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A170945
Least number k such that the decimal representation of 1/k has period Fibonacci(n).
1
3, 11, 27, 41, 73, 53, 43, 103, 1321, 497867, 323, 467, 11311, 20141, 12169, 232159532264041847249
OFFSET
2,1
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).
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.
EXAMPLE
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
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:
CROSSREFS
Cf. A000045, A039834 (signed Fibonacci numbers), A002329, A072859 (periodic sequences), A003060
Sequence in context: A101612 A123928 A186301 * A164897 A212982 A164845
KEYWORD
nonn,base
AUTHOR
Michel Lagneau, Feb 19 2010
EXTENSIONS
Edited by T. D. Noe, Apr 14 2010
STATUS
approved