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Code Golf

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How many steps does it take from n to 1 by subtracting the greatest divisor?

Inspired by this question over at Mathematics.

The Problem

Let n be a natural number ≥ 2. Take the biggest divisor of n – which is different from n itself – and subtract it from n. Repeat until you get 1.

The Question

How many steps does it take to reach 1 for a given number n ≥ 2.

Detailed Example

Let n = 30.

The greatest divisor of:

1. 30 is 15 --> 30 - 15 = 15
2. 15 is 5 --> 15 - 5 = 10
3. 10 is 5 --> 10 - 5 = 5
4. 5 is 1 --> 5 - 1 = 4
5. 4 is 2 --> 4 - 2 = 2
6. 2 is 1 --> 2 - 1 = 1

It takes 6 steps to reach 1.

Input

  • Input is an integer n, where n ≥ 2.
  • Your program should support input up to the language's maximum integer value.

Output

  • Simply output the number of steps, like 6.
  • Leading/trailing whitespaces or newlines are fine.

Examples

f(5) --> 3
f(30) --> 6
f(31) --> 7
f(32) --> 5
f(100) --> 8
f(200) --> 9
f(2016^155) --> 2015

Requirements

  • You can get input from STDIN, command line arguments, as function parameters or from the closest equivalent.
  • You can write a program or a function. If it is an anonymous function, please include an example of how to invoke it.
  • This is so shortest answer in bytes wins.
  • Standard loopholes are disallowed.

This series can be found on OEIS as well: A064097

A quasi-logarithm defined inductively by a(1) = 0 and a(p) = 1 + a(p-1) if p is prime and a(n*m) = a(n) + a(m) if m,n > 1.

Answer*

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  • 1
    \$\begingroup\$ One thing I always forget.... brute force is often the golfiest approach \$\endgroup\$ Commented May 9, 2016 at 13:34
  • \$\begingroup\$ What do you mean with special casing 1? \$\endgroup\$ Commented May 9, 2016 at 13:40
  • 1
    \$\begingroup\$ @Adnan the prime factorization of 1 is [], which causes an error when I take the first element. I have to special case it to make it return 1 again so the .u fixed-point ends. I found a better way than .x try-except which is what saved me those 2 bytes. \$\endgroup\$ Commented May 9, 2016 at 13:41
  • \$\begingroup\$ It only needs to accept numbers >= 2 (>1). \$\endgroup\$ Commented May 9, 2016 at 22:49
  • \$\begingroup\$ @SolomonUcko you're misunderstanding, the .u fixed-point will eventually reach 1 for all input, at which point it will have to be special cased. \$\endgroup\$ Commented May 9, 2016 at 22:53

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