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A123193
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Natural numbers with number of divisors equal to a Fibonacci number.
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10
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1, 2, 3, 4, 5, 7, 9, 11, 13, 16, 17, 19, 23, 24, 25, 29, 30, 31, 37, 40, 41, 42, 43, 47, 49, 53, 54, 56, 59, 61, 66, 67, 70, 71, 73, 78, 79, 81, 83, 88, 89, 97, 101, 102, 103, 104, 105, 107, 109, 110, 113, 114, 121, 127, 128, 130, 131, 135, 136, 137, 138, 139, 149, 151, 152, 154, 157, 163, 165, 167, 169, 170, 173, 174, 179, 181, 182, 184, 186, 189, 190, 191, 193, 195, 197, 199, 211, 222, 223, 227, 229, 230, 231, 232, 233, 238, 239, 241, 246, 248, 250
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OFFSET
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1,2
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COMMENTS
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How dense is this sequence? There are 7 members up to 10^1, 42 up to 10^2, 364 up to 10^3, 3379 up to 10^4, 31864 up to 10^5, 303623 up to 10^6, 2907125 up to 10^7, 27893864 up to 10^8, and 268099330 up to 10^9. - Charles R Greathouse IV, Sep 16 2015
Partial answer: a(n) << n log n/(log log n)^k for any k. Proof: Since 0 is a Fibonacci number, and Fibonacci numbers are periodic mod any number, 2^(k+1) divides infinitely many Fibonacci numbers. Take some positive Fibonacci number F divisible by 2^(k+1). By Landau's theorem there are >> x (log log x)^k/log x odd squarefree numbers divisible by k+1 primes up to x. Multiply each by 2^(F/2^(k+1)-1) which leaves the density unchanged since the expression is constant in k, and note that the products have exactly F divisors. - Charles R Greathouse IV, Sep 16 2015
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LINKS
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MATHEMATICA
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lim = 250; t = Fibonacci /@ Range@ lim; Select[Range@ lim, MemberQ[t, DivisorSigma[0, #]] &] (* Michael De Vlieger, Sep 16 2015 *)
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PROG
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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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