Least m such that n! + m is prime
Least m such that n! + m is prime is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Find the minimum positive integer m such that n factorial plus m is prime.
- E.G.
0! = 1. The next prime greater than 1 is 2. 2 - 1 = 1, so a(0) = 1. 1! = 1. The next prime greater than 1 is 2. 2 - 1 = 1, so a(1) = 1. 2! = 2. The next prime greater than 2 is 3. 3 - 2 = 1, so a(2) = 1. 3! = 6. The next prime greater than 6 is 7. 7 - 6 = 1, so a(3) = 1. 4! = 24. The next prime greater than 24 is 29. 29 - 24 = 5, so a(4) = 5.
and so on...
- Task
- Find and display the first fifty terms in the series. (0! through 49!)
- Find and display the position and value of the first m greater than 1000.
- Stretch
- Find and display the position and value of each the first m greater than 2000, 3000, 4000 ... 10,000.
- See also
ALGOL 68
Uses ALGOL 68G's LONG LONG INT which has programmer specifiable precision, 500 digits is sufficient for the basic task (the Millar Rabin routine needs extra digits, even though 107! has around 170 digits).
BEGIN # find the least m such that n! + m is prime for various n #
PR precision 500 PR # set the precision of LONG LOMG INT #
PR read "primes.incl.a68" PR # include priee utilities #
LONG LONG INT factorial n := 1;
INT m := 0;
FOR n FROM 0 WHILE m < 1000 DO
IF n > 0 THEN
factorial n *:= n
FI;
m := 1;
WHILE NOT is probably prime( factorial n + m ) DO
m +:= 2
OD;
IF n < 50 THEN
print( ( " ", whole( m, -4 ) ) );
IF ( n + 1 ) MOD 10 = 0 THEN print( ( newline ) ) FI
ELIF m > 1000 THEN
print( ( "First m > 1000: ", whole( m, 0 ), " for ", whole( n, 0 ), "!", newline ) )
FI
OD
END- Output:
1 1 1 1 5 7 7 11 23 17 11 1 29 67 19 43 23 31 37 89 29 31 31 97 131 41 59 1 67 223 107 127 79 37 97 61 131 1 43 97 53 1 97 71 47 239 101 233 53 83 First m > 1000: 1069 for 107!
C
#include <stdio.h>
#include <gmp.h>
#include <locale.h>
#define LIMIT 10000
int main() {
mpz_t fact, p;
mpz_init_set_ui(fact, 1);
mpz_init(p);
int i, diffs[50], t = 1000;
unsigned long n, m;
for (n = 0; ; ++n) {
if (n > 0) mpz_mul_ui(fact, fact, n);
mpz_nextprime(p, fact);
mpz_sub(p, p, fact);
m = mpz_get_ui(p);
setlocale(LC_NUMERIC, "");
if (n < 50) diffs[n] = m;
if (n == 49) {
printf("Least positive m such that n! + m is prime; first 50:\n");
for (i = 0; i < 50; ++i) {
printf("%3d ", diffs[i]);
if (!((i+1)%10)) printf("\n");
}
printf("\n");
} else if (m > t) {
do {
printf("First m > %'6d is %'6ld at position %ld\n", t, m, n);
t += 1000;
} while (m > t);
if (t > LIMIT) break;
}
}
mpz_clear(fact);
mpz_clear(p);
return 0;
}
- Output:
Same as Wren example.
J
(4&p:-])!i.5 10x
1 1 1 1 5 7 7 11 23 17
11 1 29 67 19 43 23 31 37 89
29 31 31 97 131 41 59 1 67 223
107 127 79 37 97 61 131 1 43 97
53 1 97 71 47 239 101 233 53 83
1 i.~1000 < (4&p:-])!i.200x
107
Julia
""" rosettacode.orgwiki/Least_m_such_that_n!_%2B_m_is_prime """
using Primes
function least_m_fact_to_prime(number_to_print, delta_limit)
fact, p, m, n, t = big"1", big"0", big"0", 0, 1000
diffs = zeros(BigInt, number_to_print)
while true
if n > 0
fact *= n
p = nextprime(fact + 1)
m = p - fact
if n < number_to_print
diffs[n] = m
end
if n == number_to_print - 1
println("Least positive m such that n! + m is prime; first $number_to_print:")
for (i, k) in enumerate(diffs)
print(lpad(k, 5), i % 10 == 0 ? "\n" : "")
end
elseif m > t
while true
print("\nFirst m > $t is $m at position $n.")
t += 1000
if m <= t
break
end
end
if t > delta_limit
return
end
end
end
n += 1
end
end
least_m_fact_to_prime(50, 10_000)
- Output:
Least positive m such that n! + m is prime; first 50:
1 1 1 5 7 7 11 23 17 11
1 29 67 19 43 23 31 37 89 29
31 31 97 131 41 59 1 67 223 107
127 79 37 97 61 131 1 43 97 53
1 97 71 47 239 101 233 53 83 0
First m > 1000 is 1069 at position 107.
First m > 2000 is 3391 at position 192.
First m > 3000 is 3391 at position 192.
First m > 4000 is 4943 at position 284.
First m > 5000 is 5233 at position 384.
First m > 6000 is 6131 at position 388.
First m > 7000 is 9067 at position 445.
First m > 8000 is 9067 at position 445.
First m > 9000 is 9067 at position 445.
First m > 10000 is 12619 at position 599.
First m > 11000 is 12619 at position 599.
First m > 12000 is 12619 at position 599.
Nim
import std/strformat
import integers
const Lim = 10000
const Step = 1000
var n = 0
var lim = 1000
var f = newInteger(1)
echo "Least positive m such that n! + m is prime; first 50:"
while true:
var m = nextPrime(f) - f
if n < 50:
stdout.write &"{m:3}"
stdout.write if n mod 10 == 9: '\n' else: ' '
if n == 49: echo()
else:
while m > lim:
echo &"First m > {lim:5} is {m:5} at position {n}."
inc lim, Step
if lim > Lim: break
inc n
f *= n
- Output:
Least positive m such that n! + m is prime; first 50: 1 1 1 1 5 7 7 11 23 17 11 1 29 67 19 43 23 31 37 89 29 31 31 97 131 41 59 1 67 223 107 127 79 37 97 61 131 1 43 97 53 1 97 71 47 239 101 233 53 83 First m > 1000 is 1069 at position 107. First m > 2000 is 3391 at position 192. First m > 3000 is 3391 at position 192. First m > 4000 is 4943 at position 284. First m > 5000 is 5233 at position 384. First m > 6000 is 6131 at position 388. First m > 7000 is 9067 at position 445. First m > 8000 is 9067 at position 445. First m > 9000 is 9067 at position 445. First m > 10000 is 12619 at position 599. First m > 11000 is 12619 at position 599. First m > 12000 is 12619 at position 599.
Perl
use strict;
use warnings;
use ntheory qw<next_prime factorial vecfirstidx>;
my($n,@least_m) = 0;
do {
my $f = factorial($n++);
push @least_m, next_prime($f) - $f;
} until $least_m[-1] > 10000;
print "Least positive m such that n! + m is prime; first fifty:\n";
print sprintf(('%4d')x50, @least_m[0..49]) =~ s/.{40}\K(?=.)/\n/gr . "\n\n";
for my $n (map { 1000 * $_ } 1..10) {
my $key = vecfirstidx { $_ > $n } @least_m;
printf "First m > $n is %d at position %d\n", $least_m[$key], $key;
}
- Output:
Least positive m such that n! + m is prime; first fifty: 1 1 1 1 5 7 7 11 23 17 11 1 29 67 19 43 23 31 37 89 29 31 31 97 131 41 59 1 67 223 107 127 79 37 97 61 131 1 43 97 53 1 97 71 47 239 101 233 53 83 First m > 1000 is 1069 at position 107 First m > 2000 is 3391 at position 192 First m > 3000 is 3391 at position 192 First m > 4000 is 4943 at position 284 First m > 5000 is 5233 at position 384 First m > 6000 is 6131 at position 388 First m > 7000 is 9067 at position 445 First m > 8000 is 9067 at position 445 First m > 9000 is 9067 at position 445 First m > 10000 is 12619 at position 599
Phix
with javascript_semantics
atom t0 = time()
requires("1.0.3") -- mpz_nextprime() added
constant LIMIT = iff(platform()=JS?1000:10000)
include mpfr.e
mpz {fact, p} = mpz_inits(2,1)
sequence diffs = {}
integer n=0, m, t = 1000
while t<=LIMIT do
progress("position: %d\r",{n})
if n>0 then mpz_mul_si(fact, fact, n) end if
mpz_nextprime(p, fact)
mpz_sub(p, p, fact);
m = mpz_get_integer(p);
if length(diffs)<50 then
diffs &= m
if length(diffs)=50 then
printf(1,"Least positive m such that n! + m is prime; first 50:\n%s\n",
{join_by(diffs,1,10," ", fmt:="%3d")})
end if
elsif m>t then
string e = elapsed(time()-t0,0," (%s)")
do
printf(1,"First m > %,6d is %,6d at position %,d%s\n", {t, m, n, e})
e = ""
t += 1000
until t>m
end if
n += 1
end while
?elapsed(time()-t0)
- Output:
Least positive m such that n! + m is prime; first 50: 1 1 1 1 5 7 7 11 23 17 11 1 29 67 19 43 23 31 37 89 29 31 31 97 131 41 59 1 67 223 107 127 79 37 97 61 131 1 43 97 53 1 97 71 47 239 101 233 53 83 First m > 1,000 is 1,069 at position 107 (0.2s) First m > 2,000 is 3,391 at position 192 (3.4s) First m > 3,000 is 3,391 at position 192 First m > 4,000 is 4,943 at position 284 (27.3s) First m > 5,000 is 5,233 at position 384 (2 minutes and 15s) First m > 6,000 is 6,131 at position 388 First m > 7,000 is 9,067 at position 445 (4 minutes and 56s) First m > 8,000 is 9,067 at position 445 First m > 9,000 is 9,067 at position 445 First m > 10,000 is 12,619 at position 599 (26 minutes and 15s) First m > 11,000 is 12,619 at position 599 First m > 12,000 is 12,619 at position 599 "26 minutes and 15s"
For comparison, the Julia entry above took 18 mins 38s on the same box, and Perl an even more impressive 10 mins 44s.
Quackery
from, index, and end are defined at Loops/Increment loop index within loop body#Quackery.
prime is defined at Miller–Rabin primality test#Quackery.
[ 1 swap times [ i^ 1+ * ] ] is ! ( n --> n )
[] 50 times
[ i^ !
1+ from
[ index prime if
[ index i^ ! -
join end ] ] ]
[] swap
witheach
[ number$ nested join ]
49 wrap$- Output:
1 1 1 1 5 7 7 11 23 17 11 1 29 67 19 43 23 31 37 89 29 31 31 97 131 41 59 1 67 223 107 127 79 37 97 61 131 1 43 97 53 1 97 71 47 239 101 233 53 83
Raku
my @f = lazy flat 1, [\×] 1..*;
sink @f[700]; # pre-reify for concurrency
my @least-m = lazy (^∞).hyper(:2batch).map: {(1..*).first: -> \n {(@f[$_] + n).is-prime}};
say "Least positive m such that n! + m is prime; first fifty:\n"
~ @least-m[^50].batch(10)».fmt("%3d").join: "\n";
for (1..10).map: * × 1e3 {
my $key = @least-m.first: * > $_, :k;
printf "\nFirst m > $_ is %d at position %d\n", @least-m[$key], $key;
}
- Output:
Least positive m such that n! + m is prime; first fifty: 1 1 1 1 5 7 7 11 23 17 11 1 29 67 19 43 23 31 37 89 29 31 31 97 131 41 59 1 67 223 107 127 79 37 97 61 131 1 43 97 53 1 97 71 47 239 101 233 53 83 First m > 1000 is 1069 at position 107 First m > 2000 is 3391 at position 192 First m > 3000 is 3391 at position 192 First m > 4000 is 4943 at position 284 First m > 5000 is 5233 at position 384 First m > 6000 is 6131 at position 388 First m > 7000 is 9067 at position 445 First m > 8000 is 9067 at position 445 First m > 9000 is 9067 at position 445 First m > 10000 is 12619 at position 599
RPL
« 49 0 « ←n FACT DUP NEXTPRIME SWAP - »
→ max ←n m
« m '←n' 0 max 1 SEQ
max '←n' STO
DO '←n' INCR DROP m EVAL
UNTIL 1000 > END
←n
» » 'TASK' STO
- Output:
2: { 1 1 1 1 5 7 7 11 23 17 11 1 29 67 19 43 23 31 37 89 29 31 31 97 131 41 59 1 67 223 107 127 79 37 97 61 131 1 43 97 53 1 97 71 47 239 101 233 53 83 }
1: 107
Needs over an hour on an iOS emulator (iHP48).
Ruby
require 'openssl'
def next_prime(n) = ((n+1)..).detect{|n| OpenSSL::BN.new(n).prime?}
def fact(n) = (1..n).inject(:*) || 1
enum_diffs = (0..).lazy.map do |n|
f = fact(n)
next_prime(f) - f
end
enum_diffs.first(50).each_slice(10){|s| puts "%4d"*s.size % s}
puts "\nFirst m > 1000 is %d at position %d." % enum_diffs.with_index.detect{|d,_id| d>1000}
- Output:
1 1 1 1 5 7 7 11 23 17 11 1 29 67 19 43 23 31 37 89 29 31 31 97 131 41 59 1 67 223 107 127 79 37 97 61 131 1 43 97 53 1 97 71 47 239 101 233 53 83 First m > 1000 is 1069 at position 107.
Sidef
with (50) {|N|
say "Least positive m such that n! + m is prime (first #{N}):"
^N -> map {|n|
var f = n!; 1..Inf -> first {|k| f+k -> is_prime }
}.each_slice(10, {|*s|
say s.map{ '%3s' % _ }.join(' ')
})
}
say ''; var prev = 0
for n in (1..5 -> map { 1e3*_ }) {
var m = (prev..Inf -> lazy.map{|k|
var f = k!; [k, f.next_prime - f]
}.first {|k|
k.tail >= n
})
say "First m > #{n} is #{m.tail} at position #{m.head}"
prev = m.head
}
- Output:
Least positive m such that n! + m is prime (first 50): 1 1 1 1 5 7 7 11 23 17 11 1 29 67 19 43 23 31 37 89 29 31 31 97 131 41 59 1 67 223 107 127 79 37 97 61 131 1 43 97 53 1 97 71 47 239 101 233 53 83 First m > 1000 is 1069 at position 107 First m > 2000 is 3391 at position 192 First m > 3000 is 3391 at position 192 First m > 4000 is 4943 at position 284 First m > 5000 is 5233 at position 384
Wren
import "./gmp" for Mpz
import "./fmt" for Fmt
var fact = Mpz.one
var p = Mpz.new()
var diffs = List.filled(50, 0)
var n = 0
var t = 1000
var limit = 10000
while (true) {
if (n > 0) fact.mul(n)
p.nextPrime(fact)
var m = p.sub(fact).toNum
if (n < 50) diffs[n] = m
if (n == 49) {
System.print("Least positive m such that n! + m is prime; first 50:")
Fmt.tprint("$3d ", diffs, 10)
System.print()
} else if (m > t) {
while (true) {
Fmt.print("First m > $,6d is $,6d at position $d", t, m, n)
t = t + 1000
if (m <= t) break
}
if (t > limit) return
}
n = n + 1
}
- Output:
Least positive m such that n! + m is prime; first 50: 1 1 1 1 5 7 7 11 23 17 11 1 29 67 19 43 23 31 37 89 29 31 31 97 131 41 59 1 67 223 107 127 79 37 97 61 131 1 43 97 53 1 97 71 47 239 101 233 53 83 First m > 1,000 is 1,069 at position 107 First m > 2,000 is 3,391 at position 192 First m > 3,000 is 3,391 at position 192 First m > 4,000 is 4,943 at position 284 First m > 5,000 is 5,233 at position 384 First m > 6,000 is 6,131 at position 388 First m > 7,000 is 9,067 at position 445 First m > 8,000 is 9,067 at position 445 First m > 9,000 is 9,067 at position 445 First m > 10,000 is 12,619 at position 599 First m > 11,000 is 12,619 at position 599 First m > 12,000 is 12,619 at position 599