A170945 Least number k such that the decimal representation of 1/k has period Fibonacci(n).

3, 11, 27, 41, 73, 53, 43, 103, 1321, 497867, 323, 467, 11311, 20141, 12169, 232159532264041847249

Offset: 2

Author

Michel Lagneau, Feb 19 2010

Keywords

• nonn
• base

Comments

The period of 1/k is the least integer p such that 10^p = 1 (mod k). The integer p is also known as the multiplicative order of 10 (mod k).

Examples

			p(k) is the period of 1/k, we obtain with k=3,11,27,41,73,53,43,103 p(3)=1,p(11)=2,p(27)=3,p(41)=5,p(73)=8, p(53)=13,p(43)=21, p(103)=34
		

References

• Mohammad K. Azarian, The Generating Function for the Fibonacci Sequence, Missouri Journal of Mathematical Sciences, Vol. 2, No. 2, Spring 1990, pp. 78-79. Zentralblatt MATH, Zbl 1097.11516.
• 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

• Dario Alejandro Alpern, Factorization using the Elliptic Curve Method
• C. K. Caldwell, Fibonacci Numbers

Crossrefs

Cf. A000045, A039834 (signed Fibonacci numbers), A002329, A072859 (periodic sequences), A003060

Programs

Maple

For the great numbers (p > 70), the maple program is very slow. That's what we use an process of two steps: factoring 10^p-1 with elliptic curve method (see the first web site), and then, for each factor q(k), k=1,2,...,r computation the periods of 1/q(k) and keep the period q(i) such that q(i) = Fibonacci number. The 17th term required 3h 2m for the computing of (10^1597) -1 T:=array(0..100);U:=array(0..100); n0:=0:n1:=1:T[1] = 1:for i from 2 to 30 do: n2:=n0+n1:T[i]:=n2:n0:=n1:n1:=n2:od:U[1]:=3:U[2]:=3:for q from 3 to 10 do: p0:=T[q]: indic:=0:for n from 1 to 2000 do:for p from 1 to 150 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 10 do:print( U[n]):od:

Extensions

Edited by T. D. Noe, Apr 14 2010