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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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  • \$\begingroup\$ that's a really nice trick with filter. \$\endgroup\$ Commented May 9, 2016 at 20:49
  • 3
    \$\begingroup\$ I don't understand why this outputs the number of times the function ran. Is this an undocumented feature of f? \$\endgroup\$ Commented May 10, 2016 at 20:35
  • \$\begingroup\$ @corsiKa f without a second argument iterates over all positive integers starting from 1 and returns the first value that gives true on the inner statement. This value happens to be unused in this program, so it returns the number of times that it ran. Not undocumented, just unorthodox :) If it helps, you can think of this as a for loop like: for(int i=1; some_condition_unrelated_to_i; i++) { change_stuff_that_affects_condition_but_not_i;} \$\endgroup\$ Commented May 13, 2016 at 15:30
  • \$\begingroup\$ @corsiKa It's documented in the character reference on the right side of the online interpreter. With only one argument (f <l:T> <none>), f is First input where A(_) is truthy over [1, 2, 3, 4...]. \$\endgroup\$ Commented May 13, 2016 at 15:31
  • \$\begingroup\$ Ah I understand it now. It uses that input but never uses the input in the calculation. That explains @Maltysen comment of "that's a really nice trick" because you only care about the iteration count not using that count anywhere in your filter. I love those ah-ha moments!: ) \$\endgroup\$ Commented May 13, 2016 at 15:48

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