# Excessive Integers

For a positive integer n with the prime factorization n = p1^e1 * p2^e2 * ... pk^ek where p1,...,pk are primes and e1,...,ek are positive integers, we can define two functions:

• Ω(n) = e1+e2+...+ek the number of prime divisors (counted with multiplicity) (A001222)
• ω(n) = k the number of distinct prime divisors. (A001221)

With those two functions we define the excess e(n) = Ω(n) - ω(n) (A046660). This can be considered as a measure of how close a number is to being squarefree.

### Challenge

For a given positive integer n return e(n). ### Examples

For n = 12 = 2^2 * 3 we have Ω(12) = 2+1 and ω(12) = 2 and therefore e(12) = Ω(12) - ω(12) = 1. For any squarefree number n we obivously have e(n) = 0. The first few terms are

1       0
2       0
3       0
4       1
5       0
6       0
7       0
8       2
9       1
10      0
11      0
12      1
13      0
14      0
15      0


Some more details in the OEIS wiki.

• Maybe clarify that ^ is power – Luis Mendo Aug 28 '16 at 12:08
• This is not necessary in my opinion. This symbol is used here and all over the internet, as well as on many calculators and in many programming languages. – flawr Aug 28 '16 at 12:11

# MATL, 7 5 bytes

Yfd~s


### Explanation

Yf    % Implicit input. Obtain prime factors, sorted and with repetitions
d     % Consecutive differences
~     % Logical negate: zeros become 1, nonzeros become 0
s     % Sum. Implicit display

• I was not aware how factor works in MATL, really cool=) – flawr Aug 28 '16 at 12:19
• @flawr Do you mean YF (in the 7-byte version of the code) or Yf (5-byte)? The latter is as in MATLAB – Luis Mendo Aug 28 '16 at 12:20
• The function for the exponents, the 5 byte now is even more clever=) – flawr Aug 28 '16 at 12:22

# Brachylog, 11 bytes

$pPd:Pr:la-  Try it online! ### Explanation $pP            P is the list of prime factors of the Input
Pd           Remove all duplicates in P
:Pr        Construct the list [P, P minus duplicates]
:la     Apply "length" to the two elements of that list
-    Output is the subtraction of the first element by the second one


# Mathematica, 23 bytes

PrimeOmega@#-PrimeNu@#&


Very boring. FactorInteger already takes up 13 bytes, and I can't see much that can be done with the remaining 10.

# Jelly, 5 bytes

ÆfI¬S


Try it online!

Verify all testcases.

Port of Luis Mendo's answer in MATL.

ÆfI¬S

Æf     Implicit input. Obtain prime factors, sorted and with repetitions
I    Consecutive differences
¬   Logical negate: zeros become 1, nonzeros become 0
S  Sum. Implicit display

• For the previous approach, ÆF’SṪ would have worked I think – Sp3000 Aug 28 '16 at 12:31
• @Sp3000 You should post that – Leaky Nun Aug 28 '16 at 12:32
• @LeakyNun I was trying to port it myself, but the definition of ¬ confused me. I didn't know it vectorized – Luis Mendo Aug 28 '16 at 12:42
• @LuisMendo Indeed the Jelly docs are messy. – Leaky Nun Aug 28 '16 at 12:45

# 05AB1E, 6 bytes

Òg¹fg-


Explanation

Òg      # number of prime factors with duplicates
-  # minus
¹fg   # number of prime factors without duplicates


Try it online!

# J, 11 10 bytes

Saved 1 byte thanks to Jonah.

1#.1-~:@q:

• 1#.1-~:@q: for 10 bytes. nice idea using ~: btw. – Jonah Feb 4 '19 at 17:32

# Pyth, 7 bytes

-lPQl{P


Try it online.

# C, 74 bytes

f(n,e,r,i){r=0;for(i=2;n>1;i++,r+=e?e-1:e)for(e=0;n%i<1;e++)n/=i;return r;}


Ideone it!

# Python 2, 57 56 bytes

f=lambda n,k=2:n/k and[f(n,k+1),(n/k%k<1)+f(n/k)][n%k<1]


Thanks to @JonathanAllan for golfing off 1 byte!

Test it on Ideone.

• Ah nice - you can save a byte by using n/k%k<1 – Jonathan Allan Aug 28 '16 at 22:40
• Right, n is already divisible by k at that point. Thanks! – Dennis Aug 28 '16 at 23:01

(c%x)n|x>n=c|mod n x>0=c%(x+1)$n|y<-div n x=(c+0^mod y x)%x$y
0%2

• if that is one function: who is the input variable? who is the output? thank u... – RosLuP Aug 30 '16 at 21:09
• (%) takes 3 input variables : an accumulator (c), an integer (x) and an integer (n) . It returns the excess of (n) when c is set to 0 and x to 2. So (0%2) is a partial function that takes n and returns its excess – Damien Aug 31 '16 at 5:58

# 05AB1E, 4 bytes

Ò¥_O


Explanation:

Ò       # Get all prime factors with duplicates from the (implicit) input
# (automatically sorted from lowest to highest)
¥      # Get all deltas
_     # Check if it's equal to 0 (0 becomes 1; everything else becomes 0)
O    # Take the sum (and output implicitly)


# Python 2, 1009998 96 bytes

n=input()
i=2
f=[]
while i<n:
if n%i:i+=1
else:n/=i;f+=i,
if-~n:f+=n,
print len(f)-len(set(f))


Most of the code is taken up by a golfed version of this SO answer, which stores the prime factors of the input in f. Then we simply use set manipulation to calculate the excess factors.

Thanks to Leaky Nun for saving 1 3 bytes!

# Brachylog, 11 bytes

$p@b:la:-a+  Try it online! Verify all testcases. (The wrapper is longer than the function...) $p@b:la:-a+

$p prime factorization @b group blocks of equal elements :la length of each :-a each minus 1 + sum  # S.I.L.O.S, 113 bytes readIO t=2 lbla e=0 GOTO b lblc i/t e+1 lblb m=i m%t n=1 n-m if n c d=e d/d e-d r+e A=i A-1 t+1 if A a printInt r  Try it online! A port of my answer in C. • So so close to beating C :( – Rohan Jhunjhunwala Aug 28 '16 at 15:55 ## Javascript (ES6), 5351 46 bytes e=(n,i=2)=>i<n?n%i?e(n,i+1):e(n/=i,i)+!(n%i):0  Saved 5 bytes thanks to Neil Example: e=(n,i=2)=>i<n?n%i?e(n,i+1):e(n/=i,i)+!(n%i):0 // computing e(n) for n in [1, 30] for(var n = 1, list = []; n <= 30; n++) { list.push(e(n)); } console.log(list.join(',')); • You can save 5 bytes by calculating r recursively: f=(n,i=2)=>i<n?n%i?f(n,i+1):f(n/=i,i)+!(n%i):0. – Neil Aug 28 '16 at 17:09 # Bash, 77 bytes IFS=$'\n '
f=(factor $1) g=(uniq<<<"${f[*]}")
echo $((${#f[*]}-${#g[*]}))  Complete program, with input in $1 and output to stdout.

We set IFS to begin with a newline, so that the expansion "\${f[*]}" is newline-separated. We use arithmetic substitution to print the difference between the number of words in the factorisation with the result of filtering through uniq. The number itself is printed as a prefix by factor, but it is also present after filtering, so falls out in the subtraction.

# Python, (with sympy) 66 bytes

import sympy;lambda n:sum(x-1for x in sympy.factorint(n).values())


sympy.factorint returns a dictionary with factors as keys and their multiplicities as values, so the sum of the values is Ω(n) and the number of values is ω(n), so the sum of the decremented values is what we want.

# CJam, 11 bytes

ri_mf,\mF,-


Try it online!

Explanation

ri_         Get an integer from input and duplicate it
mf,      Get the number of prime factors (with repetition)
\     Swap top 2 elements on the stack
mF,  Get the number of prime factors (with exponents)
- Subtract


C, 158

#define G(a,b) if(a)goto b
#define R return
f(n,i,j,o,w){if(!n)R 0;o=w=i=j=0;a:i+=2;b:if(n%i==0){++j;G(n/=i,b);}o+=!!j;w+=j;i+=(i==2);j=0;G(i*i<n,a);R w-o;}


In the start there is the goto instruction... even if this is more long than yours it is more readable and right [if i not consider n too much big...] one Language that has 10000 library functions is wrost than a Language that with few, 20 or 30 library functions can do all better [because we can not remember all these functions]

#define F for
#define P printf

main(i,r)
{F(i=0; i<100; ++i)
r=f(i,0,0,0,0),P("[%u|%u]",i,r);
R  0;
}

/*
158
[0|0][1|0][2|0][3|0][4|1][5|0][6|0][7|0][8|2]
[9|0][10|0][11|0][12|1][13|0][14|0][15|0][16|3]
[17|0][18|0][19|0][20|1][21|0][22|0][23|0][24|2][25|1][26|0][27|0] [28|1]
[29|0][30|0][31|0][32|4][33|0][34|0][35|0][36|1]
[37|0][38|0][39|0][40|2][41|0]
*/


# GNU sed + coreutils, 55 bytes

(including +1 for -r flag)

s/^/factor /e
s/ ([^ ]+)(( \1)*)/\2/g
s/[^ ]//g
y/ /1/


Input in decimal, on stdin; output in unary, on stdout.

## Explanation

#!/bin/sed -rf

# factor the number
s/^/factor /e
# remove first of each number repeated 0 or more times
s/ ([^ ]+)(( \1)*)/\2/g
# count just the spaces
s/[^ ]//g
y/ /1/


# APL(NARS) 35 chars, 70 bytes

{⍵=1:0⋄k-⍨+/+/¨{w=⍵⊃v}¨⍳k←≢v←∪w←π⍵}


the function π find the factorization in prime of its argument; there is few to say it seems clear, but for me does more operations (from factorization) than the minimum...the range of count characters is out golfing languages because it seems too much count, but less than not golfing languages... test:

  f←{⍵=1:0⋄k-⍨+/+/¨{w=⍵⊃v}¨⍳k←≢v←∪w←π⍵}
f¨1..15
0 0 0 1 0 0 0 2 1 0 0 1 0 0 0