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A146330 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 5. 2

%I

%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

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

%C For primes in this sequence see A146350.

%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 s = 10; aa = {}; Do[k = ContinuedFraction[(1 + Sqrt[n])/2, 1000]; If[Length[k] < 190, AppendTo[aa, 0], m = 1; While[k[[s ]] != k[[s + m]] || k[[s + m]] != k[[s + 2 m]] || k[[s + 2 m]] != k[[s + 3 m]] || k[[s + 3 m]] != k[[s + 4 m]], m++ ]; s = s + 1; While[k[[s ]] != k[[s + m]] || k[[s + m]] != k[[s + 2 m]] || k[[s + 2 m]] != k[[s + 3 m]] || k[[s + 3 m]] != k[[s + 4 m]], m++ ]; s = s + 1; While[k[[s ]] != k[[s + m]] || k[[s + m]] != k[[s + 2 m]] || k[[s + 2 m]] != k[[s + 3 m]] || k[[s + 3 m]] != k[[s + 4 m]], m++ ]; AppendTo[aa, m]], {n, 1, 500}]; bb = {}; Do[If[aa[[n]] == 5, AppendTo[bb, n]], {n, 1, Length[aa]}]; bb

%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

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Last modified August 25 03:55 EDT 2019. Contains 326318 sequences. (Running on oeis4.)