%I #18 Mar 31 2020 03:03:51

%S 41,149,157,181,269,397,425,493,565,697,761,941,1013,1037,1325,1565,

%T 1781,1825,2081,2153,2165,2173,2465,2477,2693,2725,3181,3221,3533,

%U 3869,4253,4409,5165,5213,5273,5297,5741,5837,6485,6757,6949,7045,7325,7465,8021,8069

%N Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 5.

%C For primes in this sequence see A146350.

%H Amiram Eldar, <a href="/A146330/b146330.txt">Table of n, a(n) for n = 1..10000</a>

%e a(1) = 41 because continued fraction of (1+sqrt(41))/2 = 3, 1, 2, 2, 1, 5, 1, 2, 2, 1, 5, 1, 2, 2, 1, 5, 1, 2, ... has period (1,2,2,1,5) length 5.

%p isA146330 := proc(n) RETURN(A146326(n) = 5) ; end:

%p for n from 2 to 2000 do if isA146330(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]]] == 5 &] (* _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 259 and 1026 removed by _R. J. Mathar_, Sep 06 2009

%E More terms from _Amiram Eldar_, Mar 31 2020