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

1,9

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

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

Sequence in context: A346972 A326485 A023528 * A105698 A105699 A023526

Adjacent sequences: A236305 A236306 A236307 * A236309 A236310 A236311

Keyword

nonn

Author

Zhi-Wei Sun, Apr 21 2014

Status

approved