cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A003285 Period of continued fraction for square root of n (or 0 if n is a square).

Original entry on oeis.org

0, 1, 2, 0, 1, 2, 4, 2, 0, 1, 2, 2, 5, 4, 2, 0, 1, 2, 6, 2, 6, 6, 4, 2, 0, 1, 2, 4, 5, 2, 8, 4, 4, 4, 2, 0, 1, 2, 2, 2, 3, 2, 10, 8, 6, 12, 4, 2, 0, 1, 2, 6, 5, 6, 4, 2, 6, 7, 6, 4, 11, 4, 2, 0, 1, 2, 10, 2, 8, 6, 8, 2, 7, 5, 4, 12, 6, 4, 4, 2, 0, 1, 2, 2, 5, 10, 2, 6, 5, 2, 8, 8, 10, 16, 4, 4, 11, 4, 2, 0, 1, 2, 12
Offset: 1

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Comments

Any string of five consecutive terms m^2 - 2 through m^2 + 2 for m > 2 in the sequence has the corresponding periods 4,2,0,1,2. - Lekraj Beedassy, Jul 17 2001
For m > 1, a(m^2+m) = 2 and the continued fraction is m, 2, 2*m, 2, 2*m, 2, 2*m, ... - Arran Fernandez, Aug 14 2011
Apparently the generating function of the sequence for the denominators of continued fraction convergents to sqrt(n) is always rational and of form p(x)/[1 - C*x^m + (-1)^m * x^(2m)], or equivalently, the denominators satisfy the linear recurrence b(n+2m) = C*b(n+m) - (-1)^m * b(n), where a(n) is equal to m for each nonsquare n, or 0. See A006702 for the conjecture regarding C. The same conjectures apply to the sequences of the numerators of continued fraction convergents to sqrt(n). - Ralf Stephan, Dec 12 2013
If a(n)=1, n is of form k^2+1 (A002522 except the initial term 1). See A013642 for a(n)=2, A013643 for a(n)=3, A013644 for a(n)=4, A010337 for a(n)=5, A020347 for a(n)=6, A010338 for a(n)=7, A020348 for a(n)=8, A010339 for a(n)=9, and furthermore A020349-A020439. - Ralf Stephan, Dec 12 2013
From William Krier, Dec 12 2024: (Start)
a(m^2-4) = 4 for even m>=6 since sqrt(m^2-4) = [m-1; 1, (m-4)/2, 1, 2*(m-1)].
a(m^2-4) = 6 for odd m>=5 since sqrt(m^2-4) = [m-1; 1, (m-3)/2, 2, (m-3)/2, 1, 2*(m-1)].
a(m^2+4) = 2 for even m>=2 since sqrt(m^2+4) = [m; m/2, 2*m].
a(m^2+4) = 5 for odd m>=3 since sqrt(m^2+4) = [m; (m-1)/2, 1, 1, (m-1)/2, 2*m]. (End)

References

  • A. Brousseau, Number Theory Tables. Fibonacci Association, San Jose, CA, 1973, p. 197.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    f:= n ->  if issqr(n) then 0
       else nops(numtheory:-cfrac(sqrt(n),'periodic','quotients')[2]) fi:
    map(f, [$1..100]); # Robert Israel, Sep 02 2015
  • Mathematica
    a[n_] := ContinuedFraction[Sqrt[n]] // If[Length[ # ] == 1, 0, Length[Last[ # ]]]&
    pcf[n_]:=Module[{s=Sqrt[n]},If[IntegerQ[s],0,Length[ContinuedFraction[s][[2]]]]]; Array[pcf,110] (* Harvey P. Dale, Jul 15 2017 *)
  • PARI
    a(n)=if(issquare(n),return(0));my(s=sqrt(n),x=s,f=floor(s),P=[0],Q=[1],k);while(1,k=#P;P=concat(P,f*Q[k]-P[k]);Q=concat(Q,(n-P[k+1]^2)/Q[k]);k++;for(i=1,k-1,if(P[i]==P[k]&&Q[i]==Q[k],return(k-i)));x=(P[k]+s)/Q[k];f=floor(x)) \\ Charles R Greathouse IV, Jul 31 2011
    
  • PARI
    isok(n, p) = {localprec(p); my(cf = contfrac(sqrt(n))); setsearch(Set(cf), 2*cf[1]);}
    a(n) = {if (issquare(n), 0, my(p=100); while (! isok(n, p), p+=100); localprec(p); my(cf = contfrac(sqrt(n))); for (k=2, #cf, if (cf[k] == 2*cf[1], return (k-1))););} \\ Michel Marcus, Jul 07 2021
    
  • Python
    from sympy.ntheory.continued_fraction import continued_fraction_periodic
    def a(n):
        cfp = continued_fraction_periodic(0, 1, d=n)
        return 0 if len(cfp) == 1 else len(cfp[1])
    print([a(n) for n in range(1, 104)]) # Michael S. Branicky, Aug 22 2021