Excessive Integers

Question

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.

• Luis Mendo: Maybe clarify that ^ is power
• flawr: 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.

Answer 1

MATL, (削除) 7 (削除ここまで) 5 bytes

Yfd~s

Try it online! Or verify all test cases.

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

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

Answer 2

Brachylog, 11 bytes

$pPd:Pr:la-

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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

Answer 3

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.

Answer 4

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

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

Answer 5

J, (削除) 11 (削除ここまで) 10 bytes

Saved 1 byte thanks to Jonah.

1#.1-~:@q:

• Jonah: 1#.1-~:@q: for 10 bytes. nice idea using ~: btw.

Answer 6

05AB1E, 6 bytes

Òg1fg-

Explanation

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

Try it online!

Answer 7

Haskell, 65 bytes

(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

• user58988: if that is one function: who is the input variable? who is the output? thank u...
• Damien: (%) 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

Answer 8

05AB1E, 4 bytes

Ò\_O

Port of @LuisMendo's MATL answer.

Try it online or verify the first 15 test cases.

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)

Answer 9

Husk, (削除) ⁵ (削除ここまで) 4 bytes

ΣẊ=p

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-1 byte from Dominic Van Essen.

• Dominic van Essen: 4 bytes...

Answer 10

Pyth, 7 bytes

-lPQl{P

Try it online.

Answer 11

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!

Answer 12

Python 2, (削除) 57 (削除ここまで) 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.

• Jonathan Allan: Ah nice - you can save a byte by using n/k%k<1
• Dennis: Right, n is already divisible by k at that point. Thanks!

Answer 13

Python 2, (削除) 100 (削除ここまで) (削除) 99 (削除ここまで) (削除) 98 (削除ここまで) 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 (削除) 1 (削除ここまで) 3 bytes!

Answer 14

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

Answer 15

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.

• Rohan Jhunjhunwala: So so close to beating C :(

Answer 16

Javascript (ES6), (削除) 53 (削除ここまで) (削除) 51 (削除ここまで) 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(','));

• Neil: 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.

Answer 17

Bash, 77 bytes

IFS=$'\n '
f=(`factor 1ドル`)
g=(`uniq<<<"${f[*]}"`)
echo $((${#f[*]}-${#g[*]}))

Complete program, with input in 1ドル 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.

Answer 18

Japt, 6 bytes

k äÎxÌ

Try it

k äÎxÌ :Implicit input of integer
k :Prime factors
 ä :Consecutive pairs reduced by
 Î : Signs of differences (0 or -1)
 x :Sum of
 Ì : Signs of differences with -1 (0 -> 1 and -1 -> 0)

Answer 19

Factor + math.primes.factors math.unicode, 40 bytes

[ group-factors unzip Σ swap length - ]

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 ! 12
group-factors ! { { 2 2 } { 3 1 } }
unzip ! { 2 3 } { 2 1 }
Σ ! { 2 3 } 3
swap ! 3 { 2 3 }
length ! 3 2
- ! 1

Answer 20

Pyt, 4 bytes

ḋ−¬Ʃ

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Port of Leaky Nun's Jelly answer

Answer 21

Haskell, 87 bytes

import Data.Numbers.Primes
import Data.List
x=(\x->length x-length(nub x)).primeFactors

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Answer 22

Kamilalisp, 15 bytes

[-&[⍴] #0 ⊙]@⋔⌹

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Answer 23

Nekomata, 3 bytes

ƒ←∑

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ƒ Factor; returns a list of unique prime factors and a list of their exponents
 ← Decrement the exponents
 ∑ Sum

Answer 24

Ruby -rprime, 31 bytes

->n{n.prime_division.sum{_2-1}}

Attempt This Online!

Answer 25

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.

Answer 26

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

Answer 27

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]
 */

Answer 28

GNU sed + coreutils, 55 bytes

(including +1 for -r flag)

s/^/factor /e
s/ ([^ ]+)(( 1円)*)/2円/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円)*)/2円/g
# count just the spaces
s/[^ ]//g
y/ /1/

Answer 29

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 

Answer 30

Arturo, 47 bytes

$=>[k:tally factors.prime&(∑values k)-size k]

Try it!