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A152012
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Indices of Fibonacci numbers having exactly one primitive prime factor.
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6
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3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 26, 28, 29, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 43, 45, 47, 48, 51, 52, 54, 56, 60, 62, 63, 65, 66, 72, 74, 75, 76, 82, 83, 93, 94, 98, 105, 106, 108, 111, 112, 119, 121, 122, 123, 124, 125, 131
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OFFSET
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1,1
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COMMENTS
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It is known that Fibonacci number A000045(n) has a primitive prime factor for all n, except n=0, 1, 2, 6 and 12. This sequence lists such indices n that A000045(n) has exactly one primitive prime factor (equal A001578(n)). Sister sequence A152013 provides indices of Fibonacci numbers with at least 2 prime factors. The current sequence A152012 and its sister sequence A152013 along with the finite set {0,1,2,6,12} form a partition of the natural numbers.
Numbers n such that A086597(n) = 1.
For prime p, all prime factors of Fibonacci(p) are primitive. Hence, the only primes in this sequence are the primes numbers in A001605, which gives the indices of prime Fibonacci numbers.
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LINKS
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MATHEMATICA
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primitivePrimeFactors[n_] := Cases[FactorInteger[Fibonacci[n]][[All, 1]], p_ /; And @@ (GCD[p, #] == 1 & /@ Array[Fibonacci, n-1])]; Reap[For[n=3, n <= 200, n++, If[Length[primitivePrimeFactors[n]] == 1, Print[n]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Dec 12 2014 *)
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PROG
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(PARI) isok(pf, vp) = sum(i=1, #pf, vecsearch(vp, pf[i]) == 0) == 1;
lista(nn) = {vp = []; for (n=3, nn, pf = factor(fibonacci(n))[, 1]; if (isok(pf, vp), print1(n, ", ")); vp = vecsort(concat(vp, pf), , 8); ); } \\ Michel Marcus, Nov 29 2014
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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