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A065129
a(n) is the least m such that m/A003285(m) = n, or 0 if no such m exists.
1
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
1,2
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
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.
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} ]
CROSSREFS
Sequence in context: A002210 A145500 A247572 * A074758 A228163 A029671
KEYWORD
nonn
AUTHOR
Naohiro Nomoto, Nov 14 2001
EXTENSIONS
a(11)-a(37) from Nathaniel Johnston, May 10 2011
Terms a(38) and beyond from Chai Wah Wu, Jan 27 2021
STATUS
approved