A065129 a(n) is the least m such that m/A003285(m) = n, or 0 if no such m exists.

0, 2, 6, 8, 5, 12, 28, 32, 18, 10, 0, 24, 0, 0, 30, 0, 17, 0, 38, 40, 42, 0, 276, 48, 125, 26, 0, 56, 406, 0, 496, 128, 66, 68, 140, 72, 37, 0, 0, 80, 0, 84, 0, 176, 90, 0, 1222, 192, 294, 50, 102, 104, 636, 432, 110, 0, 0, 928, 708, 120, 0, 248, 252, 0, 65, 132

Offset: 1

Author

Naohiro Nomoto, Nov 14 2001

Keywords

• nonn

Comments

Conjecture: A003285(m) = even or A004613, if m is divisible by A003285(m). [This sentence appears to be saying that all odd terms of this sequence are in A004613.]

Because A003285(m) < 3.76*sqrt(m)*log(m) (see Stanton et al.), it is enough to check m such that m <= (3.76*n*log(m))^2. For n <= 36 it even suffices to check m <= 5916*n. - Nathaniel Johnston, May 10 2011

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..406
• R. G. Stanton, C. Sudler, and H. C. Williams, An upper bound for the period of the simple continued fraction for sqrt(D), Pacific Journal of Mathematics, Vol. 67, No. 2 (1976), pp. 525-536.

Crossrefs

Cf. A003285, A004613.

Programs

Maple

with(numtheory): A065129 := proc(n) local m: if(n=1)then return 0:fi: for m from n by n to 5916*n do if(frac(sqrt(m))<>0)then if(n*nops(cfrac(sqrt(m),'periodic','quotients')[2])=m)then return m: fi: fi: od: return 0: end: seq(A065129(n),n=1..10); # Nathaniel Johnston, May 10 2011

Mathematica

Do[k = 2; While[ k / Length[ Last[ ContinuedFraction[ Sqrt[k]]]] != n, k++ ]; Print[k], {n, 2, 10} ]

Extensions

a(11)-a(37) from Nathaniel Johnston, May 10 2011

Terms a(38) and beyond from Chai Wah Wu, Jan 27 2021