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A146338
Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 15.
3
193, 281, 481, 1417, 1861, 1933, 2089, 2141, 2197, 2437, 2741, 2837, 3037, 3065, 3121, 3413, 3589, 3625, 3785, 3925, 3977, 4001, 4637, 4777, 4877, 5317, 5821, 5941, 6025, 6641, 6653, 6749, 7673, 8117, 8177, 8345, 10069, 10273, 10457, 11197, 11281, 11549, 11821
OFFSET
1,1
COMMENTS
For primes in this sequence see A146360.
LINKS
EXAMPLE
a(1) = 193 because continued fraction of (1+sqrt(193))/2 = 7, 2, 4, 6, 1, 2, 1, 1, 1, 1, 2, 1, 6, 4, 2, 13, 2, 4, 6, 1, 2, 1, 1, 1, 1, 2, 1, 6, 4, 2, 13, ... has period (2, 4, 6, 1, 2, 1, 1, 1, 1, 2, 1, 6, 4, 2, 13) length 15.
MAPLE
A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic', 'quotients') ; nops(%[2]) ; else 0 ; fi; end:
isA146338 := proc(n) RETURN(A146326(n) = 15) ; end:
for n from 2 to 4000 do if isA146338(n) then printf("%d, \n", n) ; fi; od: # R. J. Mathar, Sep 06 2009
MATHEMATICA
Select[Range[10^4], !IntegerQ @ Sqrt[#] && Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] == 15 &] (* Amiram Eldar, Mar 31 2020 *)
KEYWORD
nonn
AUTHOR
Artur Jasinski, Oct 30 2008
EXTENSIONS
Extended beyond 3 terms by R. J. Mathar, Sep 06 2009
More terms from Amiram Eldar, Mar 31 2020
STATUS
approved