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

%I A065129 #36 Jan 01 2024 11:55:39
%S A065129 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,
%T A065129 0,56,406,0,496,128,66,68,140,72,37,0,0,80,0,84,0,176,90,0,1222,192,
%U A065129 294,50,102,104,636,432,110,0,0,928,708,120,0,248,252,0,65,132
%N A065129 a(n) is the least m such that m/A003285(m) = n, or 0 if no such m exists.
%C A065129 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.]
%C A065129 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
%H A065129 Chai Wah Wu, <a href="/A065129/b065129.txt">Table of n, a(n) for n = 1..406</a>
%H A065129 R. G. Stanton, C. Sudler, and H. C. Williams, <a href="http://projecteuclid.org/euclid.pjm/1102817507">An upper bound for the period of the simple continued fraction for sqrt(D)</a>, Pacific Journal of Mathematics, Vol. 67, No. 2 (1976), pp. 525-536.
%p A065129 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
%t A065129 Do[k = 2; While[ k / Length[ Last[ ContinuedFraction[ Sqrt[k]]]] != n, k++ ]; Print[k], {n, 2, 10} ]
%Y A065129 Cf. A003285, A004613.
%K A065129 nonn
%O A065129 1,2
%A A065129 _Naohiro Nomoto_, Nov 14 2001
%E A065129 a(11)-a(37) from _Nathaniel Johnston_, May 10 2011
%E A065129 Terms a(38) and beyond from _Chai Wah Wu_, Jan 27 2021