A031749 Numbers k such that the least term in the periodic part of the continued fraction for sqrt(k) is 71.

5043, 20168, 45375, 80664, 126035, 181488, 247023, 322640, 408339, 504120, 609983, 725928, 851955, 988064, 1134255, 1290528, 1456883, 1633320, 1819839, 2016440, 2223123, 2439888, 2666735, 2903664, 3150675, 3407768, 3674943, 3952200, 4239539

Offset: 1

Author

David W. Wilson

Keywords

• nonn

Comments

The continued fraction expansion of sqrt((j*m)^2+t*m) for m >= 1 where t divides 2*j has the form [j*m, 2*j/t, 2*j*m, 2*j/t, 2*j*m, ...]. Thus numbers of the form (71*m)^2 + 2*m for m >= 1 are in the sequence. Are there any others? - Chai Wah Wu, Jun 18 2016

The term 25776072 is not of the form (71*m)^2 + 2*m. - Chai Wah Wu, Jun 19 2016

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A031424.

Programs

Mathematica

lt71Q[n_]:=Module[{s=Sqrt[n]},If[IntegerQ[s],0,Min[ContinuedFraction[s] [[2]]]] == 71]; Select[Range[43*10^5],lt71Q] (* Harvey P. Dale, Apr 11 2017 *)

Python

from sympy import continued_fraction_periodic
A031749_list = [n for n, d in ((n, continued_fraction_periodic(0,1,n)[-1]) for n in range(1,10**5)) if isinstance(d, list) and min(d) == 71] # Chai Wah Wu, Jun 09 2017