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A146340
Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 17.
3
521, 617, 709, 1433, 1597, 2549, 2909, 2965, 3161, 3581, 3821, 4013, 4285, 4649, 5501, 5585, 5693, 5813, 6197, 6409, 7825, 7853, 8093, 8125, 8573, 8917, 9281, 9665, 9677, 9925, 10265, 10597, 10973, 11273, 12085, 12805, 13061, 13109, 13613, 13957, 14677
OFFSET
1,1
COMMENTS
For primes in this sequence see A146362.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..150 from Harvey P. Dale)
EXAMPLE
a(1) = 521 because continued fraction of (1+sqrt(521))/2 = 11, 1, 10, 2, 5, 4, 2, 1, 1, 1, 1, 2, 4, 5, 2, 10, 1, 21, 1, 10, 2, 5, 4, 2, 1, 1, 1, 1, 2, 4, 5, 2, 10, 1, 21, 1, 10, 2, 5, ... has period (1, 10, 2, 5, 4, 2, 1, 1, 1, 1, 2, 4, 5, 2, 10, 1, 21) length 17.
MAPLE
A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic', 'quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146340 := proc(n) RETURN(A146326(n) = 17) ; end: for n from 2 do if isA146340(n) then printf("%d, \n", n) ; fi; od: # R. J. Mathar, Sep 06 2009
MATHEMATICA
cf17Q[n_]:=Module[{s=(1+Sqrt[n])/2}, If[IntegerQ[s], 1, Length[ ContinuedFraction[ s][[2]]]]==17]; Select[Range[5000], cf17Q] (* Harvey P. Dale, Dec 20 2017 *)
KEYWORD
more,nonn
AUTHOR
Artur Jasinski, Oct 30 2008
EXTENSIONS
998 and 1006 removed, sequence extended by R. J. Mathar, Sep 06 2009
More terms from Harvey P. Dale, Dec 20 2017
STATUS
approved