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

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.

• clarify the input requirement in languages with native arbitrary precision integers? May 9, 2016 at 22:20
• @Sparr I would say, you should at least support up to 2^32 - 1. The rest is up to you and your system. Hope, this is what you meant with your question. May 9, 2016 at 23:16
• I like how the title sums it all up May 10, 2016 at 21:33

# Jelly, 9 bytes

ÆṪÐĿÆFL€S


### Background

The definition of sequence A064097 implies that

$$\small{a(p)-a(p-1)=1}\text{ for all primes }p\text{ and }a(km)=a(k)+a(m)\text{ for all }k,m\in \mathbb{N}$$

$$\small{\varphi(n)=n\prod_{p|n}\left(1-{1\over p}\right)=n\prod_{p|n}{{p-1}\over p}}$$

where $$\\varphi\$$ denotes Euler's totient function and $$\p\$$ varies only over prime numbers.

Combining both, we deduce the property

$$\small{a(\varphi(n))=a\left(n\prod_{p|n}{{p-1}\over p}\right)=a(n)+\sum_{p|n}\left(a(p-1)-a(p)\right)=a(n)-\sum_{p|n}{1=a(n)-\omega(n)}}$$

where $$\\omega\$$ denotes the number of distinct prime factors of $$\n\$$.

Applying the resulting formula $$\k + 1\$$ times, where $$\k\$$ is large enough so that $$\\varphi^{k+1}(n)=1\$$, we get

\small{ \begin{aligned} 0=a\left(\varphi^{k+1}(n)\right)=a\left(\varphi^k(n)\right)-\omega\left(\varphi^k(n)\right)&=a\left(\varphi^{k-1}(n)\right)-\omega\left(\varphi^{k-1}(n)\right)-\omega\left(\varphi^k(n)\right) \\ &= \cdots = a(n)-\sum_{j=0}^k\omega\left(\varphi^j(n)\right) \end{aligned} }

From this property, we obtain the formula

$$\small{a(n)=\sum_{j=0}^k\omega\left(\varphi^j(n)\right)=\sum_{j=0}^{+\infty}\omega\left(\varphi^j(n)\right)}$$

where the last equality holds because $$\\omega(1)=0\$$.

### How it works

ÆṪÐĿÆFL€S  Main link. Argument: n

ÐĿ       Repeatedly apply the link to the left until the results are no longer
unique, and return the list of unique results.
ÆṪ           Apply Euler's totient function.
Since φ(1) = 1, This computes φ-towers until 1 is reached.
ÆF     Break each resulting integer into [prime, exponent] pairs.
L€   Compute the length of each list.
This counts the number of distinct prime factors.

• Now that is a super clever approach ! May 13, 2016 at 15:59
• With the latest version of Jelly, this is 7 bytes, meaning it's still in (joint) first Nov 2, 2020 at 16:42

# 05AB1E, 13 11 bytes

Code:

[DÒ¦P-¼D#]¾


Explanation:

[        ]   # An infinite loop and...
D#        break out of the loop when the value is equal to 1.
D           # Duplicate top of the stack (or in the beginning: duplicate input).
Ò          # Get the prime factors, in the form [2, 3, 5]
¦         # Remove the first prime factor (the smallest one), in order to get
the largest product.
P        # Take the product, [3, 5] -> 15, [] -> 1.
-       # Substract from the current value.
¼      # Add one to the counting variable.
¾  # Push the counting variable and implicitly print that value.


Uses CP-1252 encoding. Try it online!.

• Remove the first prime factor (the smallest one), in order to get the largest product How clever! :-) May 9, 2016 at 13:58
• I see, you are the language developer May 11, 2016 at 12:56
• @SargeBorsch Yes, that is correct :) May 11, 2016 at 16:10
• [¼Ñü-¤ÄD#]¾ - I was close to shaving off a byte with pairwise, oh well... Jan 18, 2017 at 18:57
• Down to 9: [Ð#Ò¦P-]N. You don't need the counter variable. May 22, 2019 at 13:19

# Pyth, 11 bytes

fq1=-Q/QhPQ


Test suite

A straightforward repeat-until-true loop.

Explanation:

fq1=-Q/QhPQ
Implicit: Q = eval(input())
f              Apply the following function until it is truthy,
incrementing T each time starting at 1:
PQ    Take the prime factorization of Q
h      Take its first element, the smallest factor of Q
/Q       Divide Q by that, giving Q's largest factor
-Q         Subtract the result from Q
=           Assign Q to that value
q1            Check if Q is now 1.

• that's a really nice trick with filter. May 9, 2016 at 20:49
• I don't understand why this outputs the number of times the function ran. Is this an undocumented feature of f? May 10, 2016 at 20:35
• @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;} May 13, 2016 at 15:30
• @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...]. May 13, 2016 at 15:31
• 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!: ) May 13, 2016 at 15:48

# Retina, 12

• 14 bytes saved thanks to @MartinBüttner
(1+)(?=\1+$)  This assumes input given in unary and output given in decimal. If this is not acceptable then we can do this for 6 bytes more: # Retina, 18 • 8 bytes saved thanks to @MartinBüttner .+$*
(1+)(?=\1+$)  Try it online - 1st line added to run all testcases in one go. Sadly this uses unary for the calculations, so input of 2016155 is not practical. • The first stage (2 lines) simply converts the decimal input to unary as a string of 1s • The second stage (1 line) calculates the largest factor of n using regex matching groups and lookbehinds to and effectively subtracts it from n. This regex will match as many times as is necessary to reduce the number as far as possible. The number of regex matches will be the number of steps, and is output by this stage. • I don't think you need the \b. May 9, 2016 at 20:31 • You can save a lot more like this though and technically you don't need the first stage either. May 9, 2016 at 20:36 • @MartinBüttner Fantastic! Very elegant - thanks! May 9, 2016 at 20:53 ## Python 2, 50 49 bytes f=lambda n,k=1:2/n or n%(n-k)and f(n,k+1)or-~f(k)  This isn't going to finish that last test case any time soon... Alternatively, here's a 48-byte which returns True instead of 1 for n=2: f=lambda n,k=1:n<3or n%(n-k)and f(n,k+1)or-~f(k)  # Jelly, 10 bytes ÆfḊPạµÐĿi2  Try it online! or verify most test cases. The last test cases finishes quickly locally. ### How it works ÆfḊPạµÐĿi2 Main link. Argument: n (integer) Æf Factorize n, yielding a list of primes, [] for 1, or [0] for 0. Ḋ Dequeue; remove the first (smallest) element. P Take the product. This yields the largest proper divisor if n > 1, 1 if n < 2. ạ Yield the abs. value of the difference of the divisor (or 1) and n. µ Convert the chain to the left into a link. ÐĿ Repeatedly execute the link until the results are no longer unique. Collect all intermediate results in a list. For each starting value of n, the last results are 2 -> 1 -> 0 (-> 1). i2 Compute the 1-based index of 2.  # Pyth - 1514 13 bytes Special casing 1 is really killing me. tl.u-N/Nh+PN2  tl One minus the length of .u Cumulative fixed point operator implicitly on input -N N - /N N / h Smallest prime factor +PN2 Prime factorization of lambda var, with two added to work with 1  • One thing I always forget.... brute force is often the golfiest approach May 9, 2016 at 13:34 • What do you mean with special casing 1? May 9, 2016 at 13:40 • @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. May 9, 2016 at 13:41 • It only needs to accept numbers >= 2 (>1). May 9, 2016 at 22:49 • @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. May 9, 2016 at 22:53 # Julia, 565045 39 bytes f(n)=n>1&&f(n-n÷first(factor(n))[1])+1  This is a recursive function that accepts an integer and returns an integer. Ungolfed: function f(n) if n < 2 # No decrementing necessary return 0 else # As Dennis showed in his Jelly answer, we don't need to # divide by the smallest prime factor; any prime factor # will do. Since factor returns a Dict which isn't # sorted, first doesn't always get the smallest, and # that's okay. return f(n - n ÷ first(factor(n))[1]) + 1 end end  Try it online! (includes all test cases) Saved 6 bytes thanks to Martin Büttner and 11 thanks to Dennis! # JavaScript (ES6), *44 38 Edit 6 bytes saved thanks @l4m2 (* 4 striked is still 4) Recursive function f=(n,d=n)=>n>1?n%--d?f(n,d):f(n-d)+1:0  Less golfed f=(n, d=n-1)=>{ if (n>1) if(n % d != 0) return f(n, d-1) // same number, try a smaller divisor else return f(n-d)+1 // reduce number, increment step, repeat else return 0 }  Test f=(n,d=n)=>n>1?n%--d?f(n,d):f(n-d)+1:0 console.log=x=>O.textContent+=x+'\n'; [5,30,31,32,100,200].forEach(x=>console.log(x+' -> '+f(x))) <pre id=O></pre> • Nice, but I think you should spend the two bytes needed to make f(1) == 0. – Neil May 11, 2016 at 11:39 • @Neil thinking again: no. "Let n be a natural number ≥ 2 ..." May 11, 2016 at 14:18 • I need new glasses. – Neil May 12, 2016 at 13:44 • Why not f=(n,d=n)=>n>1?n%--d?f(n,d):f(n-d)+1:0? – l4m2 Apr 13, 2018 at 3:50 • @l4m2 right, why not? Thanks Apr 13, 2018 at 14:55 # Regex 🐝 (ECMAScript or better), 12 bytes (x+)(?=\1+$)


Takes its input in unary, as a sequence of x characters whose length represents the number. Returns its output as the number of matches.

This of course has already been done in Digital Trauma's Retina answer, but here it is demonstrated to work on a much wider variety of regex engines.

Try it online! - ECMAScript
Try it online! - Perl
Try it online! - Java
Try it online! - PCRE
Try it online! - Boost
Try it online! - Python
Try it online! - Ruby
Try it online! - .NET

(x+)(?=\2+$) # \2 = largest proper divisor of tail; tail -= \2; # return \2 as a match  Try it online! # Regex 🐘 (.NET), 15 bytes ((x+)(?=\2+$))*


Try it online!

Returns its output as the capture count of group \1.

                   # tail = input number; no need to anchor, as all inputs return a result
(
(x+)(?=\2+$) # \2 = largest proper divisor of tail; tail -= \2 )* # Loop the above as many times as possible, minimum zero, and push each # match onto the \1 capture stack  # Regex (Perl / Java / PCRE2 v10.34 or later), 26 bytes ((?=(\2?+(x*))x(x\3)+$)x)*


Returns its output as the length of the match.

Try it online! - Perl
Try it online! - Java
Attempt This Online! - PCRE2 v10.40+

We can't just do a direct port of the .NET version, because we need to subtract at least 1 from tail on each iteration (as a regex loop terminates after a zero-width match), and with most inputs, the iteration count exceeds the cumulative result of subtracting the largest divisor before the last iteration has finished. So instead, we let \2 be the sum of the subtracted divisors minus the iteration count:

                    # tail = input number; no need to anchor, as all inputs return a result
(
(?=
(\2?+(x*))  # \3 = {conjectured largest proper divisor of tail-\2} - 1;
# \2 = {\2, or 0 if \2 is unset} + \3; tail -= \2; note that the
# subtraction of 1 from this divisor compensates for the "tail -= 1"
# on each iteration (below)
x(x\3)+$# assert tail - 1 is divisible by \3 + 1 ) x # head += 1; tail -=1 )* # Loop the above as many times as possible, minimum zero  (The PCRE version requirement is because PCRE2 v10.33 and earlier automatically makes any group containing a nested backreference atomic, thus \3 always captures the entire remaining tail, making the regex always return 0.) # Regex (.NET), 28 bytes (?=((x+)(?=\2+$))*)(?<-1>x)*


Try it online!

A simple port of the 15 byte .NET version, that returns its output as the length of the match instead.

## Regex (Perl / Java / PCRE2 v10.34 or later / .NET), 29 bytes

Obsoleted by the 29 byte regex below, which supports a superset of regex engines.

# Regex (Perl / Java / PCRE / Pythonregex / Ruby / .NET), 33 29 bytes

((?=.*(?=\3$|^)(x+)(\2+$))x)*


Try it online! - Perl
Try it online! - Java
Try it online! - PCRE1
Try it online! - PCRE2 v10.33

Attempt This Online! - PCRE2 v10.40+
Try it online! - Python import regex
Try it online! - Ruby

Try it online! - .NET

For regex engines that lack nested backreferences but support forward-declared backreferences, we switch to a different approach. In this version, each iteration, while inside a lookahead, stores the new value of tail in the capture group \3, so that the next iteration can then recall that value rather directly. Since this directly sets \3 instead of building it up incrementally, there's no need to emulate a nested backreference.

                     # tail = N = input number;
# no need to anchor, as all inputs return a result
(
(?=
.*(?=\3$|^) # tail = \3, or N if \3 is unset (x+)(\2+$)   # \2 = largest proper divisor of tail; \3 = tail - \2
)
x                # Increment the return value
)*                   # Loop the above as many times as possible, minimum zero


# Regex (Perl / PCRE / Pythonregex), 63 bytes

(?=(xx+?)\1*(?=\1$)((?R)))(?=(x+)\3*(?=\3$)((?R)))\2\4|x\B(?R)|


This implements the recursive definition of the function:

$$\a(1) = 0\$$
$$\a(p) = 1 + a(p-1)\$$ if $$\p\$$ is prime
$$\a(nm) = a(n) + a(m)\$$ if $$\m,n > 1\$$

Try it online! - Perl
Try it online! - PCRE1
Try it online! - PCRE2 v10.33
Attempt This Online! - PCRE2 v10.40+
Try it online! - Python import regex

                # tail = input number; no need to anchor, as all inputs return a result
(?=
(xx+?)\1*(?=\1$) # Assert tail isn't prime; # tail = \1 = smallest prime factor of tail ((?R)) # \2 = recursive result of f(tail) ) (?= (x+)\3*(?=\3$)    # tail = \3 = largest proper divisor of tail
((?R))            # \4 = recursive result of f(tail)
)
\2\4                  # return \2 + \4 as the match
|       # or if tail is prime:
x                     # tail -= 1; return result += 1
\B                    # Assert tail > 0
(?R)                  # return 1 + recursive result of f(tail)
|       # or if tail == 1:
# return 0 at the match


Note that (?0) could have been used as a synonym for (?R).

# Regex (Perl / PCRE / Boost / Pythonregex), 65 bytes

((?=(xx+?)\2*(?=\2$)((?1)))(?=(x+)\4*(?=\4$)((?1)))\3\5|x\B(?1)|)


The intent of the 63 byte version was to create a version that works under Boost, but as it turned out Boost has a bug (which I've reported) in which it ignores top-level alternatives in a top-level recursive call (i.e. (?R)), trying only the first top-level alternative. The workaround is to nest the entire regex in a capture group, so that Boost sees the other alternatives. Either (?R) or (?1) would work, but the latter is used for slightly better efficiency.

Try it online! - Perl
Try it online! - PCRE2
Try it online! - Boost
Try it online! - Python import regex

# Regex (Ruby), 76 71 bytes

(?=(xx+?)\1*(?=\1$)(\g<0>))(?=(x+)\3*(?=\3$)(\g<0>))\k<2+0>\4|x\B\g<0>|


Try it online!

This is a port to Ruby's style of recursion; (?R) is replaced with \g<0>. Also, the backreference \2 needs to be replaced with \k<2+0> to get its value at the current level of recursion, rather than its global value which may have been overwritten at deeper levels of recursion.

• That’s awesome. :) Jul 12, 2022 at 17:06

## Mathematica, 36 bytes

f@1=0;f@n_:=f[n-Divisors[n][[-2]]]+1


An unnamed function takes the same bytes:

If[#<2,0,#0[#-Divisors[#][[-2]]]+1]&


This is a very straightforward implementation of the definition as a recursive function.

# Python 3, 75, 70, 67 bytes.

g=lambda x,y=0:y*(x<2)or[g(x-z,y+1)for z in range(1,x)if x%z<1][-1]


This is a pretty straight forward recursive solution. It takes a VERY long time for the high number test cases.

• Same bytecount: f=lambda n,k=0:k*(n<2)or f(n-max(d*(n%d<1)for d in range(1,n)),k+1) Jul 19, 2022 at 22:19

# Matlab, 58 bytes

function p=l(a),p=0;if(a-1),p=1+l(a-a/min(factor(a)));end


# Vyxals, 6 bytes

⁽∆ṫ↔v†


Try it Online!

Port of Jelly.

⁽∆ṫ↔v†
⁽  ↔    # Repeatedly apply until the result does not change, collecting intermediate results:
∆ṫ     #  Euler's totient
v†  # For each, get the number of distinct prime factors
# s flag sums

• I found a way to do this in 6 bytes. Jul 12, 2022 at 23:49

# Vyxall, 6 5 bytes

≬K¯÷ẋ


Try it Online!

This is the same as the 6 byte answer below except the length is taken by the l flag rather than a L element.

# Vyxal, 7 6 bytes

≬K¯÷ẋL


Try it Online!

≬    # 3-element lambda:
K  # Push a list of the divisors, from 1 to the number itself in increasing order
¯  # Deltas - returns a list of the consecutive differences in the list.
# The resulting list has a length of 1 less than the one fed to it.
÷  # Unwrap the list onto the stack. For a non-empty list, this is effectively
# equivalent to t (Tail - get the last item). But for an empty list, the
# result is effectively whatever was already on the stack, i.e. the the
# number whose list of divisors was taken, i.e., 1, the only one that yields
# an empty deltas list. This makes 1, instead of 0, the fixed point.
ẋ    # Repeat the lambda on the number at the top of the stack (which is initially
# the input) until the result no longer changes, returning a list of the
# results. The last result will be 1, our fixed point.
L    # Length


This is very slow for most numbers of more than 60 decimal digits or so, since it has to generate a full list of divisors (even if it only uses the largest two). Prime factorization is still fast enough at that level, but unless the number has only one distinct prime factor, the list of divisors will be orders of magnitude longer.

• @Steffan Nice work. And now I've beaten yours by 1 byte again :-) Jul 13, 2022 at 5:41

## Japt, 12 bytes

@!(UµUk Å×}a


Test it online!

### How it works

@   !(Uµ Uk Å  ×   }a
XYZ{!(U-=Uk s1 r*1 }a
// Implicit: U = input integer
XYZ{               }a  // Return the smallest non-negative integer X which returns
// a truthy value when run through this function:
Uk            //   Take the prime factorization of U.
s1         //   Slice off the first item.
//   Now we have all but the smallest prime factor of U.
r*1     //   Reduce the result by multiplication, starting at 1.
//   This takes the product of the array, which is the
//   largest divisor of U.
U-=              //   Subtract the result from U.
!(                 //   Return !U (which is basically U == 0).
//   Since we started at 0, U == 1 after 1 less iteration than
//   the desired result. U == 0 works because the smallest
//   divisor of 1 is 1, so the next term after 1 is 0.
// Implicit: output result of last expression


This technique was inspired by the 05AB1E answer. A previous version used ²¤ (push a 2, slice off the first two items) in place of Å because it's one byte shorter than s1  (note trailing space); I only realized after the fact that because this appends a 2 to the end of the array and slices from the beginning, it in fact fails on any odd composite number, though it works on all given test cases.

# ><>, 32 bytes

<\?=2:-$@:$/:
1-$:@@:@%?!\ ;/ln  Expects the input number, n, on the stack. This program builds the complete sequence on the stack. As the only number that can lead to 1 is 2, building the sequence stops when 2 is reached. This also causes the size of the stack to equal the number of steps, rather than the number of steps +1. # Ruby, 43 bytes f=->x{x<2?0:1+f[(1..x).find{|i|x%(x-i)<1}]}  Find the smallest number i such that x divides x-i and recurse until we reach 1. # Haskell, 67 bytes Here's the code: a&b|b<2=0|a==b=1+2&(b-1)|mod b a<1=1+2&(b-div b a)|1<2=(a+1)&b (2&)  And here's one reason why Haskell is awesome: f = (2&) (-->) :: Eq a => a -> a -> Bool (-->) = (==) h=[f(5) --> 3 ,f(30) --> 6 ,f(31) --> 7 ,f(32) --> 5 ,f(100) --> 8 ,f(200) --> 9 ,f(2016^155) --> 2015 ]  Yes, in Haskell you can define --> to be equivalent to ==. # Matlab, 107 bytes a=input('');b=factor(a-isprime(a));c=log2(a);while(max(b)>1),b=max(factor(max(b)-1));c=c+1;end,disp(fix(c))  • Non-competing, this is not the iterative translation of my last submission, just another direct algerbraic method, it sums up all binary-logs of all prime factors, kinda ambiguous to illustrate. • I will golf this more when i have time. # MATL, 17 16 bytes tttYfl)/-tq]vnq  Try it Online Explanation  % Implicitly grab input  % Do while loop ttt % Make three copies of top stack element Yf % Compute all prime factors l) % Grab the smallest one / % Divide by this to get the biggest divisor - % Subtract the biggest divisor t % Duplicate the result q % Subtract one (causes loop to terminate when the value is 1). This % is functionally equivalent to doing 1> (since the input will always be positive) % with fewer bytes ] % End do...while loop v % Vertically concatenate stack contents (consumes entire stack) n % Determine length of the result q % Subtract 1 from the length % Implicitly display result  # C99, 62 61 bytes 1 byte golfed off by @Alchymist. f(a,c,b)long*c,a,b;{for(*c=0,b=a;a^1;a%--b||(++*c,b=a-=b));}  Call as f(x,&y), where x is the input and y is the output. • If you test a%--b then you can avoid the b-- at the end. A whole one byte saving. May 11, 2016 at 15:06 # Julia, 39 36 bytes \(n,k=2)=n%k>0?n>1&&n\-~k:\(n-n/k)+1  Try it online! # Clojure, 116 104 bytes (fn[n](loop[m n t 1](let[s(- m(last(filter #(=(rem m %)0)(range 1 m))))](if(< s 2)t(recur s (inc t))))))  -12 bytes by filtering a range to find multiples, then using last one to get the greatest one Naïve solution that basically just solves the problem as it's described by the OP. Unfortunately, finding the greatest divisor alone takes up like half the bytes used. At least I should have lots of room to golf from here. Pregolfed and test: (defn great-divider [n] ; Filter a range to find multiples, then take the last one to get the largest (last (filter #(= (rem n %) 0) (range 1 n)))) (defn sub-great-divide [n] (loop [m n step 1] (let [g-d (great-divider m) ; Find greatest divisor of m diff (- m g-d)] ; Find the difference (println m " is " g-d " --> " m " - " g-d " = " diff) (if (< diff 2) step (recur diff (inc step)))))) (sub-great-divide 30) 30 is 15 --> 30 - 15 = 15 15 is 5 --> 15 - 5 = 10 10 is 5 --> 10 - 5 = 5 5 is 1 --> 5 - 1 = 4 4 is 2 --> 4 - 2 = 2 2 is 1 --> 2 - 1 = 1 6  • @insertusernamehere No, unfortunately, because those are all valid identifiers. I've removed all the possible whitespace. If I want to golf it further, I'll need to rework the algorithm. Jan 18, 2017 at 17:38 # Perl 6, 35 bytes {+({$_ -first $_%%*,[R,] ^$_}...1)}


Try it online!

### How it works

{                                 }   # A bare block lambda.
[R,] ^$_ # Construct range from arg minus 1, down to 0. first$_%%*,                  # Get first element that is a divisor of the arg.
\$_ -                              # Subtract it from the arg.
{                        }...1     # Do this iteratively, until 1 is reached.
+(                              )    # Return the number of values generated this way.


# Stax, 9 7 bytes

Åiw^Å♫┌


Run and debug it

These two operations are identical.

1. Take the biggest divisor of n – which is different from n itself – and subtract it from n.
2. Let d be the smallest prime divisor of n. Calculate n * (d - 1) / d.

I like #2 better, so let's count how many times we can do that one instead. So at each step, we decrement the smallest prime factor. Let's work an example starting with 30.

Step    n   Factors             Operation
0       30  [2, 3, 5]           Replace 2 with 1
1       15  [1, 3, 5] = [3, 5]  Replace 3 with 2
2       10  [2, 5]              Replace 2 with 1
3       5   [1, 5] = [5]        Replace 5 with 4
4       4   [4] = [2, 2]        Replace 2 with 1
5       2   [1, 2] = [2]        Replace 2 with 1
6       1   [1] = []            End of the line


From the example we can see a recursive definition for steps(x) that solves the problem. Presented here in python-ish pseudo-code.

def steps(n):
pf = primeFactors(n)
return sum(steps(d - 1) for d in pf) + 1


The stax program, unpacked into ascii looks like this.

Z       push a zero under the input
|fF     for each prime factor of the input, run the rest of the program
v     decrement the prime factor
G     run the *whole* program from the beginning
+^    add to the running total, then increment


Run this one

• Looks like we have a new winner. A short question: Is there a meta agreement that characters count as bytes on Stax? I would say these are 7 characters and 13 bytes. Aug 17, 2018 at 17:17
• You can read about the character encoding stax uses here. If you doubt the byte count, you can download the program using the down arrow button, and count the bytes yourself. It will show you that the program is 7 bytes. To verify, you can re-upload the program and run it to verify the behavior is the same. Aug 17, 2018 at 19:19

# Jelly, 8 bytes

ÆḌṀạƊƬL’


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ÆḌṀạƊƬL’ - Main link. Takes n on the left