A146364 a(n) = smallest primes whose continued fraction have different period.

2, 5, 7, 17, 19, 31, 41, 43, 73, 89, 103, 139, 151, 179, 191, 193, 211, 241, 271, 331, 337, 379, 409, 421, 433, 463, 487, 491, 521, 541, 571, 601, 619, 631, 673, 739, 751, 769, 823, 919, 929, 937, 1033, 1039, 1051, 1201, 1249, 1291, 1321, 1399, 1439, 1471, 1531, 1579, 1609, 1699, 1747, 1753, 1759

Offset: 1

Author

Artur Jasinski, Oct 30 2008

Keywords

• nonn

Comments

This sequence is sorted A146363.

Links

• Robert Israel, Table of n, a(n) for n = 1..1500

Crossrefs

Cf. A000290, A078370, A146326-A146345, A146348-A146360, A146363.

Programs

Maple

g:= proc(n) local c;
      c:= NumberTheory:-ContinuedFraction((1+sqrt(n))/2);
      nops(Term(c,periodic)[2]);
end proc:
R:= NULL: S:= {}: count:= 0:
p:= 1:
while count < 100 do
  p:= nextprime(p);
  v:= g(p);
  if not member(v,S) then
    R:= R,p; count:= count+1; S:= S union {v};
    if count mod 20 = 0 then printf("%d %d\n",count,p) fi
  fi
od:
R; # Robert Israel, Jun 14 2024

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}]; Sort[bb] (*Artur Jasinski*)

Extensions

More terms from Robert Israel, Jun 14 2024