A350545 - OEIS

%I #14 Feb 07 2022 22:02:57

%S 3,41,13,58,61,193,157,337,586,821,601,421,1117,1153,1069,1669,2137,

%T 2053,1381,3733,3541,1621,4657,2389,4561,6577,3061,4261,5209,6121,

%U 6781,8317,7621,6661,6301,7561,7549,15817,9241,9349,12853,8269,11701,16729,14449,23017,31573

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

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

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

%e 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}.

%t n=30;prm={};fin={};k=2;While[Length@prm<n||Sort[PrimePi@prm][[;;n]]!=Range@n,If[!IntegerQ@Sqrt@k,l=Length@Last@ContinuedFraction@Sqrt@k,k++];If[PrimeQ@l,If[FreeQ[prm,l],AppendTo[fin,{k,PrimePi@l}];AppendTo[prm,l]]];k++];First/@SortBy[fin,Last][[;;n]]

%Y Cf. A003285. A subsequence of A013646.

%K nonn

%O 1,1

%A _Giorgos Kalogeropoulos_, Jan 04 2022