Find prime numbers of the form n*n*n+2
- Task
- Find prime numbers of the form n3+2, where 0 < n < 200
Ada
<lang Ada>with Ada.Text_Io;
procedure Find_Primes is
type Number is new Long_Integer range 0 .. Long_Integer'Last; package Number_Io is new Ada.Text_Io.Integer_Io (Number);
function Is_Prime (A : Number) return Boolean is
D : Number;
begin
if A < 2 then return False; end if;
if A in 2 .. 3 then return True; end if;
if A mod 2 = 0 then return False; end if;
if A mod 3 = 0 then return False; end if;
D := 5;
while D * D < A loop
if A mod D = 0 then
return False;
end if;
D := D + 2;
if A mod D = 0 then
return False;
end if;
D := D + 4;
end loop;
return True;
end Is_Prime;
P : Number;
begin
Ada.Text_Io.Put_Line (" N N**3+2");
Ada.Text_Io.Put_Line ("------------");
for N in Number range 1 .. 199 loop
P := N**3 + 2;
if Is_Prime (P) then
Number_Io.Put (N, Width => 3); Ada.Text_Io.Put (" ");
Number_Io.Put (P, Width => 7);
Ada.Text_Io.New_Line;
end if;
end loop;
end Find_Primes;</lang>
- Output:
N N**3+2 ------------ 1 3 3 29 5 127 29 24391 45 91127 63 250049 65 274627 69 328511 71 357913 83 571789 105 1157627 113 1442899 123 1860869 129 2146691 143 2924209 153 3581579 171 5000213 173 5177719 189 6751271
ALGOL 68
<lang algol68>BEGIN # Find n such that n^3 + 2 is a prime for n < 200 #
FOR n TO 199 DO
INT candidate = ( n * n * n ) + 2;
# there will only be 199 candidates, so a primality check by trial #
# division should be OK #
BOOL is prime := TRUE;
FOR f FROM 2 TO ENTIER sqrt( candidate )
WHILE is prime := candidate MOD f /= 0
DO SKIP OD;
IF is prime THEN
# n^3 + 2 is prime #
print( ( whole( n, -4 ), ": ", whole( candidate, -8 ), newline ) )
FI
OD
END</lang>
- Output:
1: 3 3: 29 5: 127 29: 24391 45: 91127 63: 250049 65: 274627 69: 328511 71: 357913 83: 571789 105: 1157627 113: 1442899 123: 1860869 129: 2146691 143: 2924209 153: 3581579 171: 5000213 173: 5177719 189: 6751271
ALGOL W
<lang algolw>begin % Find n such that n^3 + 2 is a prime for n < 200 %
for n := 1 until 199 do begin
integer candidate;
logical isPrime;
candidate := ( n * n * n ) + 2;
% there will only be 199 candidates, so a primality check by trial %
% division should be OK %
isPrime := true;
for f := 2 until entier( sqrt( candidate ) ) do begin
isPrime := candidate rem f not = 0;
if not isPrime then goto endPrimalityCheck
end for_f ;
endPrimalityCheck:
if isPrime then begin
% n^3 + 2 is prime %
write( i_w := 4, s_w := 0, n, ": ", i_w := 8, candidate )
end if_isPrime
end for_n
end.</lang>
- Output:
1: 3 3: 29 5: 127 29: 24391 45: 91127 63: 250049 65: 274627 69: 328511 71: 357913 83: 571789 105: 1157627 113: 1442899 123: 1860869 129: 2146691 143: 2924209 153: 3581579 171: 5000213 173: 5177719 189: 6751271
AWK
<lang AWK>
- syntax: GAWK -f FIND_PRIME_NUMBERS_OF_THE_FORM_NNN2.AWK
BEGIN {
start = 1
stop = 200
for (n=start; n<=stop; n++) {
p = n*n*n + 2
if (is_prime(p)) {
printf("%3d %'10d\n",n,p)
count++
}
}
printf("Prime numbers %d-%d of the form n*n*n+2: %d\n",start,stop,count)
exit(0)
} function is_prime(x, i) {
if (x <= 1) {
return(0)
}
for (i=2; i<=int(sqrt(x)); i++) {
if (x % i == 0) {
return(0)
}
}
return(1)
} </lang>
- Output:
1 3 3 29 5 127 29 24,391 45 91,127 63 250,049 65 274,627 69 328,511 71 357,913 83 571,789 105 1,157,627 113 1,442,899 123 1,860,869 129 2,146,691 143 2,924,209 153 3,581,579 171 5,000,213 173 5,177,719 189 6,751,271 Prime numbers 1-200 of the form n*n*n+2: 19
C
<lang c>#include <stdio.h>
- include <stdbool.h>
- include <locale.h>
bool isPrime(int n) {
int d;
if (n < 2) return false;
if (!(n%2)) return n == 2;
if (!(n%3)) return n == 3;
d = 5;
while (d*d <= n) {
if (!(n%d)) return false;
d += 2;
if (!(n%d)) return false;
d += 4;
}
return true;
}
int main() {
int n, p;
const int limit = 200;
setlocale(LC_ALL, "");
for (n = 1; n < limit; ++n) {
p = n*n*n + 2;
if (isPrime(p)) {
printf("n = %3d => n³ + 2 = %'9d\n", n, p);
}
}
return 0;
}</lang>
- Output:
n = 1 => n³ + 2 = 3 n = 3 => n³ + 2 = 29 n = 5 => n³ + 2 = 127 n = 29 => n³ + 2 = 24,391 n = 45 => n³ + 2 = 91,127 n = 63 => n³ + 2 = 250,049 n = 65 => n³ + 2 = 274,627 n = 69 => n³ + 2 = 328,511 n = 71 => n³ + 2 = 357,913 n = 83 => n³ + 2 = 571,789 n = 105 => n³ + 2 = 1,157,627 n = 113 => n³ + 2 = 1,442,899 n = 123 => n³ + 2 = 1,860,869 n = 129 => n³ + 2 = 2,146,691 n = 143 => n³ + 2 = 2,924,209 n = 153 => n³ + 2 = 3,581,579 n = 171 => n³ + 2 = 5,000,213 n = 173 => n³ + 2 = 5,177,719 n = 189 => n³ + 2 = 6,751,271
C++
<lang cpp>#include <iomanip>
- include <iostream>
bool is_prime(unsigned int n) {
if (n < 2)
return false;
if (n % 2 == 0)
return n == 2;
if (n % 3 == 0)
return n == 3;
for (unsigned int p = 5; p * p <= n; p += 4) {
if (n % p == 0)
return false;
p += 2;
if (n % p == 0)
return false;
}
return true;
}
int main() {
for (unsigned int n = 1; n < 200; n += 2) {
auto p = n * n * n + 2;
if (is_prime(p))
std::cout << std::setw(3) << n << std::setw(9) << p << '\n';
}
}</lang>
- Output:
1 3 3 29 5 127 29 24391 45 91127 63 250049 65 274627 69 328511 71 357913 83 571789 105 1157627 113 1442899 123 1860869 129 2146691 143 2924209 153 3581579 171 5000213 173 5177719 189 6751271
CLU
<lang clu>is_prime = proc (n: int) returns (bool)
if n<2 then return(false) end
if n//2=0 then return(n=2) end
if n//3=0 then return(n=3) end
d: int := 5
while d*d <= n do
if n//d=0 then return(false) end
d := d+2
if n//d=0 then return(false) end
d := d+4
end
return(true)
end is_prime
n3plus2_primes = iter (max: int) yields (int,int)
for n: int in int$from_to(1, max) do
p: int := n**3 + 2
if is_prime(p) then yield(n,p) end
end
end n3plus2_primes
start_up = proc ()
po: stream := stream$primary_output()
for n, p: int in n3plus2_primes(200) do
stream$puts(po, "n = ")
stream$putright(po, int$unparse(n), 3)
stream$puts(po, " => n^3 + 2 = ")
stream$putright(po, int$unparse(p), 7)
stream$putl(po, "")
end
end start_up</lang>
- Output:
n = 1 => n^3 + 2 = 3 n = 3 => n^3 + 2 = 29 n = 5 => n^3 + 2 = 127 n = 29 => n^3 + 2 = 24391 n = 45 => n^3 + 2 = 91127 n = 63 => n^3 + 2 = 250049 n = 65 => n^3 + 2 = 274627 n = 69 => n^3 + 2 = 328511 n = 71 => n^3 + 2 = 357913 n = 83 => n^3 + 2 = 571789 n = 105 => n^3 + 2 = 1157627 n = 113 => n^3 + 2 = 1442899 n = 123 => n^3 + 2 = 1860869 n = 129 => n^3 + 2 = 2146691 n = 143 => n^3 + 2 = 2924209 n = 153 => n^3 + 2 = 3581579 n = 171 => n^3 + 2 = 5000213 n = 173 => n^3 + 2 = 5177719 n = 189 => n^3 + 2 = 6751271
Delphi
<lang Delphi> program Find_prime_numbers_of_the_form_n_n_n_plus_2;
{$APPTYPE CONSOLE}
uses
System.SysUtils, PrimTrial;
function Commatize(n: NativeInt): string; var
fmt: TFormatSettings;
begin
fmt := TFormatSettings.Create('en-Us');
Result := double(n).ToString(ffNumber, 64, 0, fmt);
end;
const
limit = 200;
begin
for var n := 1 to limit - 1 do
begin
var p := n * n * n + 2;
if isPrime(p) then
writeln('n = ', n: 3, ' => n^3 + 2 = ', Commatize(p): 9);
end;
readln;
end.</lang>
- Output:
n = 1 => n^3 + 2 = 3 n = 3 => n^3 + 2 = 29 n = 5 => n^3 + 2 = 127 n = 29 => n^3 + 2 = 24,391 n = 45 => n^3 + 2 = 91,127 n = 63 => n^3 + 2 = 250,049 n = 65 => n^3 + 2 = 274,627 n = 69 => n^3 + 2 = 328,511 n = 71 => n^3 + 2 = 357,913 n = 83 => n^3 + 2 = 571,789 n = 105 => n^3 + 2 = 1,157,627 n = 113 => n^3 + 2 = 1,442,899 n = 123 => n^3 + 2 = 1,860,869 n = 129 => n^3 + 2 = 2,146,691 n = 143 => n^3 + 2 = 2,924,209 n = 153 => n^3 + 2 = 3,581,579 n = 171 => n^3 + 2 = 5,000,213 n = 173 => n^3 + 2 = 5,177,719 n = 189 => n^3 + 2 = 6,751,271
F#
This task uses Extensible Prime Generator (F#).
<lang fsharp>
[1..2..200]|>Seq.filter(fun n->isPrime(2+pown n 3))|>Seq.iter(fun n->printfn "n=%3d -> %d" n (2+pown n 3))
</lang>
- Output:
n= 1 -> 3 n= 3 -> 29 n= 5 -> 127 n= 29 -> 24391 n= 45 -> 91127 n= 63 -> 250049 n= 65 -> 274627 n= 69 -> 328511 n= 71 -> 357913 n= 83 -> 571789 n=105 -> 1157627 n=113 -> 1442899 n=123 -> 1860869 n=129 -> 2146691 n=143 -> 2924209 n=153 -> 3581579 n=171 -> 5000213 n=173 -> 5177719 n=189 -> 6751271
Factor
Using the parity optimization from the Wren entry:
<lang factor>USING: formatting kernel math math.functions math.primes math.ranges sequences tools.memory.private ;
1 199 2 <range> [
dup 3 ^ 2 + dup prime? [ commas "n = %3d => n³ + 2 = %9s\n" printf ] [ 2drop ] if
] each</lang> Or, using local variables:
<lang factor>USING: formatting kernel math math.primes math.ranges sequences tools.memory.private ;
[let
199 :> limit
1 limit 2 <range> [| n |
n n n * * 2 + :> p
p prime?
[ n p commas "n = %3d => n³ + 2 = %9s\n" printf ] when
] each
]</lang>
- Output:
n = 1 => n³ + 2 = 3 n = 3 => n³ + 2 = 29 n = 5 => n³ + 2 = 127 n = 29 => n³ + 2 = 24,391 n = 45 => n³ + 2 = 91,127 n = 63 => n³ + 2 = 250,049 n = 65 => n³ + 2 = 274,627 n = 69 => n³ + 2 = 328,511 n = 71 => n³ + 2 = 357,913 n = 83 => n³ + 2 = 571,789 n = 105 => n³ + 2 = 1,157,627 n = 113 => n³ + 2 = 1,442,899 n = 123 => n³ + 2 = 1,860,869 n = 129 => n³ + 2 = 2,146,691 n = 143 => n³ + 2 = 2,924,209 n = 153 => n³ + 2 = 3,581,579 n = 171 => n³ + 2 = 5,000,213 n = 173 => n³ + 2 = 5,177,719 n = 189 => n³ + 2 = 6,751,271
Fermat
<lang fermat>for n=1,199 do if Isprime(n^3+2)=1 then !!(n,n^3+2) fi od</lang>
- Output:
1 3 3 29 5 127 29 24391 45 91127 63 250049 65 274627 69 328511 71 357913 83 571789 105 1157627 113 1442899 123 1860869 129 2146691 143 2924209 153 3581579 171 5000213 173 5177719 189 6751271
Forth
<lang forth>: prime? ( n -- flag )
dup 2 < if drop false exit then dup 2 mod 0= if 2 = exit then dup 3 mod 0= if 3 = exit then 5 begin 2dup dup * >= while 2dup mod 0= if 2drop false exit then 2 + 2dup mod 0= if 2drop false exit then 4 + repeat 2drop true ;
- main
200 1 do
i i i * * 2 + dup prime? if
i 3 .r 9 .r cr
else
drop
then
2 +loop ;
main bye</lang>
- Output:
1 3 3 29 5 127 29 24391 45 91127 63 250049 65 274627 69 328511 71 357913 83 571789 105 1157627 113 1442899 123 1860869 129 2146691 143 2924209 153 3581579 171 5000213 173 5177719 189 6751271
FreeBASIC
Use the code from Primality by trial division#FreeBASIC as an include. <lang freebasic>#include"isprime.bas"
for n as uinteger = 1 to 200
if isprime(n^3+2) then
print n, n^3+2
end if
next n</lang>
- Output:
1 3 3 29 5 127 29 24391 45 91127 63 250049 65 274627 69 328511 71 357913 83 571789 105 1157627 113 1442899 123 1860869 129 2146691 143 2924209 153 3581579 171 5000213 173 5177719 189 6751271
Fōrmulæ
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.
Programs in Fōrmulæ are created/edited online in its website, However they run on execution servers. By default remote servers are used, but they are limited in memory and processing power, since they are intended for demonstration and casual use. A local server can be downloaded and installed, it has no limitations (it runs in your own computer). Because of that, example programs can be fully visualized and edited, but some of them will not run if they require a moderate or heavy computation/memory resources, and no local server is being used.
In this page you can see the program(s) related to this task and their results.
Go
<lang go>package main
import "fmt"
func isPrime(n int) bool {
switch {
case n < 2:
return false
case n%2 == 0:
return n == 2
case n%3 == 0:
return n == 3
default:
d := 5
for d*d <= n {
if n%d == 0 {
return false
}
d += 2
if n%d == 0 {
return false
}
d += 4
}
return true
}
}
func commatize(n int) string {
s := fmt.Sprintf("%d", n)
if n < 0 {
s = s[1:]
}
le := len(s)
for i := le - 3; i >= 1; i -= 3 {
s = s[0:i] + "," + s[i:]
}
if n >= 0 {
return s
}
return "-" + s
}
func main() {
const limit = 200
for n := 1; n < limit; n++ {
p := n*n*n + 2
if isPrime(p) {
fmt.Printf("n = %3d => n³ + 2 = %9s\n", n, commatize(p))
}
}
}</lang>
- Output:
n = 1 => n³ + 2 = 3 n = 3 => n³ + 2 = 29 n = 5 => n³ + 2 = 127 n = 29 => n³ + 2 = 24,391 n = 45 => n³ + 2 = 91,127 n = 63 => n³ + 2 = 250,049 n = 65 => n³ + 2 = 274,627 n = 69 => n³ + 2 = 328,511 n = 71 => n³ + 2 = 357,913 n = 83 => n³ + 2 = 571,789 n = 105 => n³ + 2 = 1,157,627 n = 113 => n³ + 2 = 1,442,899 n = 123 => n³ + 2 = 1,860,869 n = 129 => n³ + 2 = 2,146,691 n = 143 => n³ + 2 = 2,924,209 n = 153 => n³ + 2 = 3,581,579 n = 171 => n³ + 2 = 5,000,213 n = 173 => n³ + 2 = 5,177,719 n = 189 => n³ + 2 = 6,751,271
jq
Works with gojq, the Go implementation of jq
Using a definition of `is_prime` such as can be found at Safe_primes_and_unsafe_primes: <lang jq> range(1;200) | pow(.; 3) + 2 | select(is_prime) </lang>
- Output:
3 29 127 24391 91127 250049 274627 328511 357913 571789 1157627 1442899 1860869 2146691 2924209 3581579 5000213 5177719 6751271
Julia
<lang julia># Formatting output as in Go example. using Primes, Formatting
isncubedplus2prime(x) = begin fx = x * x * x + 2; (isprime(fx), fx) end
tostring(x, fx) = "n = " * lpad(x, 3) * " => n³ + 2 =" * lpad(format(fx, commas=true), 10)
function filterprintresults(x_to_bool_and_fx, start, stop, stringify=(x, fx)->"$x $fx", doprint=true)
ncount = 0
println("Filtering $x_to_bool_and_fx for integers between $start and $stop:\n")
for n in start+1:stop-1
isone, result = x_to_bool_and_fx(n)
if isone
doprint && println(stringify(n, result))
ncount += 1
end
end
println("\nThe total found was $ncount.")
end
filterprintresults(isncubedplus2prime, 0, 200, tostring)
</lang>
- Output:
n = 1 => n³ + 2 = 3 n = 3 => n³ + 2 = 29 n = 5 => n³ + 2 = 127 n = 29 => n³ + 2 = 24,391 n = 45 => n³ + 2 = 91,127 n = 63 => n³ + 2 = 250,049 n = 65 => n³ + 2 = 274,627 n = 69 => n³ + 2 = 328,511 n = 71 => n³ + 2 = 357,913 n = 83 => n³ + 2 = 571,789 n = 105 => n³ + 2 = 1,157,627 n = 113 => n³ + 2 = 1,442,899 n = 123 => n³ + 2 = 1,860,869 n = 129 => n³ + 2 = 2,146,691 n = 143 => n³ + 2 = 2,924,209 n = 153 => n³ + 2 = 3,581,579 n = 171 => n³ + 2 = 5,000,213 n = 173 => n³ + 2 = 5,177,719 n = 189 => n³ + 2 = 6,751,271 The total found was 19.
One-liner version
<lang julia>using Primes; println(filter(isprime, map(x -> x^3 + 2, 1:199)))</lang>
- Output:
[3, 29, 127, 24391, 91127, 250049, 274627, 328511, 357913, 571789, 1157627, 1442899, 1860869, 2146691, 2924209, 3581579, 5000213, 5177719, 6751271]
Mathematica/Wolfram Language
<lang Mathematica>Select[Range[199]^3 + 2, PrimeQ]</lang>
- Output:
{3, 29, 127, 24391, 91127, 250049, 274627, 328511, 357913, 571789, 1157627, 1442899, 1860869, 2146691, 2924209, 3581579, 5000213, 5177719, 6751271}
Nim
<lang Nim>import strutils
func isPrime(n: Positive): bool =
if n < 2: return false if n mod 2 == 0: return n == 2 if n mod 3 == 0: return n == 3 var d = 5 while d * d <= n: if n mod d == 0: return false inc d, 2 if n mod d == 0: return false inc d, 4 result = true
for n in 1..<200:
let p = n * n * n + 2 if p.isPrime: echo ($n).align(3), " → ", p</lang>
- Output:
1 → 3 3 → 29 5 → 127 29 → 24391 45 → 91127 63 → 250049 65 → 274627 69 → 328511 71 → 357913 83 → 571789 105 → 1157627 113 → 1442899 123 → 1860869 129 → 2146691 143 → 2924209 153 → 3581579 171 → 5000213 173 → 5177719 189 → 6751271
PARI/GP
<lang parigp>for(N=1,200,if(isprime(N^3+2),print(N," ",N^3+2)))</lang>
- Output:
1 3 3 29 5 127 29 24391 45 91127 63 250049 65 274627 69 328511 71 357913 83 571789 105 1157627 113 1442899 123 1860869 129 2146691 143 2924209 153 3581579 171 5000213 173 5177719 189 6751271
Perl
<lang perl>use strict; use warnings; use feature 'say';
- basic task results
say join ' ', grep { is_prime $_ } map { $_**3 + 2 } grep { 0 != $_%2 } 1..199;
- generalize a bit, how many primes over a range of exponents and offsets?
use Math::AnyNum ':all'; # in order to handle large values say ' ' . sprintf '%4d'x11 , 1..10; for my $e (1..10) {
printf '%2d ', $e;
for my $o (1..10) {
printf '%4d', scalar grep { is_prime $_ } map { $_**$e + $o } 1..199;
}
print "\n";
}</lang>
- Output:
3 29 127 24391 91127 250049 274627 328511 357913 571789 1157627 1442899 1860869 2146691 2924209 3581579 5000213 5177719 6751271
1 2 3 4 5 6 7 8 9 10
1 46 45 44 44 43 43 42 42 42 42
2 34 17 29 34 12 19 49 19 24 32
3 1 19 26 25 23 17 18 0 28 20
4 30 13 20 1 7 12 28 7 6 11
5 1 12 14 14 11 7 15 17 12 3
6 1 5 19 11 3 2 24 0 7 11
7 1 10 8 8 7 9 7 6 9 8
8 7 7 7 1 2 5 9 5 1 8
9 1 5 7 7 5 5 6 0 9 6
10 1 3 3 9 3 1 5 3 2 1
Phix
function pn3p2(integer n) integer n3p2 = power(n,3)+2 return iff(is_prime(n3p2)?{n,n3p2}:0) end function sequence res = filter(apply(tagset(199,1,2),pn3p2),"!=",0) printf(1,"Found %d primes of the form n^3+2:\n",length(res)) papply(true,printf,{1,{"n = %3d => n^3+2 = %,9d\n"},res})
- Output:
Found 19 primes of the form n^3+2: n = 1 => n^3+2 = 3 n = 3 => n^3+2 = 29 n = 5 => n^3+2 = 127 n = 29 => n^3+2 = 24,391 n = 45 => n^3+2 = 91,127 n = 63 => n^3+2 = 250,049 n = 65 => n^3+2 = 274,627 n = 69 => n^3+2 = 328,511 n = 71 => n^3+2 = 357,913 n = 83 => n^3+2 = 571,789 n = 105 => n^3+2 = 1,157,627 n = 113 => n^3+2 = 1,442,899 n = 123 => n^3+2 = 1,860,869 n = 129 => n^3+2 = 2,146,691 n = 143 => n^3+2 = 2,924,209 n = 153 => n^3+2 = 3,581,579 n = 171 => n^3+2 = 5,000,213 n = 173 => n^3+2 = 5,177,719 n = 189 => n^3+2 = 6,751,271
Plain English
<lang plainenglish>To run: Start up. Put 1 into a counter. Loop. Put the counter into a number. Raise the number to 3. Add 2 to the number. If the number is prime, write the counter then " " then the number on the console. Add 2 to the counter. If the counter is greater than 200, break. Repeat. Write "Done." on the console. Wait for the escape key. Shut down.</lang>
- Output:
1 3 3 29 5 127 29 24391 45 91127 63 250049 65 274627 69 328511 71 357913 83 571789 105 1157627 113 1442899 123 1860869 129 2146691 143 2924209 153 3581579 171 5000213 173 5177719 189 6751271 Done.
Raku
<lang perl6># 20210315 Raku programming solution
say ((1…199)»³ »+»2).grep: *.is-prime</lang>
- Output:
(3 29 127 24391 91127 250049 274627 328511 357913 571789 1157627 1442899 1860869 2146691 2924209 3581579 5000213 5177719 6751271)
REXX
Since REXX doesn't have a isPrime function, this REXX program generates a number of primes such that some
numbers can be tested for primality directly, other numbers have to be tested by trial division for primality.
A suitable number was calculated to generate a number of primes such that about half of the computing time used to
test the numbers for primality could be directly determined their primality, the other half of the computing time used
would use trial division.
Since the task's requirements are pretty straight-forward and easy, a little extra code was added for presentation
(title and title separator line, the count of primes found, and commatization of the numbers).
<lang rexx>/*REXX program finds and displays n primes of the form: n**3 + 2. */
parse arg LO HI hp . /*obtain optional argument from the CL.*/
if LO== | LO=="," then LO= 0 /*Not specified? Then use the default.*/
if HI== | HI=="," then HI= 200 /* " " " " " " */
if hp== | hp=="," then hp= 19 /* " " " " " " */
h= max(iSqrt(HI**3), hp**3) /*a high prime to generate primes to. */
w= length( commas(HI**3) ) + 3
call genP /*build array of semaphores for primes.*/
say right('n', 20) ' (n**3 + 2)' /*display a title for the output list. */
say left(, 20 + w + 20, '─') /*display a sep for the output list. */
finds= 0 /*# of triplet strange primes (so far).*/
do j=LO+1 by 2 to HI-1 /*look for primes of form of: n**3 + 2 */
x= j**3 + 2
if x<=@.# then if \!.x then iterate /*Not a semaphore prime? Then skip it.*/
else nop /*the NOP matches up with the "THEN".*/
else do k=2 while @.k**2<=x /*perform a primality test by division.*/
if x//@.k==0 then iterate j
end /*k*/
finds= finds + 1 /*bump # primes found of form: n**3+2 */
say right(commas(j), 20) right( commas(x), w)
end /*j*/
say left(, 20 + w + 20, '─'); say /*display a sep for the output list. */ say 'Found ' commas(finds) ' primes in the form of: n**3 + 2' exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ? /*──────────────────────────────────────────────────────────────────────────────────────*/ iSqrt: procedure; parse arg x; r=0; q=1; do while q<=x; q=q*4; end
do while q>1; q=q%4; _=x-r-q; r=r%2; if _>=0 then do;x=_;r=r+q; end; end
return r
/*──────────────────────────────────────────────────────────────────────────────────────*/ genP: !.= 0; pad= left(, 9) /*placeholders for primes; width of #'s*/
@.1=2; @.2=3; @.3=5; @.4=7; @.5=11 /*define some low primes. */
!.2=1; !.3=1; !.5=1; !.7=1; !.11=1 /* " " " " flags. */
#=5; s.#= @.# **2 /*number of primes so far; prime². */
/* [↓] generate more primes ≤ h. */
do j=@.#+2 by 2 to h /*find odd primes from here on. */
parse var j -1 _; if _==5 then iterate /*J divisible by 5? (right dig)*/
if j// 3==0 then iterate /*" " " 3? */
if j// 7==0 then iterate /*" " " 7? */
/* [↑] the above five lines saves time*/
do k=5 while s.k<=j /* [↓] divide by the known odd primes.*/
if j // @.k == 0 then iterate j /*Is J ÷ X? Then not prime. ___ */
end /*k*/ /* [↑] only process numbers ≤ √ J */
#= #+1; @.#= j; s.#= j*j; !.j= 1 /*bump # of Ps; assign next P; P²; P# */
end /*j*/; return</lang>
- output when using the default inputs:
n (n**3 + 2)
────────────────────────────────────────────────────
1 3
3 29
5 127
29 24,391
45 91,127
63 250,049
65 274,627
69 328,511
71 357,913
83 571,789
105 1,157,627
113 1,442,899
123 1,860,869
129 2,146,691
143 2,924,209
153 3,581,579
171 5,000,213
173 5,177,719
189 6,751,271
────────────────────────────────────────────────────
Found 19 primes in the form of: n**3 + 2
Ring
<lang ring> load "stdlib.ring"
see "working..." + nl
for n = 1 to 200 step 2
pr = pow(n,3)+2
if isprime(pr)
see "n = " + n + " => n³+2 = " + pr + nl
ok
next
see "done..." + nl </lang>
- Output:
working... n = 1 => n³+2 = 3 n = 3 => n³+2 = 29 n = 5 => n³+2 = 127 n = 29 => n³+2 = 24391 n = 45 => n³+2 = 91127 n = 63 => n³+2 = 250049 n = 65 => n³+2 = 274627 n = 69 => n³+2 = 328511 n = 71 => n³+2 = 357913 n = 83 => n³+2 = 571789 n = 105 => n³+2 = 1157627 n = 113 => n³+2 = 1442899 n = 123 => n³+2 = 1860869 n = 129 => n³+2 = 2146691 n = 143 => n³+2 = 2924209 n = 153 => n³+2 = 3581579 n = 171 => n³+2 = 5000213 n = 173 => n³+2 = 5177719 n = 189 => n³+2 = 6751271 done...
Rust
<lang rust>// 202100327 Rust programming solution
use primes::is_prime;
fn main() {
let mut count = 0; let begin = 0; let end = 200;
println!("Find prime numbers of the form");
println!(" n => n³ + 2 ");
for n in begin+1..end-1 {
let m = n*n*n+2;
if is_prime(m) {
println!("{:4} => {}", n, m);
count += 1;
}
}
println!("Found {} such prime numbers where {} < n < {}.", count,begin,end);
}</lang>
- Output:
Find prime numbers of the form n => n³ + 2 1 => 3 3 => 29 5 => 127 29 => 24391 45 => 91127 63 => 250049 65 => 274627 69 => 328511 71 => 357913 83 => 571789 105 => 1157627 113 => 1442899 123 => 1860869 129 => 2146691 143 => 2924209 153 => 3581579 171 => 5000213 173 => 5177719 189 => 6751271 Found 19 such prime numbers where 0 < n < 200.
Seed7
Credit for isPrime function: [1]
<lang seed7>$ include "seed7_05.s7i";
const func boolean: isPrime (in integer: number) is func
result
var boolean: prime is FALSE;
local
var integer: upTo is 0;
var integer: testNum is 3;
begin
if number = 2 then
prime := TRUE;
elsif odd(number) and number > 2 then
upTo := sqrt(number);
while number rem testNum <> 0 and testNum <= upTo do
testNum +:= 2;
end while;
prime := testNum > upTo;
end if;
end func;
const proc: main is func
local
const integer: limit is 199;
var integer: n is 1;
var integer: p is 0;
begin
writeln(" n n**3+2");
writeln("------------");
for n range 1 to limit step 2 do
p := n ** 3 + 2;
if isPrime(p) then
writeln(n lpad 3 <& p lpad 9);
end if;
end for;
end func;</lang>
- Output:
n n**3+2 ------------ 1 3 3 29 5 127 29 24391 45 91127 63 250049 65 274627 69 328511 71 357913 83 571789 105 1157627 113 1442899 123 1860869 129 2146691 143 2924209 153 3581579 171 5000213 173 5177719 189 6751271
Sidef
<lang ruby>1..^200 -> map { _**3 + 2 }.grep {.is_prime}.say</lang>
- Output:
[3, 29, 127, 24391, 91127, 250049, 274627, 328511, 357913, 571789, 1157627, 1442899, 1860869, 2146691, 2924209, 3581579, 5000213, 5177719, 6751271]
Swift
<lang swift>import Foundation
func isPrime(_ n: Int) -> Bool {
if n < 2 {
return false
}
if n % 2 == 0 {
return n == 2
}
if n % 3 == 0 {
return n == 3
}
var p = 5
while p * p <= n {
if n % p == 0 {
return false
}
p += 2
if n % p == 0 {
return false
}
p += 4
}
return true
}
for n in 1...200 {
let p = n * n * n + 2
if isPrime(p) {
print(String(format: "%3d%9d", n, p))
}
}</lang>
- Output:
1 3 3 29 5 127 29 24391 45 91127 63 250049 65 274627 69 328511 71 357913 83 571789 105 1157627 113 1442899 123 1860869 129 2146691 143 2924209 153 3581579 171 5000213 173 5177719 189 6751271
Wren
If n is even then n³ + 2 is also even, so we only need to examine odd values of n here. <lang ecmascript>import "/math" for Int import "/trait" for Stepped import "/fmt" for Fmt
var limit = 200 for (n in Stepped.new(1...limit, 2)) {
var p = n*n*n + 2
if (Int.isPrime(p)) Fmt.print("n = $3d => n³ + 2 = $,9d", n, p)
}</lang>
- Output:
n = 1 => n³ + 2 = 3 n = 3 => n³ + 2 = 29 n = 5 => n³ + 2 = 127 n = 29 => n³ + 2 = 24,391 n = 45 => n³ + 2 = 91,127 n = 63 => n³ + 2 = 250,049 n = 65 => n³ + 2 = 274,627 n = 69 => n³ + 2 = 328,511 n = 71 => n³ + 2 = 357,913 n = 83 => n³ + 2 = 571,789 n = 105 => n³ + 2 = 1,157,627 n = 113 => n³ + 2 = 1,442,899 n = 123 => n³ + 2 = 1,860,869 n = 129 => n³ + 2 = 2,146,691 n = 143 => n³ + 2 = 2,924,209 n = 153 => n³ + 2 = 3,581,579 n = 171 => n³ + 2 = 5,000,213 n = 173 => n³ + 2 = 5,177,719 n = 189 => n³ + 2 = 6,751,271
XPL0
<lang XPL0>func IsPrime(N); \Return 'true' if N is a prime number int N, I; [if N <= 1 then return false; for I:= 2 to sqrt(N) do
if rem(N/I) = 0 then return false;
return true; ];
int N; [for N:= 1 to 199 do
[if IsPrime(N*N*N+2) then
[IntOut(0, N);
ChOut(0, 9\tab\);
IntOut(0, N*N*N+2);
CrLf(0);
]
];
]</lang>
- Output:
1 3 3 29 5 127 29 24391 45 91127 63 250049 65 274627 69 328511 71 357913 83 571789 105 1157627 113 1442899 123 1860869 129 2146691 143 2924209 153 3581579 171 5000213 173 5177719 189 6751271