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%I #26 Mar 31 2020 03:01:26
%S 193,281,481,1417,1861,1933,2089,2141,2197,2437,2741,2837,3037,3065,
%T 3121,3413,3589,3625,3785,3925,3977,4001,4637,4777,4877,5317,5821,
%U 5941,6025,6641,6653,6749,7673,8117,8177,8345,10069,10273,10457,11197,11281,11549,11821
%N Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 15.
%C For primes in this sequence see A146360.
%H Amiram Eldar, <a href="/A146338/b146338.txt">Table of n, a(n) for n = 1..10000</a>
%e 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.
%p A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end:
%p isA146338 := proc(n) RETURN(A146326(n) = 15) ; end:
%p for n from 2 to 4000 do if isA146338(n) then printf("%d,\n",n) ; fi; od: # _R. J. Mathar_, Sep 06 2009
%t Select[Range[10^4], !IntegerQ @ Sqrt[#] && Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] == 15 &] (* _Amiram Eldar_, Mar 31 2020 *)
%Y Cf. A000290, A078370, A146326-A146345, A146348-A146360.
%K nonn
%O 1,1
%A _Artur Jasinski_, Oct 30 2008
%E Extended beyond 3 terms by _R. J. Mathar_, Sep 06 2009
%E More terms from _Amiram Eldar_, Mar 31 2020
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