A146338 Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 15.

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

Amiram Eldar, Table of n, a(n) for n = 1..10000

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 *)

Crossrefs

Cf. A000290, A078370, A146326-A146345, A146348-A146360.

Sequence in context: A142925 A060333 A020352 * A201858 A146360 A050964

Adjacent sequences: A146335 A146336 A146337 * A146339 A146340 A146341

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