Infinitely many primes

Question

Since Euclid, we have known that there are infinitely many primes. The argument is by contradiction: If there are only finitely many, let's say \$p_1,p_2,...,p_n\$, then surely \$m:=p_1\cdot p_2\cdot...\cdot p_n+1\$ is not divisible by any of these primes, so its prime factorization must yield a new prime that was not in the list. So the assumption that only finitely primes exist is false.

Now let's assume that \2ドル\$ is the only prime. The method from above yields \2ドル+1=3\$ as a new (possible) prime. Applying the method again yields \2ドル\cdot 3+1=7\$, and then \2ドル\cdot 3\cdot 7+1=43\$, then \2ドル\cdot 3\cdot 7\cdot 43+1=13\cdot 139\$, so both \13ドル\$ and \139ドル\$ are new primes, etc. In the case where we get a composite number, we just take the least new prime. This results in A000945.

Challenge

Given a prime \$p_1\$ and an integer \$n\$ calculate the \$n\$-th term \$p_n\$ of the sequence defined as follows:

$$p_n := \min(\operatorname{primefactors}(p_1\cdot p_2\cdot ... \cdot p_{n-1} + 1))$$

These sequences are known as Euclid-Mullin-sequences.

Examples

For \$p_1 = 2\$:

1 2
2 3
3 7
4 43
5 13
6 53
7 5
8 6221671
9 38709183810571

For \$p_1 = 5\$ (A051308):

1 5
2 2
3 11
4 3
5 331
6 19
7 199
8 53
9 21888927391

For \$p_1 = 97\$ (A051330)

1 97
2 2
3 3
4 11
5 19
6 7
7 461
8 719
9 5

Answer 1

JavaScript (ES6), (削除) 45 (削除ここまで) 44 bytes

Takes input as (n)(p1), where \$n\$ is 0-indexed.

n=>g=(p,d=2)=>n?~p%d?g(p,d+1):--n?g(p*d):d:p

Try it online!

Commented

n => // n = 0-based index of the requested term
 g = ( // g is a recursive function taking:
 p, // p = current prime product
 d = 2 // d = current divisor
 ) => //
 n ? // if n is not equal to 0:
 ~p % d ? // if d is not a divisor of ~p (i.e. not a divisor of p + 1):
 g(p, d + 1) // increment d until it is
 : // else:
 --n ? // decrement n; if it's still not equal to 0:
 g(p * d) // do a recursive call with the updated prime product
 : // else:
 d // stop recursion and return d
 : // else:
 p // don't do any recursion and return p right away

Answer 2

05AB1E, 6 bytes

This produces and infinite output stream.

λλP>fW

Try it online! (link includes a slightly modified version, λ£λP>fW, which instead outputs the first \$n\$ terms)

Explanation

Very straightforward. Given \$p_1\$ and \$n\$, the program does the following:

• Starts with \$p_1\$ as an initial parameter for the infinite stream (which is generated using the first λ) and begins a recursive environment which generates a new term after each interation and appends it to the stream.
• The second λ, now being used inside the recursive environment, changes its functionality: Now, it retrieves all previously generated elements (i.e. the list \$[\lambda_0, \lambda_1, \lambda_2, \ldots, \lambda_{n-1}]\$), where \$n\$ represents the current iteration number.
• The rest is trivial: P takes the product (\$\lambda_0\lambda_1\lambda_2 \cdots \lambda_{n-1}\$), > adds one to this product, and fW retrieves the minimum prime factor.

Answer 3

J, 15 bytes

-10 bytes thanks to miles!

Returning the sequence up to n (zero-indexed) – thanks to @miles

(,0({q:)1+*/)^:

Try it online!

J, 25 bytes

Returns the n th item

_2{((],0{[:q:1+*/@])^:[])

Try it online!

• miles: (,0({q:)1+*/)^: for 15 bytes, returning the sequence up to n (zero-indexed)
• Jonah: Very nice. @miles what exactly is happening there grammatically? we put a verb and a conjunction together and get a dyadic verb back. I thought verb conj produced an adverb.
• miles: @Jonah it's a trick I learned from golfing. I think it's one of the older parsing rules that's still valid
• Jonah: @miles I just realized it is an adverb (or adnoun). It modifies the noun to its left, which "attaches" to the right of the ^:, and then that becomes a verb that applies to the right arg. I think that's what's happening grammatically.

Answer 4

Python 2, 56 bytes

i=input();k=1
while 1:
 k*=i;print i;i=2
 while~k%i:i+=1

Try it online!

---

Commented

i=input() # the initial prime
k=1 # the product of all previous primes
while 1: # infinite loop
 k*=i # update the product of primes
 print i # output the last prime
 i=2 # starting at two ...
 while~k%i: # find the lowest number that divides k+1
 i+=1
 # this our new prime

Try it online!

• Anthony: I just started with Python, but do you need int(input()) otherwise i is a str?
• ovs: In Python 3 this would be true as input() always returns strings. In Python 2 input() tries to evaluate the input. I'm using Python 2 in this case because the resulting code is slightly shorter. For real code you should try to use Python 3 as it is the newer and more supported version of Python.
• sintax: How does this terminate after n steps?
• ovs: @sintax it outputs the sequence for a given p1 indefinitely, as allowed by the default sequence rules.

Answer 5

Jelly, 8 bytes

P‘ÆfṂṭμ¡

A full program (using zero indexing) accepting \$P_0\$ and \$n\$ which prints a Jelly representation of the list of \$P_0\$ to \$P_n\$ inclusive. (As a dyadic Link, with n=0 we'll be given back an integer, not a list.)

Try it online!

How?

P‘ÆfṂṭμ¡ - Link: integer, p0; integer n
 μ¡ - repeat the monadic chain to the left n times, starting with x=p0:
P - product of x (p0->p0 or [p0,...,pm]->pm*...*p0)
 ‘ - increment
 Æf - prime factors
 Ṃ - minimum
 ṭ - tack
 - implicit print

Answer 6

05AB1E, 8 bytes

GDˆ ̄P>fß

First input is \$n\$, second is prime \$p\$.

Try it online or very some more test cases (test suite lacks the test cases for \$n\geq9\$, because for \$p=2\$ and \$p=5\$ the builtin f takes too long).

Explanation:

G # Loop (implicit input) n-1 amount of times:
 Dˆ # Add a copy of the number at the top of the stack to the global array
 # (which will take the second input p implicitly the first iteration)
 ̄ # Push the entire global array
 P # Take the product of this list
 > # Increase it by 1
 f # Get the prime factors of this number (without counting duplicates)
 ß # Pop and only leave the smallest prime factor
 # (after the loop: implicitly output the top of the stack as result)

• Mr. Xcoder: I had λλP>fW (6 bytes) with output as an infinite list and λ£λP>fW (7 bytes) for the first \$n\$ terms. However getting the \$n^{\text{th}}\$ should be 9 bytes... If only we had a flag like £ but for the last element!
• Kevin Cruijssen: @Mr.Xcoder "If only we had a flag like £ but for the last element!", like .£? ;) EDIT: Actually, it doesn't work exactly like £ for lists.. using a list like [1,2] with .£ results in two loose items with the last 1 and 2 items (i.e. 12345 becomes [5,45] instead of [45,3] or [3,45], with 12S.£)..
• Mr. Xcoder: Umm, no, I don't see how λ.£ should work. I used flag as in additional function associated with λ (see this conversation with Adnan). I basically want some flag è such that when running λè...} it would generate the n-th element rather than the the infinite stream (just like λ£ would work for generating the first n elements).
• Kevin Cruijssen: @Mr.Xcoder Ah sorry, you've used the £ for the recursive environment. Yeah, then λ.£ is indeed not gonna work, my bad. Nice 6-byter regardless. Now you just have to wait for @flawr's response whether it's allowed or not (it probably is).

Answer 7

Japt, (削除) 12 (削除ここまで) 11 bytes

Struggled to get this one right so may have missed something that can be golfed.

Takes n as the first input and p1, as a singleton array, as the second. Returns the first n terms. Change h to g to return the nth 0-indexed term instead.

×ばつÄ k Î}hV

Try it

×ばつÄ k Î}hV :Implicit input of integer U=n & array V=[p1]
@ :Function taking an array as an argument via parameter Z
 ×ばつ : Reduce Z by multiplication
 Ä : Add 1
 k : Prime factors
 Î : First element
 } :End function
 hV :Run that function, passing V as Z, and
 : push the result to V.
 : Repeat until V is of length U

Answer 8

Retina, 56 bytes

,|$
$*
"$&"{~`.+¶
$$¶_
)`\b(__+?)1円*$
$.1$*
1A`
.$
\*
,

Try it online! Takes input as the number of new terms to add on the first line and the seed term(s) on the second line. Note: Gets very slow since it uses unary factorisation so it needs to create a string of the relevant length. Explanation:

,|$
$*

Replace the commas in the seed terms with *s and append a *. This creates a Retina expression for a string of length of the product of the values.

"$&"{
)`

Repeat the loop the number of times given by the first input.

~`.+¶
$$¶_

Temporarily replace the number on the first line with a $ and prepend a _ to the second line, then evaluate the result as a Retina program, thus appending a string of _s of length 1 more than the product of the values.

\b(__+?)1円*$
$.1$*

Find the smallest nontrivial factor of the number in decimal and append a * ready for the next loop.

1A`

Delete the iteration input.

.$

Delete the last *.

\*
,

Replace the remaining *s with ,s.

Answer 9

JavaScript (Node.js), 54 bytes

f=(p,n,P=p,F=n=>-~P%n?F(n+1):n)=>--n?f(p=F(2),n,P*p):p

Try it online!

Ungolfed

F=(p,n=2)=> // Helper function F for finding the smallest prime factor
 p%n // If n (starting at 2) doesn't divide p:
 ?F(n+1) // Test n+1 instead
 :n // Otherwise, return n
f=(p,n,P=p)=> // Main function f:
 --n // Repeat n - 1 times:
 ?f(p=F(P+1),n,P*p) // Find the next prime factor and update the product
 :p // Return the last prime

Answer 10

bash +GNU coreutils, 89 bytes

IFS=\*;n=1ドル;shift;for((;++i<n;));{ set $@ `factor $["$*+1"]|cut -d\ -f2`;};echo ${@: -1}

TIO

Answer 11

Ruby 2.6, 51 bytes

f=->s,n{[s,l=(2..).find{|d|~s%d<1}][n]||f[l*s,n-1]}

(2..), the infinite range starting from 2, isn't supported on TIO yet.

This is a recursive function that takes a starting value s (can be a prime or composite), returns it when n=0 (edit: note that this means it's zero-indexed), returns the least number l that's greater than 1 and divides -(s+1) when n=1, and otherwise recurses with s=l*s and n=n-1.

• Value Ink: You should probably mention that you're doing zero-indexed; I replaced (2..) with 2.step (just 1 byte longer) to allow it to work on TIO and everything was off by one. Try it online!

Answer 12

APL (Dyalog Extended), 15 bytes

This is a fairly simple implementation of the algorithm which uses Extended's very helpful prime factors builtin, ⍭. Try it online!

×ばつ/⍵}⍣⎕⊢⎕

Explanation

×ばつ/⍵}⍣⎕⊢⎕
 ⊢⎕ First get the first prime of the sequence S from input.
{ }⍣⎕ Then we repeat the code another input number of times.
 ×ばつ/⍵ We take the product of S and add 1.
 ⍭ Get the prime factors of product(S)+1.
 ⊃ Get the first element, the smallest prime factor of prod(S)+1.
 ⍵, And append it to S.

Answer 13

Stax, 9 bytes

é|G╝╓c1ドル⁄4_

Run and debug it

Takes p0 and (zero-indexed) n for input. Produces pn.

Answer 14

C (gcc), (削除) 54 (削除ここまで) 53 bytes

p;f(x,n){for(p=x;--n;p*=x)for(x=1;~p%++x;);return x;}

Try it online!

-1 byte thanks to ceilingcat

Answer 15

Perl 6, (削除) 33 (削除ここまで) 32 bytes

-1 byte thanks to nwellnhof

{$_,{1+(2...-+^[*](@_)%%*)}...*}

Try it online!

Anonymous code block that takes a number and returns a lazy list.

Explanation:

{ } # Anonymous codeblock
 ...* # That returns an infinite list
 $_, # Starting with the input
 { } # Where each element is
 1+(2... ) # The first number above 2
 %%* # That cleanly divides
 [*](@_) # The product of all numbers so far
 -+^ # Plus one

Answer 16

Pari/GP, 45 bytes

(p,n)->s=p;for(i=2,n,s*=p=divisors(s+1)[2]);p

Try it online!

Answer 17

Haskell, 49 bytes

g 1
g a b=b:g(a*b)([c|c<-[2..],1>mod(a*b+1)c]!!0)

Try it online!

Returns the infinite sequence as a lazy list.

Explanation:

g 1 -- Initialise the product as 1
g a b= -- Given the product and the current number
 b: -- Return the current number, followed by
 g -- Recursively calliong the function with
 (a*b) -- The new product
 ( ) -- And get the next number as
 [c|c<-[2..], ]!!0 -- The first number above 2
 1>mod c -- That cleanly divides
 (a*b+1) -- The product plus one

Answer 18

Husk, 8 bytes

!¡ö←p→Π;

Try it online!

For input m,n outputs the n-th term in the sequence starting with m.

Drop the initial ! (for 7 bytes) to output the entire infinite sequence (this link selects the first 10 elements from the infinite list).
If both input & output are flexible enough (input: list of just the first term, output: infinite sequence) we could get down to 6 bytes.