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Natural numbers with number of divisors equal to a Fibonacci number.
10

%I #20 Sep 17 2015 03:32:37

%S 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,

%T 53,54,56,59,61,66,67,70,71,73,78,79,81,83,88,89,97,101,102,103,104,

%U 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

%N Natural numbers with number of divisors equal to a Fibonacci number.

%C Does not contain (for example) 180, so is different from A000028. - _Max Alekseyev_, Sep 20 2007

%C 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

%C 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

%H T. D. Noe, <a href="/A123193/b123193.txt">Table of n, a(n) for n = 1..10000</a>

%t lim = 250; t = Fibonacci /@ Range@ lim; Select[Range@ lim, MemberQ[t, DivisorSigma[0, #]] &] (* _Michael De Vlieger_, Sep 16 2015 *)

%o (PARI) is(n)=my(k=numdiv(n)^2); issquare(k+=(k+1)<<2)||issquare(k-8) \\ _Charles R Greathouse IV_, Sep 16 2015

%Y Complement of A123240. Different from A026416.

%K nonn

%O 1,2

%A _Giovanni Teofilatto_, Oct 04 2006