|
| |
|
|
A065129
|
|
Least m such that m/A003285(m) is 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
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 1,2
|
|
|
COMMENTS
| Conjecture: A003285(m) = even or A004613, if m is divisible by A003285(m).
Because A003285(m) < 3.76*sqrt(m)*ln(m) (see Stanton et al), it is enough to check m such that m <= (3.76*n*ln(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
| Cf. A004613.
Sequence in context: A011045 A002210 A145500 * A074758 A029671 A154970
Adjacent sequences: A065126 A065127 A065128 * A065130 A065131 A065132
|
|
|
KEYWORD
| more,nonn
|
|
|
AUTHOR
| Naohiro Nomoto (n_nomoto(AT)yabumi.com), Nov 14 2001
|
|
|
EXTENSIONS
| a(11)-a(37) from Nathaniel Johnston (nathaniel(AT)nathanieljohnston.com), May 10 2011
|
| |
|
|
