OFFSET
1,3
COMMENTS
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).
REFERENCES
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.
LINKS
T. D. Noe, Table of n, a(n) for n = 1..300
MATHEMATICA
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Herbert A. Hauptman (hauptman(AT)hwi.buffalo.edu) & Robert G. Wilson v, Jul 07 2007
EXTENSIONS
Extended by T. D. Noe, Jan 12 2009
STATUS
approved