A013646 -id:A013646 - cp's OEIS frontend

A338785 a(n) is the least number k such that continued fraction for sqrt(prime(k)) has period n.

1, 2, 13, 4, 6, 8, 21, 11, 30, 14, 18, 27, 44, 41, 29, 43, 37, 34, 68, 36, 42, 94, 147, 58, 88, 47, 186, 93, 142, 75, 110, 90, 112, 67, 178, 228, 82, 114, 100, 222, 187, 105, 191, 143, 204, 131, 180, 115, 172, 177, 197, 133, 263, 272, 353, 175, 231, 242, 322, 157

Offset: 1

Ilya Gutkovskiy, Nov 08 2020

nonn

			sqrt(prime(1))  = sqrt(2)  = 1 + 1/(2 + 1/(2 + ...)), period 1.
sqrt(prime(2))  = sqrt(3)  = 1 + 1/(1 + 1/(2 + 1/(1 + 1/(2 + ...)))), period 2.
sqrt(prime(13)) = sqrt(41) = 6 + 1/(2 + 1/(2 + 1/(12 + 1/(2 + 1/(2 + 1/(12 + ...)))))), period 3.

Robert Israel, Table of n, a(n) for n = 1..1000

Cf. A000720, A013646, A054269, A059800.

N:= 100: # for a(1)..a(N)
A:= Vector(N): count:= 0: p:= 1:
for n from 1 while count < N do
  p:= nextprime(p);
  v:= nops(numtheory:-cfrac(sqrt(p),periodic,quotients)[2]);
  if v <= N and A[v] = 0 then count:= count+1; A[v]:= n; fi
od:
convert(A,list); # Robert Israel, Nov 11 2020

Table[SelectFirst[Range[500], Length[Last[ContinuedFraction[Sqrt[Prime[#]]]]] == n &], {n, 60}]

a(n) = A000720(A059800(n)).

A351139 a(n) is the least k such that the continued fraction for sqrt(k) has periodic part [r, 1, 2, ..., n-1, n, n-1, ..., 1, 2r] for some positive integer r.

Original entry on oeis.org

3, 14, 216, 25185, 23287359, 1953082923, 81112983931776, 6667182474680388, 699567746120736710880, 855784807474766398870755, 51592564054278677032777194015, 1474855822717073602911008555048040, 23175672095781915301598668218548941215, 474577479777785868138090462593743556930231

Offset: 1

Giorgos Kalogeropoulos, Feb 02 2022

nonn

			a(3) = 216 because the continued fraction of sqrt(216) has periodic part [14; 1, 2, 3, 2, 1, 28] and this is the least number with this property.

Cf. A013646.

from itertools import count
from sympy.ntheory.continued_fraction import continued_fraction_reduce
def A351139(n):
    if n == 2:
        return 14
    for r in count(1):
        if (k := continued_fraction_reduce([r,list(range(1,n+1))+list(range(n-1,0,-1))+[2*r]])**2).is_integer:
            return k # Chai Wah Wu, Feb 09 2022

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

Original entry on oeis.org

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

Giorgos Kalogeropoulos, Jan 04 2022

nonn

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

			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}.

Cf. A003285. A subsequence of A013646.

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

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

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A338785 a(n) is the least number k such that continued fraction for sqrt(prime(k)) has period n.

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A351139 a(n) is the least k such that the continued fraction for sqrt(k) has periodic part [r, 1, 2, ..., n-1, n, n-1, ..., 1, 2r] for some positive integer r.

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

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