A059800 -id:A059800 - cp's OEIS frontend

A288184 Least odd number k such that the continued fraction for sqrt(k) has period n.

5, 3, 41, 7, 13, 19, 73, 31, 113, 43, 61, 103, 193, 179, 109, 133, 157, 139, 337, 151, 181, 253, 853, 271, 457, 211, 949, 487, 821, 379, 601, 463, 613, 331, 1061, 1177, 421, 619, 541, 589, 1117, 571, 1153, 823, 1249, 739, 1069, 631, 1021, 1051, 1201, 751

Offset: 1

Ilya Gutkovskiy, Jun 06 2017

nonn

			a(2) = 3, sqrt(3) = 1 + 1/(1 + 1/(2 + 1/(1 + 1/(2 + 1/...)))), period 2: [1, 2].

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Periodic Continued Fraction

Cf. A000035, A003285, A010337-A010339, A013642-A013644, A013646, A013943, A020347-A020439, A059800, A062769, A097853, A288185.

from sympy import continued_fraction_periodic
def A288184(n):
    d = 1
    while True:
        s = continued_fraction_periodic(0,1,d)[-1]
        if isinstance(s, list) and len(s) == n:
            return d
        d += 2 # Chai Wah Wu, Jun 07 2017

A003285(a(n)) = n, A000035(a(n)) = 1.

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

Original entry on oeis.org

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)).

cp's OEIS Frontend

A288184 Least odd number k such that the continued fraction for sqrt(k) has period n.

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