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

Author

Ilya Gutkovskiy, Jun 06 2017

Keywords

• nonn

Examples

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

Links

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

Crossrefs

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

Programs

Python

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

Formula

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