A215508 Smallest m such that the period of the continued fraction of sqrt(m) is A215485(n); records of A013646.

1, 2, 3, 41, 58, 106, 193, 337, 586, 949, 1061, 1117, 1153, 1249, 1669, 2381, 3733, 5857, 6577, 6781, 8389, 11173, 14293, 15817, 17137, 17209, 23017, 37921, 38377, 46261, 47293, 56929, 82561, 90121, 113173, 122401, 148957, 151057, 161149, 163729, 193873, 206209, 225769, 322513, 497473, 576529, 676129, 686893, 706621, 862921, 946489, 992281, 1032649, 1198081, 1597033, 1655677, 1779409, 1930021, 2299489, 2367481, 2584081, 3209281, 3528409, 3933073, 4068241, 4160521, 4283689, 4726009, 4833901

Offset: 0

Author

Patrick McKinley, Aug 13 2012

Keywords

• nonn

Comments

The continued fractions of these numbers have the "hard to get" lengths listed in sequence A215485. They fill the last gaps in the table when computing A013646.

Examples

			The lengths of the continued fractions of sqrt(1), sqrt(2), sqrt(3) and sqrt(41) are 0, 1, 2 and 3 respectively. The rest of the sequence follows A215485 similarly.
		

Links

• Patrick McKinley, Table of n, a(n) for n = 0..253

Crossrefs

Cf. A013646, A215485.

Formula

a(n) = A013646(A215485(n)). - Pontus von Brömssen, Nov 24 2024