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A236308
Number of primes q < prime(n)/2 such that the Catalan number C(q) is a primitive root modulo prime(n).
8
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
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}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 21 2014
STATUS
approved