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A065069 Numbers n such that Fibonacci(n) is not squarefree, but for all proper divisors k of n, Fibonacci(k) is squarefree.

%I A065069 #24 Dec 09 2022 19:36:01
%S A065069 6,25,56,91,110,153,406,703,752,820,915,979,1431,1892,2147,2701,2943,
%T A065069 3029,3422,4378,4556,4753,4970,5513,6394,7868,8841,9453,10712,12403,
%U A065069 13508,13546,15051,16256,17030,17267,18023,18721,19503,20827,21206
%N A065069 Numbers n such that Fibonacci(n) is not squarefree, but for all proper divisors k of n, Fibonacci(k) is squarefree.
%C A065069 These are first primitive indices m for which Fib(m) is squareful. Note that Fib(km) is divisible by Fib(m).
%C A065069 This sequence is closely related to A001602(n), which gives the index of the smallest Fibonacci number divisible by prime(n). It can be shown that the index of the first Fibonacci number divisible by prime(n)^2 is A001602(n)*prime(n). This sequence is the collection of numbers A001602(n)*prime(n) with multiples removed. For example, A001602(2)*prime(2) = 12, but all multiples of 12 will generated by 6, the first number in this sequence. The Mathematica code assumes that Fibonacci numbers do not have any square primitive prime factors -- an assumption whose truth is an open question. - _T. D. Noe_, Jul 24 2003
%C A065069 These are the primitive elements of A037917. - _Charles R Greathouse IV_, Feb 02 2014
%C A065069 Terms after a(12) are conjectures until the factorizations of F(1271), F(1273), etc. are completed. - _Charles R Greathouse IV_, Feb 02 2014
%C A065069 Three more factorizations are needed to get the next term: F(1423), F(1427), and F(1429). If these are each squarefree, a(13) = 1431. - _Charles R Greathouse IV_, Dec 09 2022
%H A065069 Hisanori Mishima, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha108.htm">Factorizations of Fibonacci numbers: n=1..100</a>, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha109.htm">n=101..200</a>, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha110.htm">n=201..300</a>, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha111.htm">n=301..400</a>, <a href="http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha112.htm">n=401..480</a>.
%H A065069 Blair Kelly, <a href="http://mersennus.net/fibonacci//">Fibonacci and Lucas Factorizations</a>
%e A065069 a(1) = 6 because 2^2 divides Fibonacci(6) but no smaller Fibonacci number.
%t A065069 << NumberTheory`NumberTheoryFunctions`; a = {}; l = 0; Do[m = n; If[k = 1; While[k < l + 1 && !IntegerQ[ n/ a[[k]]], k++ ]; k > l, If[ !SquareFreeQ[ Fibonacci[n]], a = Append[a, n]; l++; Print[n]]], {n, 1, 480} ]
%t A065069 nLimit=50000; i=3; pMax=1; iMax=1; While[p=Transpose[FactorInteger[Fibonacci[i]]][[1, -1]]; i*p<nLimit, If[p>pMax, pMax=p; iMax=i]; i++ ]; nMax=PrimePi[pMax]; fs={}; Do[p=Prime[n]; k=1; found=False; While[found=(Mod[Fibonacci[k], p]==0); !found&&k*p<nLimit, k++ ]; If[found, AppendTo[fs, p*k]], {n, nMax}]; fs=Sort[fs]; i=1; While[i<=Length[fs]-1, n=fs[[i]]; If[n>0, j=i+1; While[j<=Length[fs], If[Mod[fs[[j]], n]==0, fs[[j]]=0]; j++ ]]; i++ ]; Select[fs, #>0&&#<nLimit&]
%o A065069 (PARI) is_A065069(n)=!fordiv(n,k,k>1 && k<n!=issquarefree(fibonacci(k)) && return) && n>1 \\
%Y A065069 Cf. A037917 (all indices <= 240 for which Fib(m) is squareful).
%Y A065069 Cf. A065106, A001602, A013929 (not squarefree).
%K A065069 nonn
%O A065069 1,1
%A A065069 _Len Smiley_, Nov 07 2001
%E A065069 One more term from _Robert G. Wilson v_, Nov 08 2001
%E A065069 More terms from _T. D. Noe_, Jul 24 2003