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A146339
Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 16.
3
172, 191, 217, 232, 249, 310, 311, 329, 343, 344, 355, 369, 391, 393, 416, 428, 431, 446, 496, 513, 520, 524, 536, 537, 550, 559, 589, 647, 655, 679, 682, 686, 700, 704, 748, 760, 768, 775, 802, 816, 848, 851, 872, 927, 995, 996, 1036, 1058, 1079, 1080, 1120, 1136
OFFSET
1,1
COMMENTS
For primes in this sequence see A146361.
LINKS
EXAMPLE
a(1) = 191 because continued fraction of (1+sqrt(191))/2 = 7, 2, 2, 3, 1, 1, 4, 1, 26, 1, 4, 1, 1, 3, 2, 2, 13, 2, 2, 3, 1, 1, 4, 1, 26, 1, 4, 1, 1, 3, 2, 2, 13, 2, 2, 3, 1, 1, 4, 1, 26... has period (2, 2, 3, 1, 1, 4, 1, 26, 1, 4, 1, 1, 3, 2, 2, 13) length 16.
MAPLE
A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic', 'quotients') ; nops(%[2]) ; else 0 ; fi; end:
isA146339 := proc(n) RETURN(A146326(n) = 16) ; end:
for n from 2 to 1000 do if isA146339(n) then printf("%d, ", n) ; fi; od: # R. J. Mathar, Sep 06 2009
MATHEMATICA
Select[Range[1000], !IntegerQ @ Sqrt[#] && Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] == 16 &] (* Amiram Eldar, Mar 31 2020 *)
KEYWORD
nonn
AUTHOR
Artur Jasinski, Oct 30 2008
EXTENSIONS
311 inserted, sequence extended by R. J. Mathar, Sep 06 2009
More terms from Amiram Eldar, Mar 31 2020
STATUS
approved