Least m such that n! + m is prime: Difference between revisions
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=={{header|RPL}}== |
=={{header|RPL}}== |
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« « |
« 49 0 « ←n FACT DUP NEXTPRIME SWAP - » |
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→ max ←n m |
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| ⚫ | |||
« m '←n' 0 max 1 SEQ |
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max '←n' STO |
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« 49 DUP <span style="color:blue">A33932</span> |
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'''DO''' '←n' INCR DROP m EVAL |
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SWAP |
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''' |
'''UNTIL''' 1000 > '''END''' |
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←n |
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DUP FACT DUP NEXTPRIME SWAP - |
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| ⚫ | |||
'''UNTIL''' 1000 > '''END''' |
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» '<span style="color:blue">TASK</span>' STO |
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{{out}} |
{{out}} |
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<pre> |
<pre> |
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Revision as of 09:07, 13 November 2023
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 31. 31 - 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
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
requires("1.0.3") -- mpz_nextprime() added constant LIMIT = 10000 include mpfr.e mpz {fact, p} = mpz_inits(2,1) sequence diffs = {} integer n=0, m, t = 1000 while t<=LIMIT do 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 do printf(1,"First m > %,6d is %,6d at position %,d\n", {t, m, n}) t += 1000 until t>m end if n += 1 end while
- 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
Note this is very slow, and still running..... so I killed it there
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.
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