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A131401
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Least number dividing Fibonacci(n) but not dividing Fibonacci(m) for m < n, or 0 if there is no such number.
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2
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1, 0, 2, 3, 5, 4, 13, 7, 17, 11, 89, 6, 233, 29, 10, 47, 1597, 19, 37, 15, 26, 199, 28657, 14, 25, 521, 53, 39, 514229, 20, 557, 2207, 178, 3571, 65, 27, 73, 9349, 466, 35, 2789, 52, 433494437, 43, 85, 139, 2971215073, 64, 97, 101, 3194, 699, 953, 212, 445, 49, 74, 59
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OFFSET
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1,3
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COMMENTS
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First occurrence of n in A001177 or 0 if impossible.
Conjecture: only a(2)=0. I have not found values of a(n) < 2*106 less than 100 for n = 43, 47, 74, 82, 83 & 94.
When Fibonacci(n) is a prime number, then a(n)=Fibonacci(n). Note that a(n)=0 for n=2 because Fibonacci(1)=Fibonacci(2)=1. For n > 2, an upper bound for a(n) is Fibonacci(n). The difficulty in computing this sequence for large n is the factorization of Fibonacci(n), which is required to find the divisors of Fibonacci(n). - T. D. Noe, Jan 12 2009
In other words, the conjecture is true. For n > 2, Fibonacci(n) has at least one divisor that does not divide Fibonacci(k) for k < n. The number of such divisors is A120256(n).
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REFERENCES
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Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers, Afterword by Herbert A. Hauptman, 2. 'The Minor Modulus m(n)', Prometheus Books, NY, 2007, pp. 329-342.
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LINKS
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MATHEMATICA
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f[n_] := Block[{k = 1}, While[Mod[Fibonacci@k, n] != 0 && k < 101, k++ ]; k]; t = Table[0, {100}]; Do[ a = f@n; If[a < 101 && t[[a]] == 0, t[[a]] = n; Print[{a, n}]], {n, 106}]
nn=100; fib=Fibonacci[Range[nn]]; Join[{1, 0}, Table[dvrs=Rest[Divisors[fib[[n]]]]; k=1; While[d=dvrs[[k]]; pos=Position[fib, _?(Mod[ #, d]==0&), 1, 1]; pos!={{n}}, k++ ]; d, {n, 3, nn}]] (* T. D. Noe, Jan 12 2009 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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