Euclid-Mullin sequence
You are encouraged to solve this task according to the task description, using any language you may know.
- Definition
The Euclid–Mullin sequence is an infinite sequence of distinct prime numbers, in which each element is the least prime factor of one plus the product of all earlier elements.
The first element is usually assumed to be 2. So the second element is : (2) + 1 = 3 and the third element is : (2 x 3) + 1 = 7 as this is prime.
Although intermingled with smaller elements, the sequence can produce very large elements quite quickly and only the first 51 have been computed at the time of writing.
- Task
Compute and show here the first 16 elements of the sequence or, if your language does not support arbitrary precision arithmetic, as many as you can.
- Stretch goal
Compute the next 11 elements of the sequence.
- Reference
ALGOL 68[edit]
Uses ALGOL 68G's LONG LONG INT which has programmer specifiable precission, the default is sufficient for this task.
Although the first 16 elements will all fit in 64 bits, the product exceeds 64 bits after the ninth element.
BEGIN # find elements of the Euclid-Mullin sequence: starting from 2, #
# the next element is the smallest prime factor of 1 + the product #
# of the previous elements #
print( ( " 2" ) );
LONG LONG INT product := 2;
FROM 2 TO 16 DO
LONG LONG INT next := product + 1;
# find the first prime factor of next #
LONG LONG INT p := 3;
BOOL found := FALSE;
WHILE p * p <= next AND NOT ( found := next MOD p = 0 ) DO
p +:= 2
OD;
IF found THEN next := p FI;
print( ( " ", whole( next, 0 ) ) );
product *:= next
OD
END
- Output:
2 3 7 43 13 53 5 6221671 38709183810571 139 2801 11 17 5471 52662739 23003
AWK[edit]
# syntax: GAWK -f EUCLID-MULLIN_SEQUENCE.AWK
# converted from FreeBASIC
BEGIN {
limit = 7 # we'll stop here
arr[0] = 2
printf("%s ",arr[0])
for (i=1; i<=limit; i++) {
k = 3
while (1) {
em = 1
for (j=0; j<=i-1; j++) {
em = (em * arr[j]) % k
}
em = (em + 1) % k
if (em == 0) {
arr[i] = k
printf("%s ",arr[i])
break
}
k += 2
}
}
printf("\n")
exit(0)
}
- Output:
2 3 7 43 13 53 5 6221671
BASIC[edit]
Craft Basic[edit]
define size = 16, em = 0
dim list[size]
let list[0] = 2
print 2
for i = 1 to 15
let k = 3
do
let em = 1
for j = 0 to i - 1
let em = ( em * list[j] ) % k
next j
let em = ( em + 1 ) % k
if em = 0 then
let list[i] = k
print list[i]
break
endif
let k = k + 2
wait
loop
next i
print "done."
end
FreeBASIC[edit]
Naive and takes forever to find the largest term, but does get there in the end.
dim as ulongint E(0 to 15), k
dim as integer i, em
E(0) = 2 : print 2
for i=1 to 15
k=3
do
em = 1
for j as uinteger = 0 to i-1
em = (em*E(j)) mod k
next j
em = (em + 1) mod k
if em = 0 then
E(i)=k
print E(i)
exit do
end if
k = k + 2
loop
next i
F#[edit]
//Euclid-Mullin sequence. Nigel Galloway: October 29th., 2021
let(|Prime|_|)(n,g)=if Open.Numeric.Primes.MillerRabin.IsProbablePrime &g then Some(n*g,n*g+1I) else None
let n=Seq.unfold(fun(n,g)->match n,g with Prime n->Some(g,n) |_->let g=Open.Numeric.Primes.Extensions.PrimeExtensions.PrimeFactors g|>Seq.item 1 in Some(g,(n*g,n*g+1I)))(1I,2I)
n|>Seq.take 16|>Seq.iter(printfn "%A")
- Output:
2 3 7 43 13 53 5 6221671 38709183810571 139 2801 11 17 5471 52662739 23003
Fermat[edit]
Func Firstfac(n) =
j := 3;
up := Sqrt(n);
while j <= up do
if Divides(j,n) then Return(j) fi;
j:=j+2;
od;
Return(n).;
Array eu[16];
eu[1]:=2;
!(eu[1],' ');
for i=2 to 16 do
eu[i]:=Firstfac(1+Prod<k=1,i-1>[eu[k]]);
!(eu[i],' ');
od;
- Output:
2 3 7 43 13 53 5 6221671 38709183810571 139 2801 11 17 5471 52662739 23003
Go[edit]
This runs in about 54 seconds which, puzzlingly, is a good bit slower than Wren even though both are using GMP and the Pollard's rho algorithm. I have no idea why.
package main
import (
"fmt"
big "github.com/ncw/gmp"
"log"
)
var (
zero = big.NewInt(0)
one = big.NewInt(1)
two = big.NewInt(2)
three = big.NewInt(3)
four = big.NewInt(4)
five = big.NewInt(5)
six = big.NewInt(6)
ten = big.NewInt(10)
max = big.NewInt(100000)
)
func pollardRho(n, c *big.Int) *big.Int {
g := func(x, y *big.Int) *big.Int {
x2 := new(big.Int)
x2.Mul(x, x)
x2.Add(x2, c)
return x2.Mod(x2, y)
}
x, y, z := big.NewInt(2), big.NewInt(2), big.NewInt(1)
d := new(big.Int)
count := 0
for {
x = g(x, n)
y = g(g(y, n), n)
d.Sub(x, y)
d.Abs(d)
d.Mod(d, n)
z.Mul(z, d)
count++
if count == 100 {
d.GCD(nil, nil, z, n)
if d.Cmp(one) != 0 {
break
}
z.Set(one)
count = 0
}
}
if d.Cmp(n) == 0 {
return zero
}
return d
}
func smallestPrimeFactorWheel(n *big.Int) *big.Int {
if n.ProbablyPrime(15) {
return n
}
z := new(big.Int)
if z.Rem(n, two).Cmp(zero) == 0 {
return two
}
if z.Rem(n, three).Cmp(zero) == 0 {
return three
}
if z.Rem(n, five).Cmp(zero) == 0 {
return five
}
k := big.NewInt(7)
i := 0
inc := []*big.Int{four, two, four, two, four, six, two, six}
for z.Mul(k, k).Cmp(n) <= 0 {
if z.Rem(n, k).Cmp(zero) == 0 {
return k
}
k.Add(k, inc[i])
if k.Cmp(max) > 0 {
break
}
i = (i + 1) % 8
}
return nil
}
func smallestPrimeFactor(n *big.Int) *big.Int {
s := smallestPrimeFactorWheel(n)
if s != nil {
return s
}
c := big.NewInt(1)
s = new(big.Int).Set(n)
for n.Cmp(max) > 0 {
d := pollardRho(n, c)
if d.Cmp(zero) == 0 {
if c.Cmp(ten) == 0 {
log.Fatal("Pollard Rho doesn't appear to be working.")
}
c.Add(c, one)
} else {
// can't be sure PR will find the smallest prime factor first
if d.Cmp(s) < 0 {
s.Set(d)
}
n.Quo(n, d)
if n.ProbablyPrime(5) {
if n.Cmp(s) < 0 {
return n
}
return s
}
}
}
return s
}
func main() {
k := 19
fmt.Println("First", k, "terms of the Euclid–Mullin sequence:")
fmt.Println(2)
prod := big.NewInt(2)
z := new(big.Int)
count := 1
for count < k {
z.Add(prod, one)
t := smallestPrimeFactor(z)
fmt.Println(t)
prod.Mul(prod, t)
count++
}
}
- Output:
First 19 terms of the Euclid–Mullin sequence: 2 3 7 43 13 53 5 6221671 38709183810571 139 2801 11 17 5471 52662739 23003 30693651606209 37 1741
Java[edit]
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.BitSet;
import java.util.List;
import java.util.concurrent.ThreadLocalRandom;
public class EulerMullinSequence {
public static void main(String[] aArgs) {
primes = listPrimesUpTo(1_000_000);
System.out.println("The first 27 terms of the Euler-Mullin sequence:");
System.out.println(2);
for ( int i = 1; i < 27; i++ ) {
System.out.println(nextEulerMullin());
}
}
private static BigInteger nextEulerMullin() {
BigInteger smallestPrime = smallestPrimeFactor(product.add(BigInteger.ONE));
product = product.multiply(smallestPrime);
return smallestPrime;
}
private static BigInteger smallestPrimeFactor(BigInteger aNumber) {
if ( aNumber.isProbablePrime(probabilityLevel) ) {
return aNumber;
}
for ( BigInteger prime : primes ) {
if ( aNumber.mod(prime).signum() == 0 ) {
return prime;
}
}
BigInteger factor = pollardsRho(aNumber);
return smallestPrimeFactor(factor);
}
private static BigInteger pollardsRho(BigInteger aN) {
if ( aN.equals(BigInteger.ONE) ) {
return BigInteger.ONE;
}
if ( aN.mod(BigInteger.TWO).signum() == 0 ) {
return BigInteger.TWO;
}
final BigInteger core = new BigInteger(aN.bitLength(), random);
BigInteger x = new BigInteger(aN.bitLength(), random);
BigInteger xx = x;
BigInteger divisor = null;
do {
x = x.multiply(x).mod(aN).add(core).mod(aN);
xx = xx.multiply(xx).mod(aN).add(core).mod(aN);
xx = xx.multiply(xx).mod(aN).add(core).mod(aN);
divisor = x.subtract(xx).gcd(aN);
} while ( divisor.equals(BigInteger.ONE) );
return divisor;
}
private static List<BigInteger> listPrimesUpTo(int aLimit) {
BitSet sieve = new BitSet(aLimit + 1);
sieve.set(2, aLimit + 1);
final int squareRoot = (int) Math.sqrt(aLimit);
for ( int i = 2; i <= squareRoot; i = sieve.nextSetBit(i + 1) ) {
for ( int j = i * i; j <= aLimit; j = j + i ) {
sieve.clear(j);
}
}
List<BigInteger> result = new ArrayList<BigInteger>(sieve.cardinality());
for ( int i = 2; i >= 0; i = sieve.nextSetBit(i + 1) ) {
result.add(BigInteger.valueOf(i));
}
return result;
}
private static List<BigInteger> primes;
private static BigInteger product = BigInteger.TWO;
private static ThreadLocalRandom random = ThreadLocalRandom.current();
private static final int probabilityLevel = 20;
}
jq[edit]
Works with jq, gojq and jaq, that is, the C, Go and Rust implementations of jq. (*)
Adapted from Algol 68
(*) The precision of jq and jaq is insufficient to compute more than the first nine numbers in the sequence (i.e., beyond 38709183810571). gojq's memory consumption becomes excessive after the first 16 numbers in the sequence are produced.
# Output: the Euclid-Mullins sequence, beginning with 2
def euclid_mullins:
foreach range(1; infinite|floor) as $i ( { product: 1 };
.next = .product + 1
# find the first prime factor of .next
| .p = 3
| .found = false
| until( .p * .p > .next or .found;
.found = ((.next % .p) == 0)
| if .found then . else .p += 2 end)
| if .found then .next = .p else . end
| .product *= .next)
| .next ;
# Produce 16 terms
limit(16; euclid_mullins)
- Output:
gojq supports unbounded-precision integer arithmetic and was accordingly used to produce the following output in accordance with the basic task requirements Beyond this number, gojq's memory consumption becomes excessive.
Invocation: $ gojq -n -f euclid-mullin-sequence.algol.jq
2 3 7 43 13 53 5 6221671 38709183810571 139 2801 11 17 5471 52662739 23003
Julia[edit]
using Primes
struct EuclidMullin end
Base.length(em::EuclidMullin) = 1000 # not expected to get to 1000
Base.eltype(em::EuclidMullin) = BigInt
Base.iterate(em::EuclidMullin, t=big"1") = (p = first(first(factor(t + 1).pe)); (p, t * p))
println("First 16 Euclid-Mullin numbers: ", join(Iterators.take(EuclidMullin(), 16), ", "))
- Output:
First 16 Euclid-Mullin numbers: 2, 3, 7, 43, 13, 53, 5, 6221671, 38709183810571, 139, 2801, 11, 17, 5471, 52662739, 23003
Mathematica/Wolfram Language[edit]
list = {2};
Do[
prod = Times @@ list;
prod++;
new = Min[FactorInteger[prod][[All, 1]]];
AppendTo[list, new]
,
{21 - 1}
];
list
- Output:
The first 21 numbers of the sequence:
{2, 3, 7, 43, 13, 53, 5, 6221671, 38709183810571, 139, 2801, 11, 17, 5471, 52662739, 23003, 30693651606209, 37, 1741, 1313797957, 887}
Others may be found by adjusting the range of the Do loop but it will take a while.
Nim[edit]
import integers
let
Zero = newInteger()
One = newInteger(1)
Two = newInteger(2)
Three = newInteger(3)
Five = newInteger(5)
Ten = newInteger(10)
Max = newInteger(100000)
None = newInteger(-1)
proc pollardRho(n, c: Integer): Integer =
template g(x: Integer): Integer = (x * x + c) mod n
var
x = newInteger(2)
y = newInteger(2)
z = newInteger(1)
d = Max + 1
count = 0
while true:
x = g(x)
y = g(g(y))
d = abs(x - y) mod n
z *= d
inc count
if count == 100:
d = gcd(z, n)
if d != One: break
z = newInteger(1)
count = 0
result = if d == n: Zero else: d
template isEven(n: Integer): bool = isZero(n and 1)
proc smallestPrimeFactorWheel(n: Integer): Integer =
if n.isPrime(5): return n
if n.isEven: return Two
if isZero(n mod 3): return Three
if isZero(n mod 5): return Five
var k = newInteger(7)
var i = 0
const Inc = [4, 2, 4, 2, 4, 6, 2, 6]
while k * k <= n:
if isZero(n mod k): return k
k += Inc[i]
if k > Max: return None
i = (i + 1) mod 8
proc smallestPrimeFactor(n: Integer): Integer =
var n = n
result = smallestPrimeFactorWheel(n)
if result != None: return
var c = One
result = newInteger(n)
while n > Max:
var d = pollardRho(n, c)
if d.isZero:
if c == Ten:
quit "Pollard Rho doesn't appear to be working.", QuitFailure
inc c
else:
# Can't be sure PR will find the smallest prime factor first.
result = min(result, d)
n = n div d
if n.isPrime(2):
return min(result, n)
proc main() =
var k = 19
echo "First ", k, " terms of the Euclid–Mullin sequence:"
echo 2
var prod = newInteger(2)
var count = 1
while count < k:
let t = smallestPrimeFactor(prod + One)
echo t
prod *= t
inc count
main()
- Output:
The first 16 numbers are displayed instantly, but it took about 9 seconds on my (moderately powerful) laptop to get the next three. I gave up for the 20th number.
First 19 terms of the Euclid–Mullin sequence: 2 3 7 43 13 53 5 6221671 38709183810571 139 2801 11 17 5471 52662739 23003 30693651606209 37 1741
PARI/GP[edit]
E=vector(16)
E[1]=2
for(i=2,16,E[i]=factor(prod(n=1,i-1,E[n])+1)[1,1])
print(E)
- Output:
[2, 3, 7, 43, 13, 53, 5, 6221671, 38709183810571, 139, 2801, 11, 17, 5471, 52662739, 23003]
Perl[edit]
use strict;
use warnings;
use feature 'say';
use ntheory <factor vecprod vecmin>;
my @Euclid_Mullin = 2;
push @Euclid_Mullin, vecmin factor (1 + vecprod @Euclid_Mullin) for 2..16+11;
say "First sixteen: @Euclid_Mullin[ 0..15]";
say "Next eleven: @Euclid_Mullin[16..26]";
- Output:
First sixteen: 2 3 7 43 13 53 5 6221671 38709183810571 139 2801 11 17 5471 52662739 23003 Next eleven: 30693651606209 37 1741 1313797957 887 71 7127 109 23 97 159227
Phix[edit]
with javascript_semantics requires("1.0.1") -- (added mpz_set_v()) include mpfr.e sequence res = {} mpz {total,tmp} = mpz_inits(2,1) while length(res)<16 do mpz_add_si(tmp,total,1) mpz_set_v(tmp,mpz_pollard_rho(tmp)[1][1]) res = append(res,mpz_get_str(tmp)) mpz_mul(total,total,tmp) end while printf(1,"The first 16 Euclid-Mulin numbers: %s\n",{join(res)})
- Output:
The first 16 Euclid-Mulin numbers: 2 3 7 43 13 53 5 6221671 38709183810571 139 2801 11 17 5471 52662739 23003
While the first 16 are pretty fast, mpz_pollard_rho("723023114226131400979589798874734076807875188379971") took 3 minutes, and yielded the next element as 30693651606209, but beyond that I gave up.
Python[edit]
""" Rosetta code task: Euclid-Mullin_sequence """
from primePy import primes
def euclid_mullin():
""" generate Euclid-Mullin sequence """
total = 1
while True:
next_iter = primes.factor(total + 1)
total *= next_iter
yield next_iter
GEN = euclid_mullin()
print('First 16 Euclid-Mullin numbers:', ', '.join(str(next(GEN)) for _ in range(16)))
- Output:
First 16 Euclid-Mullin numbers: 2, 3, 7, 43, 13, 53, 5, 6221671, 38709183810571, 139, 2801, 11, 17, 5471, 52662739, 23003
Raku[edit]
use Prime::Factor;
my @Euclid-Mullin = 2, { state $i = 1; (1 + [×] @Euclid-Mullin[^$i++]).&prime-factors.min } … *;
put 'First sixteen: ', @Euclid-Mullin[^16];
- Output:
First sixteen: 2 3 7 43 13 53 5 6221671 38709183810571 139 2801 11 17 5471 52662739 23003
RPL[edit]
RPL code | Comment |
---|---|
≪ IF # 1d DUP2 AND ≠ THEN DROP # 2d ELSE IF DUP 3 DUP2 / * == THEN DROP # 3d ELSE DUP B→R √ → divm ≪ 4 5 divm FOR n IF OVER n DUP2 / * == THEN SWAP DROP n R→B SWAP divm 'n' STO END 6 SWAP - DUP STEP DROP ≫ END END ≫ ‘bDIV1’ STO ≪ DUP SIZE 1 1 ROT FOR j OVER j GET * NEXT 1 + bDIV1 + ≫ 'NXTEM' STO |
bDIV1 ( #m -- #first_divisor ) is #2 a divisor ? is #3 a divisor ? otherwise get sqrt(m) d = 4 ; for n = 5 to sqrt(m) if n divides m replace m by n and prepare loop exit d = 6 - d ; n += d NXTEM ( { #EM(1) .. #EM(n) } -- { #EM(1) .. #EM(n+1) } ) get EM(1)*..*EM(n) get least prime factor of 1+EM(1)*..*EM(n) and add to list |
- Input:
≪ { # 2 } WHILE DUP SIZE ≤ 16 REPEAT NXTEM END ≫ EVAL
The emulator's watchdog timer prevents checking the primality of EM(9) = # 38709183810571d. Even if this device stayed idle, EM(10) could not be calculated, since the product of all earlier elements is more than 64 bits long.
- Output:
1: { # 2d # 3d # 7d # 43d # 13d # 53d # 5d # 6221671d }
Sidef[edit]
func f(n) is cached {
return 2 if (n == 1)
lpf(1 + prod(1..^n, {|k| f(k) }))
}
say f.map(1..16)
say f.map(17..27)
- Output:
[2, 3, 7, 43, 13, 53, 5, 6221671, 38709183810571, 139, 2801, 11, 17, 5471, 52662739, 23003] [30693651606209, 37, 1741, 1313797957, 887, 71, 7127, 109, 23, 97, 159227]
Wren[edit]
Wren-cli[edit]
This uses the Pollard Rho algorithm to try and speed up the factorization of the 15th element but overall time still slow at around 32 seconds.
import "./big" for BigInt
var zero = BigInt.zero
var one = BigInt.one
var two = BigInt.two
var ten = BigInt.ten
var max = BigInt.new(100000)
var pollardRho = Fn.new { |n, c|
var g = Fn.new { |x, y| (x*x + c) % n }
var x = two
var y = two
var z = one
var d = max + one
var count = 0
while (true) {
x = g.call(x, n)
y = g.call(g.call(y, n), n)
d = (x - y).abs % n
z = z * d
count = count + 1
if (count == 100) {
d = BigInt.gcd(z, n)
if (d != one) break
z = one
count = 0
}
}
if (d == n) return zero
return d
}
var smallestPrimeFactorWheel = Fn.new { |n|
if (n.isProbablePrime(5)) return n
if (n % 2 == zero) return BigInt.two
if (n % 3 == zero) return BigInt.three
if (n % 5 == zero) return BigInt.five
var k = BigInt.new(7)
var i = 0
var inc = [4, 2, 4, 2, 4, 6, 2, 6]
while (k * k <= n) {
if (n % k == zero) return k
k = k + inc[i]
if (k > max) return null
i = (i + 1) % 8
}
}
var smallestPrimeFactor = Fn.new { |n|
var s = smallestPrimeFactorWheel.call(n)
if (s) return s
var c = one
s = n
while (n > max) {
var d = pollardRho.call(n, c)
if (d == 0) {
if (c == ten) Fiber.abort("Pollard Rho doesn't appear to be working.")
c = c + one
} else {
// can't be sure PR will find the smallest prime factor first
s = BigInt.min(s, d)
n = n / d
if (n.isProbablePrime(2)) return BigInt.min(s, n)
}
}
return s
}
var k = 16
System.print("First %(k) terms of the Euclid–Mullin sequence:")
System.print(2)
var prod = BigInt.two
var count = 1
while (count < k) {
var t = smallestPrimeFactor.call(prod + one)
System.print(t)
prod = prod * t
count = count + 1
}
- Output:
First 16 terms of the Euclid–Mullin sequence: 2 3 7 43 13 53 5 6221671 38709183810571 139 2801 11 17 5471 52662739 23003
Embedded[edit]
This finds the first 16 in 0.11 seconds and the next 3 in around 39 seconds. I gave up after that as it would take too long for the Pollard's Rho algorithm to find any more.
/* euclid_mullin_gmp.wren */
import "./gmp" for Mpz
var max = Mpz.from(100000)
var smallestPrimeFactorWheel = Fn.new { |n|
if (n.probPrime(15) > 0) return n
if (n.isEven) return Mpz.two
if (n.isDivisibleUi(3)) return Mpz.three
if (n.isDivisibleUi(5)) return Mpz.five
var k = Mpz.from(7)
var i = 0
var inc = [4, 2, 4, 2, 4, 6, 2, 6]
while (k * k <= n) {
if (n.isDivisible(k)) return k
k.add(inc[i])
if (k > max) return null
i = (i + 1) % 8
}
}
var smallestPrimeFactor = Fn.new { |n|
var s = smallestPrimeFactorWheel.call(n)
if (s) return s
var c = Mpz.one
s = n.copy()
while (n > max) {
var d = Mpz.pollardRho(n, 2, c)
if (d.isZero) {
if (c == 100) Fiber.abort("Pollard Rho doesn't appear to be working.")
c.inc
} else {
// can't be sure PR will find the smallest prime factor first
s.min(d)
n.div(d)
if (n.probPrime(5) > 0) return Mpz.min(s, n)
}
}
return s
}
var k = 19
System.print("First %(k) terms of the Euclid–Mullin sequence:")
System.print(2)
var prod = Mpz.two
var count = 1
while (count < k) {
var t = smallestPrimeFactor.call(prod + Mpz.one)
System.print(t)
prod.mul(t)
count = count + 1
}
- Output:
As Wren-cli plus three more:
30693651606209 37 1741