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%I #12 May 19 2019 02:09:59
%S 1,0,2,3,5,4,13,7,17,11,89,6,233,29,10,47,1597,19,37,15,26,199,28657,
%T 14,25,521,53,39,514229,20,557,2207,178,3571,65,27,73,9349,466,35,
%U 2789,52,433494437,43,85,139,2971215073,64,97,101,3194,699,953,212,445,49,74,59
%N Least number dividing Fibonacci(n) but not dividing Fibonacci(m) for m < n, or 0 if there is no such number.
%C First occurrence of n in A001177 or 0 if impossible.
%C 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.
%C 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
%C 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).
%D 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.
%H T. D. Noe, <a href="/A131401/b131401.txt">Table of n, a(n) for n = 1..300</a>
%t 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}]
%t 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 *)
%Y Cf. A060442. - _T. D. Noe_, Jan 12 2009
%K nonn
%O 1,3
%A Herbert A. Hauptman (hauptman(AT)hwi.buffalo.edu) & _Robert G. Wilson v_, Jul 07 2007
%E Extended by _T. D. Noe_, Jan 12 2009
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