A173491 a(n) is the least k such that the period of the decimal expansion of 1/k is A000204(n).
3, 27, 101, 239, 21649, 19, 3191, 35121409, 722817036322379041, 2241, 797, 967, 1230610745978027, 3373, 60787, 509538919, 15060275578609, 5779, 37397, 423557
Offset: 1
Examples
a(1)=3 because the period of 1/3 = 0.333... is 1, and 3 is the smallest number with that period. a(2)=27 because the period of 1/27 = 0.037037... is 3 = A000204(2), and 27 is the smallest number with that period. a(3)=101 because the period of 1/101 = 0.00990099... is 4 = A000204(3), and 101 is the smallest number with that period. a(4)= 239 because the period of 1/239 = 0.00418410041841... is 7 = A000204(4), and 239 is the smallest number with that period.
References
- V. E. Hoggatt, Jr., Fibonacci and Lucas Numbers. Houghton, Boston, MA, 1969.
- Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.
- S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.
Links
- R. Javonovic, Lucas Function Calculator
Programs
-
Maple
T:=array(0..100); U:=array(0..100); n0:=1: n1:=3: T[1] := 1: T[2] := 3:for i from 3 to 30 do: n2:=n0+n1: T[i]:=n2: n0:=n1: n1:=n2: od: for q from 1 to 7 do: p0:=T[q]: indic:=0: for n from 1 to 25000 do: for p from 1 to 30 while(irem(10^p, n)<>1 or gcd(n,10)<>1 ) do: od: if irem(10^p,n) = 1 and gcd(n,10) = 1 and p=p0 and indic=0 then U[q]:=n: indic:=1: else fi: od: od: for n from 1 to 7 do: print( U[n]): od:
-
Mathematica
(* This [slow] mma program gives all denominators < 50000 and disagrees with existing sequence for n = 11: a(11) = 797 instead of 29453 *) a204[n_] := a204[n] = Coefficient[Series[(2 - t )/(1 - t - t^2), {t, 0, n}], t^n] ; a7732[n_] := a7732[n] = MultiplicativeOrder[10, FixedPoint[Quotient[#, GCD[#, 10]] &, n]]; a[n_] := (k = 2; While[k++; k < 50000 && a7732[k] != a204[n] ]; k); Table[a[n], {n, 1, 15}](* Jean-François Alcover, Sep 02 2011 *)
Extensions
References to unrelated sequences removed by R. J. Mathar, Feb 24 2010
Comments