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A170945
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Least number k such that the decimal representation of 1/k has period Fibonacci(n).
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1
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3, 11, 27, 41, 73, 53, 43, 103, 1321, 497867, 323, 467, 11311, 20141, 12169, 232159532264041847249
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OFFSET
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2,1
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COMMENTS
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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).
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REFERENCES
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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.
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LINKS
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EXAMPLE
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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
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MAPLE
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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:
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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