A146363 a(n) = smallest prime p such that continued fraction of (1 + sqrt(p))/2 has period length n.

5, 2, 17, 7, 41, 19, 89, 31, 73, 43, 541, 103, 421, 179, 193, 191, 521, 139, 241, 151, 337, 491, 433, 271, 929, 211, 409, 487, 673, 379, 937, 463, 601, 331, 769, 1439, 2297, 619, 1033, 1399, 1777, 571, 1753, 823, 1993, 739, 1249, 631, 4337, 1051, 1321, 751, 1201

Offset: 1

Author

Artur Jasinski, Oct 30 2008

Keywords

• nonn

Links

• Artur Jasinski, Table of n, a(n) for n=1..1000

Crossrefs

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

Programs

Maple

A := proc(n) option remember ; local c; try c := numtheory[cfrac](1/2+sqrt(n)/2,'periodic,quotients') ; RETURN(nops(c[2]) ); catch: RETURN(-1) end try ; end: A146363 := proc(n) local p,i ; for i from 1 do p := ithprime(i) ; if A(p) = n then RETURN(p) ; fi; od; end: for n from 1 do printf("%d, ",A146363(n)) ; od: # R. J. Mathar, Nov 08 2008

Mathematica

$MaxExtraPrecision = 300; 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, 1200}]; Print[aa]; bb = {}; Do[k = 1; yes = 0&&PeimeQ[k]; Do[If[aa[[k]] == n && yes == 0, AppendTo[bb, k]; yes = 1], {k, 1, Length[aa]}], {n, 1, 22}]; bb (* Artur Jasinski *)
aa = {}; Do[n = 1; While[m != Length[ContinuedFraction[(1 + Sqrt[Prime[n]])/2][[2]]], n++ ]; AppendTo[aa, Prime[n]], {m, 1, 100}]; aa (* Artur Jasinski, Feb 03 2010 *)

Extensions

a(25) replaced by 929 and extended by R. J. Mathar, Nov 08 2008