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%I #19 Mar 31 2020 03:03:16
%S 89,109,113,137,373,389,509,653,685,797,853,925,949,997,1009,1105,
%T 1145,1165,1261,1493,1997,2309,2621,2677,2885,2941,3133,3277,3445,
%U 3653,3797,4325,4505,4745,4825,4973,5353,5429,5765,6305,6437,6845,7085,7373,7817,7873
%N Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 7.
%C For primes in this sequence see A146352.
%H Amiram Eldar, <a href="/A146332/b146332.txt">Table of n, a(n) for n = 1..10000</a>
%e a(4) = 137 because continued fraction of (1+sqrt(137))/2 = 6, 2, 1, 5, 5, 1, 2, 11, 2, 1, 5, 5, 1, 2, 11, 2, 1, 5, 5, 1, 2, 11 ... has period (2, 1, 5, 5, 1, 2, 11) length 7.
%p A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146332 := proc(n) RETURN(A146326(n) = 7) ; end: for n from 2 to 1100 do if isA146332(n) then printf("%d,",n) ; fi; od: # _R. J. Mathar_, Sep 06 2009
%t Select[Range[10^4], !IntegerQ @ Sqrt[#] && Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] == 7 &] (* _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 997 added by _R. J. Mathar_, Sep 06 2009
%E More terms from _Amiram Eldar_, Mar 31 2020
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