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%I #14 Mar 31 2020 03:00:37
%S 265,541,593,661,701,857,1061,1109,1217,1237,1709,1733,1949,2333,2509,
%T 2557,2957,3125,3229,3677,3701,4181,4373,4685,5081,5237,5309,6133,
%U 6425,6445,7013,7025,8185,8545,8693,9305,9533,9553,10333,10525,10853,10961,11125,11141
%N Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 11.
%C For primes in this sequence see A146356.
%H Amiram Eldar, <a href="/A146335/b146335.txt">Table of n, a(n) for n = 1..10000</a>
%e a(4) = 661 because continued fraction of (1+sqrt(661))/2 = 13, 2, 1, 4, 2, 8, 8, 2, 4, 1, 2, 25, 2, 1, 4, 2, 8, 8, 2, 4, 1, 2, 25, 2, 1, 4, 2, 8, 8, 2, 4, 1, 2, 25, 2, 1, 4, 2, 8, 8 ... has period (2, 1, 4, 2, 8, 8, 2, 4, 1, 2, 25) length 11.
%p A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146335 := proc(n) RETURN(A146326(n) = 11) ; end: for n from 2 to 2000 do if isA146335(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]]] == 11 &] (* _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 916 removed by _R. J. Mathar_, Sep 06 2009
%E More terms from _Amiram Eldar_, Mar 31 2020