, 13, 10, 14, 14, 13, 6, 18, 18, 14, 24, 13, 16, 9, 22, 20, 12, 23, 15

Offset: 1

Author

Zhi-Wei Sun, Apr 21 2014

Keywords

• nonn

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.

Examples

			a(13) = 1 since C(7) = 429 is a primitive root modulo prime(13) = 41.
		

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.

Crossrefs

Cf. A000040, A000108, A234972, A235709, A235712, A236306, A236966.

Programs

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}]