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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
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    \$\begingroup\$ I'm not sure I understand why 2 and 6 were removed from the list of multiples. Can please you clarify the rules? \$\endgroup\$ Commented Jun 9, 2016 at 20:23
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    \$\begingroup\$ @Mattysen, psuedofactorial(N) is the smallest number which has the numbers 1 through N as factors (The least common multiple of those numbers). That is the technical definition, but the way I wrote it was somewhat suggestive that it is similar to a factorial. \$\endgroup\$ Commented Jun 9, 2016 at 20:26
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    \$\begingroup\$ oeis.org/A003418 \$\endgroup\$ Commented Jun 10, 2016 at 0:19
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    \$\begingroup\$ Welcome to Programming Puzzles & Code Golf! This is a nice first challenge! \$\endgroup\$ Commented Jun 10, 2016 at 0:40
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    \$\begingroup\$ Your first challenge got to the top of HNQ. Nice! \$\endgroup\$ Commented Jun 10, 2016 at 0:56

45 Answers 45

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Pyke, 3 bytes

S.L

Try it here!

S - range(1, input+1) .L - lowest_common_multiple(^)
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1
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J-uby, 10 bytes

:+|:/&:lcm

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Explanation

:+ | :/ & :lcm :+ | # Get range 1..n, then :/ & :lcm # reduce with LCM
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1
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Factor, 26 bytes

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

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1
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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)
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1
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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
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1
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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
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1
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Python, 44 bytes

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

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Requires Python 3.9+

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1
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Vyxal, 3 bytes

ɾ∆Ŀ

Run it!

Explanation:

ɾ # numbers from 1 to input ∆Ŀ # least common multiple
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1
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C++, 62 bytes

#include<numeric> int f(int n){return n?std::lcm(n,f(n-1)):1;}

This is portable standard C++ (C++17 or later). It's a simple recursive function:

int f(int n) { if (n == 0) { return 1; } return std::lcm(n, f(n-1)); }

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0
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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 :/

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0
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𝔼𝕊𝕄𝕚𝕟, 7 chars / 9 bytes

Мū…⩤⁽1ï

Try it here (ES6 only).

Just a LCM of inclusive range from 1 to input.

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0
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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.

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0
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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

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0
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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
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0
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Husk, 3 bytes

F⌉ḣ

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same as APL.

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