Cubic special primes: Difference between revisions
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Cubic special primes 2-15000: 23
</pre>
=={{header|C}}==
{{trans|Wren}}
<syntaxhighlight lang="c">#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <math.h>
#include <locale.h>
bool *sieve(int limit) {
int i, p;
limit++;
// True denotes composite, false denotes prime.
bool *c = calloc(limit, sizeof(bool)); // all false by default
c[0] = true;
c[1] = true;
for (i = 4; i < limit; i += 2) c[i] = true;
p = 3; // Start from 3.
while (true) {
int p2 = p * p;
if (p2 >= limit) break;
for (i = p2; i < limit; i += 2 * p) c[i] = true;
while (true) {
p += 2;
if (!c[p]) break;
}
}
return c;
}
bool isCube(int n) {
int s = (int)cbrt((double)n);
return s*s*s == n;
}
int main() {
const int limit = 14999;
int i, j, p, pc = 0, gap = 1, count = 1, lastCubicSpecial = 3;
const char *fmt = "%'7d %'7d %'6d %'4d\n";
bool *c = sieve(limit);
for (i = 0; i < limit; ++i) {
if (!c[i]) ++pc;
}
int *primes = (int *)malloc(pc * sizeof(int));
for (i = 0, j = 0; i < limit; ++i) {
if (!c[i]) primes[j++] = i;
}
free(c);
printf("Cubic special primes under 15,000:\n");
printf(" Prime1 Prime2 Gap Cbrt\n");
setlocale(LC_NUMERIC, "");
printf(fmt, 2, 3, 1, 1);
for (i = 2; i < pc; ++i) {
p = primes[i];
gap = p - lastCubicSpecial;
if (isCube(gap)) {
printf(fmt, lastCubicSpecial, p, gap, (int)cbrt((double)gap));
lastCubicSpecial = p;
++count;
}
}
printf("\n%d such primes found.\n", count+1);
free(primes);
return 0;
}</syntaxhighlight>
{{out}}
<pre>
Same as Wren example.
</pre>
|
Revision as of 15:32, 8 April 2023
Cubic special primes 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.
- Task
n is smallest prime such that the difference of successive terms are the smallest cubics of positive integers,
where n < 15000.
11l
F is_prime(a)
I a == 2
R 1B
I a < 2 | a % 2 == 0
R 0B
L(i) (3 .. Int(sqrt(a))).step(2)
I a % i == 0
R 0B
R 1B
V p = 2
V n = 1
print(2, end' ‘ ’)
L p + n ^ 3 < 15000
I is_prime(p + n ^ 3)
p += n ^ 3
n = 1
print(p, end' ‘ ’)
E
n++
- Output:
2 3 11 19 83 1811 2027 2243 2251 2467 2531 2539 3539 3547 4547 5059 10891 12619 13619 13627 13691 13907 14419
Action!
INCLUDE "H6:SIEVE.ACT"
PROC Main()
DEFINE MAX="14999"
BYTE ARRAY primes(MAX+1)
INT i=[2],next,count=[1],n=[1]
Put(125) PutE() ;clear the screen
Sieve(primes,MAX+1)
PrintI(i)
DO
next=i+n*n*n
IF next>=MAX THEN
EXIT
FI
IF primes(next) THEN
n=1 i=next count==+1
Put(32) PrintI(i)
ELSE
n==+1
FI
OD
PrintF("%E%EThere are %I cubic special primes",count)
RETURN
- Output:
Screenshot from Atari 8-bit computer
2 3 11 19 83 1811 2027 2243 2251 2467 2531 2539 3539 3547 4547 5059 10891 12619 13619 13627 13691 13907 14419 There are 23 cubic special primes
ALGOL 68
BEGIN # find a sequence of primes where the members differ by a cube #
INT max prime = 15 000;
# sieve the primes to max prime #
PR read "primes.incl.a68" PR
[]BOOL prime = PRIMESIEVE max prime;
# construct a table of cubes, we will need at most the cube root of max prime #
[ 1 : ENTIER exp( ln( max prime ) / 3 ) ]INT cube;
FOR i TO UPB cube DO cube[ i ] := i * i * i OD;
# find the prime sequence #
print( ( "2" ) );
INT p := 2;
WHILE p < max prime DO
# find a prime that is p + a cube #
INT q := 0;
FOR i WHILE q := p + cube[ i ];
IF q > max prime THEN FALSE ELSE NOT prime[ q ] FI
DO SKIP OD;
IF q <= max prime THEN print( ( " ", whole( q, 0 ) ) ) FI;
p := q
OD;
print( ( newline ) )
END
- Output:
2 3 11 19 83 1811 2027 2243 2251 2467 2531 2539 3539 3547 4547 5059 10891 12619 13619 13627 13691 13907 14419
ALGOL W
begin % find a sequence of primes where the members differ by a cube %
integer MAX_PRIME, MAX_CUBE;
MAX_PRIME := 15000;
MAX_CUBE := entier( exp( ln( MAX_PRIME ) / 3 ) );
begin
logical array prime ( 1 :: MAX_PRIME );
integer array cube ( 1 :: MAX_CUBE );
integer p, q, c;
% sieve the primes to MAX_PRIME %
prime( 1 ) := false; prime( 2 ) := true;
for i := 3 step 2 until MAX_PRIME do prime( i ) := true;
for i := 4 step 2 until MAX_PRIME do prime( i ) := false;
for i := 3 step 2 until entier( sqrt( MAX_PRIME ) ) do begin
integer ii; ii := i + i;
if prime( i ) then for s := i * i step ii until MAX_PRIME do prime( s ) := false
end for_i;
% construct a table of cubes, we will need at most the cube root of max prime %
for i := 1 until MAX_CUBE do cube( i ) := i * i * i;
% find the prime sequence %
writeon( "2" );
p := 2;
while p < MAX_PRIME do begin
% find a prime that is p + a cube %
c := 1;
q := p + cube( c );
while if q > MAX_PRIME then false else not prime( q ) do begin
c := c + 1;
q := p + cube( c )
end while_not_prime_q ;
if q <= MAX_PRIME then writeon( i_w := 1, s_w := 0, " ", q );
p := q
end while_p_lt_MAX_PRIME
end;
write()
end.
- Output:
2 3 11 19 83 1811 2027 2243 2251 2467 2531 2539 3539 3547 4547 5059 10891 12619 13619 13627 13691 13907 14419
Arturo
[p n]: [2 1]
prints -> 2
until -> 15000 =< p + n^3 [
if? prime? p + n^3 [
'p + n^3
n: 1
prints -> p
]
else -> 'n+1
]
- Output:
2 3 11 19 83 1811 2027 2243 2251 2467 2531 2539 3539 3547 4547 5059 10891 12619 13619 13627 13691 13907 14419
AWK
# syntax: GAWK -f CUBIC_SPECIAL_PRIMES.AWK
# converted from FreeBASIC
BEGIN {
start = p = 2
stop = 15000
n = 1
printf("%5d ",p)
count = 1
do {
if (is_prime(p + n^3)) {
p += n^3
n = 1
printf("%5d%1s",p,++count%10?"":"\n")
}
else {
n++
}
} while (p + n^3 <= stop)
printf("\nCubic special primes %d-%d: %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)
}
- Output:
2 3 11 19 83 1811 2027 2243 2251 2467 2531 2539 3539 3547 4547 5059 10891 12619 13619 13627 13691 13907 14419 Cubic special primes 2-15000: 23
C
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <math.h>
#include <locale.h>
bool *sieve(int limit) {
int i, p;
limit++;
// True denotes composite, false denotes prime.
bool *c = calloc(limit, sizeof(bool)); // all false by default
c[0] = true;
c[1] = true;
for (i = 4; i < limit; i += 2) c[i] = true;
p = 3; // Start from 3.
while (true) {
int p2 = p * p;
if (p2 >= limit) break;
for (i = p2; i < limit; i += 2 * p) c[i] = true;
while (true) {
p += 2;
if (!c[p]) break;
}
}
return c;
}
bool isCube(int n) {
int s = (int)cbrt((double)n);
return s*s*s == n;
}
int main() {
const int limit = 14999;
int i, j, p, pc = 0, gap = 1, count = 1, lastCubicSpecial = 3;
const char *fmt = "%'7d %'7d %'6d %'4d\n";
bool *c = sieve(limit);
for (i = 0; i < limit; ++i) {
if (!c[i]) ++pc;
}
int *primes = (int *)malloc(pc * sizeof(int));
for (i = 0, j = 0; i < limit; ++i) {
if (!c[i]) primes[j++] = i;
}
free(c);
printf("Cubic special primes under 15,000:\n");
printf(" Prime1 Prime2 Gap Cbrt\n");
setlocale(LC_NUMERIC, "");
printf(fmt, 2, 3, 1, 1);
for (i = 2; i < pc; ++i) {
p = primes[i];
gap = p - lastCubicSpecial;
if (isCube(gap)) {
printf(fmt, lastCubicSpecial, p, gap, (int)cbrt((double)gap));
lastCubicSpecial = p;
++count;
}
}
printf("\n%d such primes found.\n", count+1);
free(primes);
return 0;
}
- Output:
Same as Wren example.
F#
// Cubic Special Primes: Nigel Galloway. March 30th., 2021
let fN=let N=[for n in [0..25]->n*n*n] in let mutable n=2 in (fun g->match List.contains(g-n)N with true->n<-g; true |_->false)
primes32()|>Seq.takeWhile((>)16000)|>Seq.filter fN|>Seq.iter(printf "%d "); printfn ""
- Output:
2 3 11 19 83 1811 2027 2243 2251 2467 2531 2539 3539 3547 4547 5059 10891 12619 13619 13627 13691 13907 14419
Factor
USING: fry io kernel lists lists.lazy math math.functions
math.primes prettyprint ;
2 [ 1 lfrom swap '[ 3 ^ _ + ] lmap-lazy [ prime? ] lfilter car ]
lfrom-by [ 15000 < ] lwhile [ pprint bl ] leach nl
- Output:
2 3 11 19 83 1811 2027 2243 2251 2467 2531 2539 3539 3547 4547 5059 10891 12619 13619 13627 13691 13907 14419
FreeBASIC
#include once "isprime.bas"
dim as integer p = 2, n = 1
print 2;" ";
do
if isprime(p + n^3) then
p += n^3
n = 1
print p;" ";
else
n += 1
end if
loop until p + n^3 >= 15000
print
end
- Output:
2 3 11 19 83 1811 2027 2243 2251 2467 2531 2539 3539 3547 4547 5059 10891 12619 13619 13627 13691 13907 14419
Go
package main
import (
"fmt"
"math"
)
func sieve(limit int) []bool {
limit++
// True denotes composite, false denotes prime.
c := make([]bool, limit) // all false by default
c[0] = true
c[1] = true
// no need to bother with even numbers over 2 for this task
p := 3 // Start from 3.
for {
p2 := p * p
if p2 >= limit {
break
}
for i := p2; i < limit; i += 2 * p {
c[i] = true
}
for {
p += 2
if !c[p] {
break
}
}
}
return c
}
func isCube(n int) bool {
s := int(math.Cbrt(float64(n)))
return s*s*s == n
}
func commas(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() {
c := sieve(14999)
fmt.Println("Cubic special primes under 15,000:")
fmt.Println(" Prime1 Prime2 Gap Cbrt")
lastCubicSpecial := 3
gap := 1
count := 1
fmt.Printf("%7d %7d %6d %4d\n", 2, 3, 1, 1)
for i := 5; i < 15000; i += 2 {
if c[i] {
continue
}
gap = i - lastCubicSpecial
if isCube(gap) {
cbrt := int(math.Cbrt(float64(gap)))
fmt.Printf("%7s %7s %6s %4d\n", commas(lastCubicSpecial), commas(i), commas(gap), cbrt)
lastCubicSpecial = i
count++
}
}
fmt.Println("\n", count+1, "such primes found.")
}
- Output:
Same as Wren example.
jq
Adapted from Julia
Works with gojq, the Go implementation of jq
For the definition of `is_prime` used here, see https://rosettacode.org/wiki/Additive_primes
# Output: an array
def cubic_special_primes:
. as $N
| sqrt as $maxidx
| {cprimes: [2], q: 0}
| until( .cprimes[-1] >= $N or .q >= $N;
label $out
| foreach range(1; $maxidx + 1) as $i (.;
.q = (.cprimes[-1] + ($i * $i * $i))
| if .q >= $N
then .emit = true
elif .q | is_prime
then .cprimes = .cprimes + [.q]
| .emit = true
else .
end;
select(.emit)) | {cprimes, q}, break $out )
| .cprimes ;
15000
| ("Cubic special primes < \(.):",
cubic_special_primes)
- Output:
[2,3,11,19,83,1811,2027,2243,2251,2467,2531,2539,3539,3547,4547,5059,10891,12619,13619,13627,13691,13907,14419]
Julia
using Primes
function cubicspecialprimes(N = 15000)
pmask = primesmask(1, N)
cprimes, maxidx = [2], isqrt(N)
while (n = cprimes[end]) < N
for i in 1:maxidx
q = n + i * i * i
if q > N
return cprimes
elseif pmask[q] # got next qprime
push!(cprimes, q)
break
end
end
end
end
println("Cubic special primes < 15000:")
foreach(p -> print(rpad(p[2], 6), p[1] % 10 == 0 ? "\n" : ""), enumerate(cubicspecialprimes()))
- Output:
Cubic special primes < 16000: 2 3 11 19 83 1811 2027 2243 2251 2467 2531 2539 3539 3547 4547 5059 10891 12619 13619 13627 13691 13907 14419
Mathematica/Wolfram Language
start = {2};
Do[
If[PrimeQ[i],
last = Last[start];
If[IntegerQ[Surd[i - last, 3]],
AppendTo[start, i]
]
]
,
{i, 3, 15000}
]
start
- Output:
{2,3,11,19,83,1811,2027,2243,2251,2467,2531,2539,3539,3547,4547,5059,10891,12619,13619,13627,13691,13907,14419}
Nim
import math, strformat
func sieve(limit: Positive): seq[bool] =
# True denotes composite, false denotes prime.
result = newSeq[bool](limit + 1) # All false by default.
result[0] = true
result[1] = true
# No need to bother with even numbers over 2 for this task.
var p = 3
while true:
let p2 = p * p
if p2 > limit: break
for i in countup(p2, limit, 2 * p):
result[i] = true
while true:
inc p, 2
if not result[p]: break
func isCube(n: int): bool =
let s = cbrt(n.toFloat).int
result = s * s * s == n
let c = sieve(14999)
echo "Cubic special primes under 15_000:"
echo " Prime1 Prime2 Gap Cbrt"
var lastCubicSpecial = 3
var count = 1
echo &"{2:7} {3:7} {1:6} {1:4}"
for n in countup(5, 14999, 2):
if c[n]: continue
let gap = n - lastCubicSpecial
if gap.isCube:
let gapCbrt = cbrt(gap.toFloat).int
echo &"{lastCubicSpecial:7} {n:7} {gap:6} {gapCbrt:4}"
lastCubicSpecial = n
inc count
echo &"\n{count + 1} such primes found."
- Output:
Cubic special primes under 15_000: Prime1 Prime2 Gap Cbrt 2 3 1 1 3 11 8 2 11 19 8 2 19 83 64 4 83 1811 1728 12 1811 2027 216 6 2027 2243 216 6 2243 2251 8 2 2251 2467 216 6 2467 2531 64 4 2531 2539 8 2 2539 3539 1000 10 3539 3547 8 2 3547 4547 1000 10 4547 5059 512 8 5059 10891 5832 18 10891 12619 1728 12 12619 13619 1000 10 13619 13627 8 2 13627 13691 64 4 13691 13907 216 6 13907 14419 512 8 23 such primes found.
Perl
use strict;
use warnings;
use feature 'say';
use ntheory 'is_prime';
my @sp = my $previous = 2;
do {
my($next,$n);
while () { last if is_prime( $next = $previous + ++$n**3 ) }
push @sp, $next;
$previous = $next;
} until $sp[-1] >= 15000;
pop @sp and say join ' ', @sp;
- Output:
2 3 11 19 83 1811 2027 2243 2251 2467 2531 2539 3539 3547 4547 5059 10891 12619 13619 13627 13691 13907 14419
Phix
See Quadrat_Special_Primes#Phix
Python
#!/usr/bin/python
def isPrime(n):
for i in range(2, int(n**0.5) + 1):
if n % i == 0:
return False
return True
if __name__ == '__main__':
p = 2
n = 1
print("2",end = " ")
while True:
if isPrime(p + n**3):
p += n**3
n = 1
print(p,end = " ")
else:
n += 1
if p + n**3 >= 15000:
break
- Output:
2 3 11 19 83 1811 2027 2243 2251 2467 2531 2539 3539 3547 4547 5059 10891 12619 13619 13627 13691 13907 14419
Raku
A two character difference from the Quadrat Special Primes entry. (And it could have been one.)
my @sqp = 2, -> $previous {
my $next;
for (1..∞).map: *³ {
$next = $previous + $_;
last if $next.is-prime;
}
$next
} … *;
say "{+$_} matching numbers:\n", $_».fmt('%5d').batch(7).join: "\n" given
@sqp[^(@sqp.first: * > 15000, :k)];
- Output:
23 matching numbers: 2 3 11 19 83 1811 2027 2243 2251 2467 2531 2539 3539 3547 4547 5059 10891 12619 13619 13627 13691 13907 14419
REXX
/*REXX program finds the smallest primes such that the difference of successive terms */
/*───────────────────────────────────────────────────── are the smallest cubic numbers. */
parse arg hi cols . /*obtain optional argument from the CL.*/
if hi=='' | hi=="," then hi= 15000 /* " " " " " " */
if cols=='' | cols=="," then cols= 10 /* " " " " " " */
call genP /*build array of semaphores for primes.*/
w= 10 /*width of a number in any column. */
title= 'the smallest primes < ' commas(hi) " such that the" ,
'difference of successive terma are the smallest cubic numbers'
if cols>0 then say ' index │'center(title, 1 + cols*(w+1) ) /*display the title.*/
if cols>0 then say '───────┼'center("" , 1 + cols*(w+1), '─') /* " " sep.*/
found= 0; idx= 1 /*initialize number of cbp and index.*/
op= 1
$= /*a list of cubic primes (so far). */
do j=0 by 0
do k=1 until !.np
np= op + k**3 /*find the next cube plus the oldPrime.*/
if np>= hi then leave j /*Is newPrime too big? Then quit. */
end /*k*/
found= found + 1 /*bump number of primes of this type. */
op= np /*assign the newPrime to the oldPrime*/
if cols<0 then iterate /*Build the list (to be shown later)? */
c= commas(np) /*maybe add commas to the number. */
$= $ right(c, max(w, length(c) ) ) /*add a cubic prime──►list, allow big#.*/
if found//cols\==0 then iterate /*have we populated a line of output? */
say center(idx, 7)'│' substr($, 2); $= /*display what we have so far (cols). */
idx= idx + cols /*bump the index count for the output*/
end /*j*/
if $\=='' then say center(idx, 7)"│" substr($, 2) /*possible display residual output.*/
if cols>0 then say '───────┴'center("" , 1 + cols*(w+1), '─') /*display foot sep. */
say
say 'Found ' commas(found) " of " title
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 ?
/*──────────────────────────────────────────────────────────────────────────────────────*/
genP: !.= 0 /*placeholders for primes (semaphores).*/
@.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 ≤ high.*/
do j=@.#+2 by 2 to hi /*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? */
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
- output when using the default inputs:
index │ the smallest primes < 15,000 such that the difference of successive terma are the smallest cubic numbers ───────┼─────────────────────────────────────────────────────────────────────────────────────────────────────────────── 1 │ 2 3 11 19 83 1,811 2,027 2,243 2,251 2,467 11 │ 2,531 2,539 3,539 3,547 4,547 5,059 10,891 12,619 13,619 13,627 21 │ 13,691 13,907 14,419 ───────┴─────────────────────────────────────────────────────────────────────────────────────────────────────────────── Found 23 of the smallest primes < 15,000 such that the difference of successive terma are the smallest cubic numbers
Ring
Also see Quadrat_Special_Primes#Ring
load "stdlib.ring"
see "working..." + nl
Primes = []
limit1 = 50
oldPrime = 2
add(Primes,2)
for n = 1 to limit1
nextPrime = oldPrime + pow(n,3)
if isprime(nextPrime)
n = 1
add(Primes,nextPrime)
oldPrime = nextPrime
else
nextPrime = nextPrime - oldPrime
ok
next
see "prime1 prime2 Gap Cbrt" + nl
for n = 1 to Len(Primes)-1
diff = Primes[n+1] - Primes[n]
for m = 1 to diff
if pow(m,3) = diff
cbrt = m
exit
ok
next
see ""+ Primes[n] + " " + Primes[n+1] + " " + diff + " " + cbrt + nl
next
see "Found " + Len(Primes) + " of the smallest primes < 15,000 such that the difference of successive terma are the smallest cubic numbers" + nl
see "done..." + nl
- Output:
working... prime1 prime2 Gap Cbrt 2 3 1 1 3 11 8 2 11 19 8 2 19 83 64 4 83 1811 1728 12 1811 2027 216 6 2027 2243 216 6 2243 2251 8 2 2251 2467 216 6 2467 2531 64 4 2531 2539 8 2 2539 3539 1000 10 3539 3547 8 2 3547 4547 1000 10 4547 5059 512 8 5059 10891 5832 18 10891 12619 1728 12 12619 13619 1000 10 13619 13627 8 2 13627 13691 64 4 13691 13907 216 6 13907 14419 512 8 Found 23 of the smallest primes < 15,000 such that the difference of successive terma are the smallest cubic numbers done...
Ruby
require 'prime'
res = [2]
until res.last > 15000 do
res << (1..).detect{|n| (res.last + n**3).prime? } ** 3 + res.last
end
puts res[..-2].join(" ")
- Output:
2 3 11 19 83 1811 2027 2243 2251 2467 2531 2539 3539 3547 4547 5059 10891 12619 13619 13627 13691 13907 14419
Sidef
func cubic_primes(callback) {
var prev = 2
callback(prev)
loop {
var curr = (1..Inf -> lazy.map { prev + _**3 }.first { .is_prime })
callback(curr)
prev = curr
}
}
say gather {
cubic_primes({|k|
break if (k >= 15000)
take(k)
})
}
- Output:
[2, 3, 11, 19, 83, 1811, 2027, 2243, 2251, 2467, 2531, 2539, 3539, 3547, 4547, 5059, 10891, 12619, 13619, 13627, 13691, 13907, 14419]
Wren
import "/math" for Int
import "/fmt" for Fmt
var primes = Int.primeSieve(14999)
System.print("Cubic special primes under 15,000:")
System.print(" Prime1 Prime2 Gap Cbrt")
var lastCubicSpecial = 3
var gap = 1
var count = 1
Fmt.print("$,7d $,7d $,6d $4d", 2, 3, 1, 1)
for (p in primes.skip(2)) {
gap = p - lastCubicSpecial
if (Int.isCube(gap)) {
Fmt.print("$,7d $,7d $,6d $4d", lastCubicSpecial, p, gap, gap.cbrt.truncate)
lastCubicSpecial = p
count = count + 1
}
}
System.print("\n%(count+1) such primes found.")
- Output:
Cubic special primes under 15,000: Prime1 Prime2 Gap Cbrt 2 3 1 1 3 11 8 2 11 19 8 2 19 83 64 4 83 1,811 1,728 12 1,811 2,027 216 6 2,027 2,243 216 6 2,243 2,251 8 2 2,251 2,467 216 6 2,467 2,531 64 4 2,531 2,539 8 2 2,539 3,539 1,000 10 3,539 3,547 8 2 3,547 4,547 1,000 10 4,547 5,059 512 8 5,059 10,891 5,832 18 10,891 12,619 1,728 12 12,619 13,619 1,000 10 13,619 13,627 8 2 13,627 13,691 64 4 13,691 13,907 216 6 13,907 14,419 512 8 23 such primes found.
XPL0
func IsPrime(N); \Return 'true' if N is prime
int N, I;
[if N <= 2 then return N = 2;
if (N&1) = 0 then \even >2\ return false;
for I:= 3 to sqrt(N) do
[if rem(N/I) = 0 then return false;
I:= I+1;
];
return true;
];
int N, C, T;
[N:= 2;
repeat C:= 1;
loop [T:= N + C*C*C;
if IsPrime(T) then
[IntOut(0, N);
ChOut(0, ^ );
N:= T;
quit;
]
else C:= C+1;
];
until N > 15000;
]
- Output:
2 3 11 19 83 1811 2027 2243 2251 2467 2531 2539 3539 3547 4547 5059 10891 12619 13619 13627 13691 13907 14419