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

1,1

Links

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

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)

Crossrefs

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

Sequence in context: A282224 A282201 A146343 * A246797 A087958 A286161

Adjacent sequences: A146360 A146361 A146362 * A146364 A146365 A146366

Keyword

nonn

Author

Artur Jasinski, Oct 30 2008

Extensions

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

Status

approved