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A065069
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Numbers n such that Fibonacci(n) is not squarefree, but for all proper divisors k of n, Fibonacci(k) is squarefree.
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5
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6, 25, 56, 91, 110, 153, 406, 703, 752, 820, 915, 979, 1431, 1892, 2147, 2701, 2943, 3029, 3422, 4378, 4556, 4753, 4970, 5513, 6394, 7868, 8841, 9453, 10712, 12403, 13508, 13546, 15051, 16256, 17030, 17267, 18023, 18721, 19503, 20827, 21206
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OFFSET
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1,1
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COMMENTS
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These are first primitive indices m for which Fib(m) is squareful. Note that Fib(km) is divisible by Fib(m).
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
Terms after a(12) are conjectures until the factorizations of F(1271), F(1273), etc. are completed. - Charles R Greathouse IV, Feb 02 2014
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
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LINKS
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EXAMPLE
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a(1) = 6 because 2^2 divides Fibonacci(6) but no smaller Fibonacci number.
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MATHEMATICA
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<< 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} ]
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&]
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PROG
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(PARI) is_A065069(n)=!fordiv(n, k, k>1 && k<n!=issquarefree(fibonacci(k)) && return) && n>1 \\
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CROSSREFS
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Cf. A037917 (all indices <= 240 for which Fib(m) is squareful).
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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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