A340281 a(n) is the smallest prime p such that the number of distinct values of the ratio (number of nonnegative m < p such that m^k == m (mod p))/(number of nonnegative m < p such that -m^k == m (mod p)) is equal to n for some nonnegative k.

2, 3, 7, 19, 31, 163, 127, 1459, 211, 883, 811, 472393, 631, 8503057, 32077, 4051, 2311, 86093443, 4951, 6347497291777, 10531, 36451, 1299079, 251048476873, 8191, 388963, 5314411, 22051, 51031, 596046447753906250001, 28351, 411782264189299, 24571, 5904901

Offset: 1

Author

Juri-Stepan Gerasimov, Jan 02 2021

Keywords

• nonn

Comments

a(n) is the least prime p such that there are n distinct terms in the p-th row of A334006.

Conjecture: a(n) is the smallest prime p such that the number of distinct values of the ratio T(p, k) = (number of nonnegative m < p such that m^k == m (mod p))/(number of nonnegative m < p such that -m^k == m (mod p)) is equal to n for some 0 <= k <= floor((p + 2)/3).

Proof: for k > 1, iff t is a k-th power residue mod p, the number of nonnegative m < p such that m^k == t (mod p) is gcd(k, p - 1). Thus, the ratio T(p, 1+x) = T(p, 1+gcd(x, p-1)) and T(p, 2*t) = T(p, (p+1)/2) = 1. For odd prime p and 0 <= k < p - 1, notice that if k is an odd number of the form 1 + gcd(x, p-1) and x != (p - 1)/2, then k <= floor((p + 2)/3). - Jinyuan Wang, Jan 23 2021

For n >= 2, a(n) is the least prime p such that p - 1 has n - 1 odd divisors. - Jinyuan Wang, Jan 23 2021

Examples

			A334006 triangle begins:
   1 | 1;
   2 | 1, 1;   : 1 distinct value
   3 | 1, 3, 1;   : 2 distinct values
   4 | 1, 2, 1, 3;
   5 | 1, 5, 1, 1, 1;   : 2 distinct values
   6 | 1, 3, 1, 3, 1, 3;
   7 | 1, 7, 1, 3, 1, 3, 1;   : 3 distinct values
		

Links

• Seth A. Troisi, Table of n, a(n) for n = 1..200

Crossrefs

Cf. A001227, A019434, A058500, A074781, A334006.

Programs

PARI

T(n, k) = sum(m=0, n-1, Mod(m, n)^k == m)/sum(m=0, n-1, -Mod(m, n)^k == m); \\ A334006
a(n) = my(p=2); while (#Set(vector(p, k, T(p,k))) != n, p = nextprime(p+1)); p; \\ Michel Marcus, Jan 21 2021

PARI

lista(nn, show=50) = my(c, v=vector(show)); v[1]=2; forprime(p=3, nn, c=1+numdiv(p\2^valuation(p-1, 2)); if(c<=show && !v[c], v[c]=p)); v; \\ Jinyuan Wang, Jan 23 2021

Extensions

More terms from Jinyuan Wang, Jan 23 2021

Typo in a(34) corrected by Seth A. Troisi, May 22 2022