A308720 - OEIS

The OEIS is supported by the many generous donors to the OEIS Foundation.

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

Other ways to Give

A308720

The maximum value in the continued fraction of sqrt(n), or 0 if there is no fractional part.

0, 0, 2, 2, 0, 4, 4, 4, 4, 0, 6, 6, 6, 6, 6, 6, 0, 8, 8, 8, 8, 8, 8, 8, 8, 0, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 0, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 0, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 0, 16, 16, 16, 16, 16, 16, 16

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,3

COMMENTS

The continued fraction expansion of sqrt(n) is periodic, and the maximal element is the last element in the period, 2*floor(sqrt(n)).

LINKS

Table of n, a(n) for n=0..71.

Oskar Perron, Die Lehre von den Kettenbrüchen, B. G. Teubner (1913), section 24, p. 87.

FORMULA

a(k^2) = 0.

a(m) = floor(sqrt(m)) for nonsquare m.

a(n) = 2 * A320471(n) for n > 0.