1,1

For primes in this sequence see A146362.

a(1) = 521 because continued fraction of (1+Sqrt[521])/2 = 11, 1, 10, 2, 5, 4, 2, 1, 1, 1, 1, 2, 4, 5, 2, 10, 1, 21, 1, 10, 2, 5, 4, 2, 1, 1, 1, 1, 2, 4, 5, 2, 10, 1, 21, 1, 10, 2, 5,...

has period (1, 10, 2, 5, 4, 2, 1, 1, 1, 1, 2, 4, 5, 2, 10, 1, 21) length 17

A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic', 'quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146340 := proc(n) RETURN(A146326(n) = 17) ; end: for n from 2 do if isA146340(n) then printf("%d, \n", n) ; fi; od: [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 06 2009]

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++ ]; AppendTo[aa, m]], {n, 1, 500}]; bb = {}; Do[If[aa[[n]] == 17, AppendTo[bb, n]], {n, 1, Length[aa]}]; bb (*Artur Jasinski*)

A000290, A078370, A146326-A146345, A146348-A146360.

Sequence in context: A147856 A094903 A139663 this_sequence A146362 A050966 A113158

Adjacent sequences: A146337 A146338 A146339 this_sequence A146341 A146342 A146343

more,nonn

Artur Jasinski (grafix(AT)csl.pl), Oct 30 2008

998 and 1006 removed, sequence extended - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 06 2009