Pseudofactorial

Question

There is a rather curious number which shows up sometimes in math problems or riddles. The pseudofactorial(N) is the least (i.e. lowest) common multiple of the numbers 1 through N; in other words, it's the lowest number which has all numbers from 1 through N as factors.

For instance pseudofactorial(7) = 3 * 4 * 5 * 7, which is the same as 7! except that 2 and 6 have been removed because they are contained in other terms.

Write a program to calculate pseudofactorial(N) and as always, shortest code wins.

Here is a short list for your use. More test cases can be found in the OEIS under A003418.

Factorial:

1. 1
2. 2
3. 6
4. 24
5. 120
6. 720
7. 5040

Pseudofactorial:

1. 1
2. 2
3. 6
4. 12
5. 60
6. 60
7. 420

Answer 1

J-uby, 10 bytes

:+|:/&:lcm

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Explanation

:+ | :/ & :lcm

:+ |            # Get range 1..n, then
     :/ & :lcm  # reduce with LCM

Answer 2

Factor, 26 bytes

[ [1,b] 1 [ lcm ] reduce ]

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Answer 3

05AB1E, 3 bytes

L.¿

Pretty similar as most golfing languages.

Try it online or verify all test cases.

Explanation:

L    # Push a list in the range [1, (implicit) input]
 .¿  # Pop and push the LCM (Least Common Multiple) of this list
     # (which is output implicitly as result)

Answer 4

Pyt, 9 bytes

Đ`Đ↔Ĺ↔⁻ł+

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Đ            implicit input; Đuplicate
 `     ł     do... while top of stack is truthy
  Đ          Đuplicate top of stack
   ↔         flip stack
    Ĺ        get ĹCM of top two on stack
     ↔       flip stack
      ⁻      decrement
        +    add (removes pesky 0); implicit print

Answer 5

Thunno, \$ 5 \log_{256}(96) \approx \$ 4.12 bytes

R1+Al

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Explanation

R      # range(0, input)
 1+    # plus one
   Al  # LCM of list

Answer 6

Python, 44 bytes

lambda n:math.lcm(*range(1,n+1))
import math

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Requires Python 3.9+

Answer 7

Vyxal, 3 bytes

ɾ∆Ŀ

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Explanation:

ɾ   # numbers from 1 to input
 ∆Ŀ # least common multiple

Answer 8

Thunno 2, 2 bytes

Rŀ

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Explanation

    # Implicit input
R   # Range [1..input]
 ŀ  # Reduced by LCM
    # Implicit output

Answer 9

Hoon, 67 bytes

|*
*
(roll (gulf 1 +<) |=({a/@ b/_1} (div (mul a b) d:(egcd a b))))

Create the list [1..n], fold over the list with lcm. Unfortunately, the Hoon stdlib doesn't have a pre-built one I can use :/

Answer 10

𝔼𝕊𝕄𝕚𝕟, 7 chars / 9 bytes

Мū…⩤⁽1ï

Try it here (ES6 only).

Just a LCM of inclusive range from 1 to input.

Answer 11

AWK, 42 bytes

{for(x=n=1;n<=$1;)if(x%n++){x++;n=1}$0=x}1

Command line usage:

awk '{for(x=n=2;n<=$1;)if(x%n++){x++;n=2}$0=x}1' <<< NUM

I didn't see an AWK solution and a duplicate of the question just got posted yesterday, so I thought I'd throw this together. It's rather slow solving for 19 or larger on my box, but it works.

Answer 12

Axiom, 35 bytes

f(x)==reduce(lcm,[i for i in 1..x])

test code and results

(25) -> [[i,f(i)] for i in [1,6,19,22,30]]
   (25)  [[1,1],[6,60],[19,232792560],[22,232792560],[30,2329089562800]]
                                                  Type: List List Integer

i just make the solution of Find the smallest positive integer which has all integers from 1 to n as factors becouse you say it is douplicate i post it here

Answer 13

8th, 23 bytes

Code

1 ' lcm rot 2 swap loop

This code leaves resulting pseudofactorial on TOS

Usage and example

ok> 7 1 ' lcm rot 2 swap loop .
420

Or more clearly

ok> : pseudofact 1 ' n:lcm rot 2 swap loop ;

ok> 7 pseudofact .
420

Answer 14

Husk, 3 bytes

F⌉ḣ

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same as APL.