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A350545 a(n) is the least k such that the continued fraction for sqrt(k) has period prime(n). 0
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
Conjecture: All terms in this sequence (except a(4) = 58 and a(9) = 586) are primes.
LINKS
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
KEYWORD
nonn
AUTHOR
_Giorgos Kalogeropoulos_, Jan 04 2022
STATUS
approved

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Last modified March 3 03:04 EST 2024. Contains 370499 sequences. (Running on oeis4.)