A123193 Natural numbers with number of divisors equal to a Fibonacci number.

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

Offset: 1

Author

Giovanni Teofilatto, Oct 04 2006

Keywords

• nonn

Comments

Does not contain (for example) 180, so is different from A000028. - Max Alekseyev, Sep 20 2007

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

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000

Crossrefs

Complement of A123240. Different from A026416.

Programs

Mathematica

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

PARI

is(n)=my(k=numdiv(n)^2); issquare(k+=(k+1)<<2)||issquare(k-8) \\ Charles R Greathouse IV, Sep 16 2015