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approved

Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 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_, Sep 06 2009]

998 and 1006 removed, sequence extended - _by _R. J. Mathar_, Sep 06 2009

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Numbers k such that continued fraction of (1+Sqrt[sqrt(k]))/2 has period 17.

a(1) = 521 because continued fraction of (1+Sqrt[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.

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*)

proposed

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proposed

Numbers k such that continued fraction of (1+Sqrt[k])/2 has period 17.

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.

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

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

approved

editing

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

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]

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

521, 617, 709, 998, 10061433, 1597, 2549, 2909, 2965, 3161, 3581, 3821, 4013, 4285, 4649

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]

A000290, A078370, A146326-A146345, A146346A146348-A146360.

more,nonn,new

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

Numbers k such that continued fraction of (1+Sqrt[k])/2 has period 17

521, 617, 709, 998, 1006

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

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, A146346-A146360.

more,nonn

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

approved