Largest Prime Exponents

Question

Given an integer n >= 2, output the largest exponent in its prime factorization. This is OEIS sequence A051903.

Example

Let n = 144. Its prime factorization is 2^4 * 3^2. The largest exponent is 4.

Test Cases

2 -> 1
3 -> 1
4 -> 2
5 -> 1
6 -> 1
7 -> 1
8 -> 3
9 -> 2
10 -> 1
11 -> 1
12 -> 2
144 -> 4
200 -> 3
500 -> 3
1024 -> 10
3257832488 -> 3

Answer 1

05AB1E, 2 bytes

ÓZ

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How?

Ó   exponents of prime factors
 Z  maximum

Answer 2

Python 2, 62 57 56 bytes

lambda n:max(k%n-n%(k/n+2)**(k%n)*n for k in range(n*n))

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Answer 3

Jelly, 3 bytes

ÆEṀ

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ÆEṀ  - Full program / Monadic link.

ÆE   - Array of exponents of prime factorization.
  Ṁ  - Maximum.

_{This also works in M. Try it online!}

Answer 4

Haskell, 61 60 50 48 46 bytes

-2 bytes thanks to xnor

f n=maximum[a|k<-[2..n],a<-[1..n],n`mod`k^a<1]

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45 bytes with an import:

import NumberTheory
maximum.map snd.factorize

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Answer 5

Ohm v2, 2 bytes

n↑

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Explanation?

No.

Answer 6

Python 2, 78 bytes

n=input()
e=m=0
f=2
while~-n:q=n%f<1;f+=1-q;e=q*-~e;m=max(m,e);n/=f**q
print m

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-5 thanks to ovs.

This answer doesn't do prime checks. Instead, it takes advantage of the fact that the highest exponent of a prime factor will be greater than or equal to the exponent of any other factor in any factorization of a number.

Answer 7

Japt -h, 9 7 bytes

k ü mÊn

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k ü mÊn     :Implicit input of integer
k           :Prime factors
  ü         :Group by value
    m       :Map
     Ê      :  Length
      n     :Sort
            :Implicit output of last element

Answer 8

Mathematica, 27 bytes

Max[Last/@FactorInteger@#]&

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Answer 9

MATL, 4 bytes

YFX>

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       % implicit input
YF     % Exponents of prime factors
X>     % maximum
       % implicit output

Answer 10

Husk, 5 bytes

▲mLgp

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• p – Gets the prime factors.
• g – Groups adjacent values.
• mL – Gets the lengths of each group.
• ▲ – Maximum.

Answer 11

Brachylog, 5 bytes

ḋḅlᵐ⌉

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Explanation

ḋ          Prime decomposition
 ḅ         Group consecutive equal values
  lᵐ       Map length
    ⌉      Maximum

Answer 12

APL (Dyalog), 19 bytes

⎕CY'dfns'
⌈/1↓2pco⎕

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How?

2pco⎕ - 2D array of prime factors and exponents

1↓ - drop the factors

⌈/ - maximum

Answer 13

Javascript 54 bytes

*assuming infinite stack (as do in code-golf challenges)

P=(n,i=2,k)=>i>n?k:n%i?k>(K=P(n,i+1))?k:K:P(n/i,i,-~k)

console.log(P(2 )== 1)
console.log(P(3 )== 1)
console.log(P(4 )== 2)
console.log(P(5 )== 1)
console.log(P(6 )== 1)
console.log(P(7 )== 1)
console.log(P(8 )== 3)
console.log(P(9 )== 2)
console.log(P(10 )== 1)
console.log(P(11 )== 1)
console.log(P(12 )== 2)
console.log(P(144 )== 4)
console.log(P(200 )== 3)
console.log(P(500 )== 3)
console.log(P(1024 )== 10)
//console.log(P(3257832488 )== 3)

Answer 14

PARI/GP, 24 bytes

n->vecmax(factor(n)[,2])

If I do not count the n-> part, it is 21 bytes.

Answer 15

Octave, 25 bytes

@(n)[~,m]=mode(factor(n))

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Explanation

factor produces the array of (possibly repeated) prime exponents The second output of mode gives the number of times that the mode (i.e. the most repeated entry) appears.

Answer 16

Pyth, 7 bytes

eShMr8P

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Answer 17

Python 2, 90 84 bytes

f=lambda n,i=2,l=[0]:(n<2)*max(map(l.count,l))or n%i and f(n,i+1,l)or f(n/i,2,l+[i])

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Answer 18

Gaia, 4 bytes

ḋ)⌠)

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• ḋ - Computes the prime factorization as [prime, exponent] pairs.

  • ⌠ - Map and collect the result with the maximal value.

  • ) - Last element (exponent).

  • ) - Last element (maximal exponent)

Gaia, 4 bytes

ḋ)¦⌉

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• ḋ - Computes the prime factorization as [prime, exponent] pairs.

  • )¦ - Map with the last element (exponent).

  • ⌉ - Gets the maximum element.

Answer 19

MY, 4 bytes

ωĖ⍐←

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Explanation?

ωĖ⍐←
ω    = argument
 Ė   = prime exponents
  ⍐  = maximum
   ← = output without a newline

Answer 20

Octave: 30 bytes

@(x)max(histc(a=factor(x),a));

1. a=factor(x) returns a vector containing the prime factors of x. This is a vector sorted in ascending order where the multiplication of all numbers in factor(x) yields x itself such that each number in the vector is prime.
2. histc(...,a) calculates a histogram on the prime factor vector where the bins are the prime factors. The histogram counts up how many times we have seen each prime number thus yielding the exponent of each prime number. We can cheat here a bit because even though factor(x) will return duplicate numbers or bins, only one of the bins will capture the total amount of times we see a prime number.
3. max(...) thus returns the largest exponent.

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Answer 21

Alice, 17 bytes

/o
\i@/w].D:.t$Kq

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Explanation

/o
\i@/...

This is just a framework for simple-ish arithmetic programs with decimal I/O. The ... is the actual program, which already has the input on the stack and leaves the output on top of the stack.

Alice actually has built-ins to get the prime factorisation of an integer (even with prime-exponent pairs), but the shortest I've come up with using those is 10 bytes longer than this.

Instead the idea is that we repeatedly divide one copy of each distinct prime factor out of the input, until we reach 1. The number of steps this takes is equal to the largest prime exponent. We'll be abusing the tape head as the counter variable.

w      Remember the current IP position. Effectively starts a loop.
  ]      Move the tape head to the right, which increments our counter.
  .D     Duplicate the current value, and deduplicate its prime factors.
         That means, we'll get a number which is the product of the value's
         unique prime factors. For example 144 = 2^4 * 3^2 would become
         6 = 2 * 3.
  :      Divide the value by its deduplicated version, which decrements the
         exponents of its prime factors.
  .t     Duplicate the result and decrement it. This value becomes 0 once we
         reach a result of 1, which is when we want to terminate the loop.
$K     Jump back to the beginning of the loop if the previous value wasn't 0.
q      Retrieve the tape head's position, i.e. the number of steps we've taken
       through the above loop.

Answer 22

Julia, 60 52 40 bytes

f(x)=maximum(collect(values(factor(x))))

-12 +correction thanks to Steadybox

Answer 23

Actually, 4 bytes

w♂NM

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w♂NM  - Full program.

w     - Pushes the prime factorization as [prime, exponent] pairs.
 ♂N   - Get the last element of each (the exponents).
   M  - Maximum.

Answer 24

Python 2, 64 bytes

-4 bytes thanks to H.PWiz.

lambda n:max(a*(n%k**a<1)for a in range(n)for k in range(2,-~n))

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Port of H.PWiz's Haskell answer. I'm only sharing this because I'm proud that I was able to understand this piece of Haskell code and translate it. :P

Answer 25

Axiom, 61 bytes

f n==(a:=factors n;reduce(max,[a.i.exponent for i in 1..#a]))

This is the first time i find it is possible define the function without the use of () parenthesis. Instead of "f(n)==" "f n==" one character less...

Answer 26

Racket, 83 79 bytes

(λ(n)(cadr(argmax cadr((let()(local-require math/number-theory)factorize)n))))

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(I'm not sure if there's a consensus on what constitutes a complete Racket solution, so I'm going with the Mathematica convention that a pure function counts.)

How it works

factorize gives the factorization as a list of pairs: (factorize 108) gives '((2 2) (3 3)). The second element of a pair is given by cadr, a shorthand for the composition of car (head of a list) with cdr (tail of a list).

I feel silly doing (cadr (argmax cadr list)) to find the maximum of the second elements, but max doesn't work on lists: (max (map cadr list)) doesn't do what we want. I'm not an expert in Racket, so maybe there's a standard better way to do this.

Racket, 93 bytes

(λ(n)(define(p d m)(if(=(gcd m d)d)(+(p d(/ m d))1)0))(p(argmax(λ(d)(p d n))(range 2 n))n))

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How it works

An alternative version that doesn't import factorize and instead does everything from scratch, more or less. The function (p m d) finds the highest power of d that divides m and then we just find highest value of (p n d) for d between 2 and n. (We don't need to restrict this to primes, since there will not be a composite power that works better than prime powers.)

Answer 27

J, 9 bytes

[:>./_&q:

Max of <./ all prime exponents _&q:

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Answer 28

APL(NARS), 15 chars, 30 bytes

{⌈/+/¨v∘=¨v←π⍵}

test:

  f←{⌈/+/¨v∘=¨v←π⍵}
  f¨2..12
1 1 2 1 1 1 3 2 1 1 2 
  f¨144 200 500 1024 3257832488
4 3 3 10 3 

comment:

{⌈/+/¨v∘=¨v←π⍵}
          v←π⍵    π12 return 2 2 3; assign to v the array of prime divisors of argument ⍵
      v∘=¨        for each element of v, build one binary array, show with 1 where are in v array, else puts 0 
                  return one big array I call B, where each element is the binary array above
   +/¨            sum each binary element array of  B
 ⌈/               get the max of all element of B (that should be the max exponet)