A350545 a(n) is the least k such that the continued fraction for sqrt(k) has period prime(n).

3, 41, 13, 58, 61, 193, 157, 337, 586, 821, 601, 421, 1117, 1153, 1069, 1669, 2137, 2053, 1381, 3733, 3541, 1621, 4657, 2389, 4561, 6577, 3061, 4261, 5209, 6121, 6781, 8317, 7621, 6661, 6301, 7561, 7549, 15817, 9241, 9349, 12853, 8269, 11701, 16729, 14449, 23017, 31573

Offset: 1

Author

Giorgos Kalogeropoulos, Jan 04 2022

Keywords

• nonn

Comments

Conjecture: All terms in this sequence (except a(4) = 58 and a(9) = 586) are primes.

Examples

			a(5) = 61 because 61 is the least integer k whose period of the continued fraction for sqrt(k) is prime(5)=11, namely {1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14}.
		

Crossrefs

Cf. A003285. A subsequence of A013646.

Programs

Mathematica

n=30;prm={};fin={};k=2;While[Length@prm
				

Formula

a(n) = A013646(prime(n)).