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

1,1

Comments

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

Links

Table of n, a(n) for n=1..47.

Formula

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

Example

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}.

Mathematica

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]]

Crossrefs

Cf. A003285. A subsequence of A013646.

Sequence in context: A110468 A327356 A295612 * A286661 A171058 A105906

Adjacent sequences: A350542 A350543 A350544 * A350546 A350547 A350548

Keyword

nonn

Author

Giorgos Kalogeropoulos, Jan 04 2022

Status

approved