Compute the Wilson numbers

Question

Given a positive integer n, compute the n^th Wilson number W(n) where

and e = 1 if n has a primitive root modulo n, otherwise e = -1. In other words, n has a primitive root if there does not exist an integer x where 1 < x < n-1 and x² = 1 mod n.

• This is code-golf so create the shortest code for a function or program that computes the n^th Wilson number for an input integer n > 0.
• You may use either 1-based or 0-based indexing. You may also choose to output the first n Wilson numbers.
• This is the OEIS sequence A157249.

Test Cases

n  W(n)
1  2
2  1
3  1
4  1
5  5
6  1
7  103
8  13
9  249
10 19
11 329891
12 32
13 36846277
14 1379
15 59793
16 126689
17 1230752346353
18 4727
19 336967037143579
20 436486
21 2252263619
22 56815333
23 48869596859895986087
24 1549256
25 1654529071288638505

Answer 1

Jelly, 8 bytes

Rg=1TP‘:

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You don't really have to compute e since you need to divide anyway.

Answer 2

Husk, 11 bytes

S÷ȯ→Π§foε⌋ḣ

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Explanation

          ḣ   Range from 1 to input
     §foε⌋    Keep only those whose gcd with the input is 1
    Π         Product
  ȯ→          Plus 1
S÷            Integer division with input

Answer 3

Mathematica, 91 bytes

If[(k=#)==1,2,(Times@@Select[Range@k,CoprimeQ[k,#]&]+If[IntegerQ@PrimitiveRoot@#,1,-1])/#]&

Answer 4

Pyth, 11 bytes

/h*Ff>2iTQS

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---

How?

• /h*Ff>2iTQS - Full program.

• S - Generate the inclusive range [1, input]

• f - Filter-keep those:

  • iTQ - whose GCD with the input.

  • >2 - Is less than two (can be replaced by either of the following: q1, !t)

• *F - Apply multiplication repeatedly. In other words, the product of the list.

• h - Increment the product by 1.

• / - Floor division with the input.

TL;DR: Get all the coprimes to the input in the range [1, input], get their product, increment it and divide it by the input.

Answer 5

Python 2, 62 bytes

n=input();k=r=1
exec'r*=k/n*k%n!=1or k/n;k+=1;'*n*n
print-~r/n

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Answer 6

J, 33 bytes

3 :'<.%&y>:*/(#~1&=@(+.&y))1+i.y'

This one is more of a request to see an improvement than anything else. I tried a tacit solution first, but it was longer than this.

explanation

This is fairly straightforward translation of Mr. Xcoder's solution into J.

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Answer 7

R, 82 bytes

function(n)(prod((1:n)[g(n,1:n)<2])+1)%/%n
g=function(a,b)ifelse(o<-a%%b,g(b,o),b)

Uses integer division rather than figuring out e like many answers here, although I did work out that e=2*any((1:n)^2%%n==1%%n)-1

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Answer 8

Jelly, 22 bytes

ḊṖṖ²%ċ1ȧ2’
gRỊTP+Ç÷_.Ċ

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-1 thanks to Dennis (and Mr. Xcoder's comment).

Answer 9

05AB1E, 8 bytes

Lʒ¿}P>I÷

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Answer 10

f=(n,i=1,p=1,g=(a,b)=>b?g(b,a%b):a)=>i<n?f(n,i+1,g(n,i)-1?p:p*i):-~p/n|0

<input type=number min=1 oninput=o.textContent=f(+this.value)><pre id=o>

Integer division strikes again.