# Pierpont primes

A Pierpont prime is a prime number of the form: 2u3v + 1 for some non-negative integers u and v .

Pierpont primes
You are encouraged to solve this task according to the task description, using any language you may know.

A Pierpont prime of the second kind is a prime number of the form: 2u3v - 1 for some non-negative integers u and v .

The term "Pierpont primes" is generally understood to mean the first definition, but will be called "Pierpont primes of the first kind" on this page to distinguish them.

• Write a routine (function, procedure, whatever) to find Pierpont primes of the first & second kinds.
• Use the routine to find and display here, on this page, the first 50 Pierpont primes of the first kind.
• Use the routine to find and display here, on this page, the first 50 Pierpont primes of the second kind
• If your language supports large integers, find and display here, on this page, the 250th Pierpont prime of the first kind and the 250th Pierpont prime of the second kind.

## ALGOL 68

As in the second Go sample, generates the 3-smooth number sequence to find Pierpoint Prime candidates. Uses a sliding window on the sequence to avoid having to keep it all in memory.

```BEGIN # find some pierpoint primes of the first kind (2^u3^v + 1)             #
#                              and second kind (2^u3^v - 1)             #
# construct a sieve of primes up to 10 000                                #
[]BOOL prime = PRIMESIEVE 10 000;
# returns the minimum of a and b                                          #
PROC min = ( INT a, b )INT: IF a < b THEN a ELSE b FI;
# returns TRUE if p is prime, FALSE otherwise                             #
PROC is prime = ( INT p )BOOL:
IF   NOT ODD p
THEN p = 2
ELIF p <= UPB prime
THEN prime[ p ] # small enough to use the sieve                      #
ELSE            # use trial division by the sieved primes            #
BOOL probably prime := TRUE;
FOR d FROM 3 BY 2 WHILE d * d <= p AND probably prime DO
IF prime[ d ] THEN probably prime := p MOD d /= 0 FI
OD;
probably prime
FI; # is prime #
# sets the elements of pp1 to the first UPB pp1 Pierpoint primes of the   #
#       first kind and pp2 to the first UPB pp2 Pierpoint primes of the   #
#      second kind                                                        #
PROC find pierpoint = ( REF[]INT pp1, pp2 )VOID:
BEGIN
# saved 3-smooth values                                           #
# - only the currently active part of the seuence is kept         #
INT three smooth limit = 33;
[ 1 : three smooth limit * 3 ]INT s3s; FOR i TO UPB s3s DO s3s[ i ] := 0 OD;
INT pp1 pos := LWB pp1 - 1, pp2 pos := LWB pp2 - 1;
INT pp1 max  = UPB pp1;
INT pp2 max  = UPB pp2;
INT p2, p3, last2, last3;
INT s pos := s3s[ 1 ] := last2 := last3 := p2 := p3 := 1;
FOR n FROM 2 WHILE pp1 pos < pp1 max OR pp2 pos < pp2 max DO
# the next 3-smooth number is the lowest of the next          #
#     multiples of 2 and 3                                    #
INT m = min( p2, p3 );
IF pp1 pos < pp1 max THEN
IF  INT first = m + 1;
is prime( first )
THEN # have a Pierpoint prime of the first kind           #
pp1[ pp1 pos +:= 1 ] := first
FI
FI;
IF pp2 pos < pp2 max THEN
IF  INT second = m - 1;
is prime( second )
THEN # have a Pierpoint prime of the second kind          #
pp2[ pp2 pos +:= 1 ] := second
FI
FI;
s3s[ s pos +:= 1 ] := m;
IF m = p2 THEN p2 := 2 * s3s[ last2 +:= 1 ] FI;
IF m = p3 THEN p3 := 3 * s3s[ last3 +:= 1 ] FI;
INT last min = IF last2 > last3 THEN last3 ELSE last2 FI;
IF last min > three smooth limit THEN
# shuffle the sequence down, over the now unused bits     #
INT new pos := 0;
FOR pos FROM last min TO s pos DO
s3s[ new pos +:= 1 ] := s3s[ pos ]
OD;
INT diff := last min - 1;
last2   -:= diff;
last3   -:= diff;
s pos   -:= diff
FI
OD
END; # find pierpoint #
# prints a table of Pierpoint primes of the specified kind                #
PROC print pierpoint = ( []INT primes, STRING kind )VOID:
BEGIN
print( ( "The first "
, whole( ( UPB primes + 1 ) - LWB primes, 0 )
, " Pierpoint primes of the "
, kind
, " kind:"
, newline
)
);
INT p count := 0;
FOR i FROM LWB primes TO UPB primes DO
print( ( " ", whole( primes[ i ], -8 ) ) );
IF ( p count +:= 1 ) MOD 10 = 0 THEN print( ( newline ) ) FI
OD;
print( ( newline ) )
END; # print pierpoint #
# find the first 50 Pierpoint primes of the first and second kinds        #
INT max pierpoint    = 50;
[ 1 : max pierpoint ]INT pfirst;
[ 1 : max pierpoint ]INT psecond;
find pierpoint( pfirst, psecond );
# show the Pierpoint primes                                               #
print pierpoint( pfirst,  "First"  );
print pierpoint( psecond, "Second" )
END```
Output:
```The first 50 Pierpoint primes of the First kind:
2        3        5        7       13       17       19       37       73       97
109      163      193      257      433      487      577      769     1153     1297
1459     2593     2917     3457     3889    10369    12289    17497    18433    39367
52489    65537   139969   147457   209953   331777   472393   629857   746497   786433
839809   995329  1179649  1492993  1769473  1990657  2654209  5038849  5308417  8503057

The first 50 Pierpoint primes of the Second kind:
2        3        5        7       11       17       23       31       47       53
71      107      127      191      383      431      647      863      971     1151
2591     4373     6143     6911     8191     8747    13121    15551    23327    27647
62207    73727   131071   139967   165887   294911   314927   442367   472391   497663
524287   786431   995327  1062881  2519423 10616831 17915903 18874367 25509167 30233087
```

## C

Translation of: D
```#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

const int PRIMES[] = {
2,   3,   5,   7,  11,  13,  17,  19,  23,  29,  31,  37,  41,  43,  47,
53,  59,  61,  67,  71,  73,  79,  83,  89,  97, 101, 103, 107, 109, 113,
127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197,
199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281,
283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379,
383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463,
467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571,
577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659,
661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761,
769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863,
877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977,
};
#define PRIME_SIZE (sizeof(PRIMES) / sizeof(int))

bool isPrime(const int n) {
int i;

if (n < 2) {
return false;
}

for (i = 0; i < PRIME_SIZE; i++) {
if (n == PRIMES[i]) {
return true;
}
if (n % PRIMES[i] == 0) {
return false;
}
}

if (n < PRIMES[PRIME_SIZE - 1] * PRIMES[PRIME_SIZE - 1]) {
return true;
}

i = PRIMES[PRIME_SIZE - 1]+2;
while (i * i < n) {
if (n % i == 0) {
return false;
}
i += 2;
}

return true;
}

#define N 50
int p[2][50];
void pierpont() {
int64_t s[8 * N];
int count = 0;
int count1 = 1;
int count2 = 0;
int i2 = 0;
int i3 = 0;
int k = 1;
int64_t n2, n3, t;
int64_t *sp = &s[1];

memset(p[0], 0, N * sizeof(int));
memset(p[1], 0, N * sizeof(int));
p[0][0] = 2;
s[0] = 1;

while (count < N) {
n2 = s[i2] * 2;
n3 = s[i3] * 3;
if (n2 < n3) {
t = n2;
i2++;
} else {
t = n3;
i3++;
}
if (t > s[k - 1]) {
*sp++ = t;
k++;

t++;
if (count1 < N && isPrime(t)) {
p[0][count1] = t;
count1++;
}

t -= 2;
if (count2 < N && isPrime(t)) {
p[1][count2] = t;
count2++;
}

count = min(count1, count2);
}
}
}

int main() {
int i;

pierpont();

printf("First 50 Pierpont primes of the first kind:\n");
for (i = 0; i < N; i++) {
printf("%8d ", p[0][i]);
if ((i - 9) % 10 == 0) {
printf("\n");
}
}
printf("\n");

printf("First 50 Pierpont primes of the second kind:\n");
for (i = 0; i < N; i++) {
printf("%8d ", p[1][i]);
if ((i - 9) % 10 == 0) {
printf("\n");
}
}
printf("\n");
}
```
Output:
```First 50 Pierpont primes of the first kind:
2        3        5        7       13       17       19       37       73       97
109      163      193      257      433      487      577      769     1153     1297
1459     2593     2917     3457     3889    10369    12289    17497    18433    39367
52489    65537   139969   147457   209953   331777   472393   629857   746497   786433
839809   995329  1179649  1492993  1769473  1990657  2654209  5038849  5308417  8503057

First 50 Pierpont primes of the second kind:
2        3        5        7       11       17       23       31       47       53
71      107      127      191      383      431      647      863      971     1151
2591     4373     6143     6911     8191     8747    13121    15551    23327    27647
62207    73727   131071   139967   165887   294911   314927   442367   472391   497663
524287   786431   995327  1062881  2519423 10616831 17915903 18874367 25509167 30233087```

## C#

```using System;
using System.Collections.Generic;
using System.Numerics;

namespace PierpontPrimes {
public static class Helper {
private static readonly Random rand = new Random();
private static readonly List<int> primeList = new List<int>() {
2,   3,   5,   7,  11,  13,  17,  19,  23,  29,  31,  37,  41,  43, 47,
53,  59,  61,  67,  71,  73,  79,  83,  89,  97, 101, 103, 107, 109, 113,
127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197,
199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281,
283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379,
383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463,
467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571,
577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659,
661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761,
769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863,
877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977,
};

public static BigInteger GetRandom(BigInteger min, BigInteger max) {
var bytes = max.ToByteArray();
BigInteger r;

do {
rand.NextBytes(bytes);
bytes[bytes.Length - 1] &= (byte)0x7F; // force sign bit to positive
r = new BigInteger(bytes);
} while (r < min || r >= max);

return r;
}

//Modified from https://rosettacode.org/wiki/Miller-Rabin_primality_test#Python
public static bool IsProbablePrime(this BigInteger n) {
if (n == 0 || n == 1) {
return false;
}

bool Check(BigInteger num) {
foreach (var prime in primeList) {
if (num == prime) {
return true;
}
if (num % prime == 0) {
return false;
}
if (prime * prime > num) {
return true;
}
}

return true;
}

if (Check(n)) {
var large = primeList[primeList.Count - 1];
if (n <= large) {
return true;
}
}

var s = 0;
var d = n - 1;
while (d.IsEven) {
d >>= 1;
s++;
}

bool TrialComposite(BigInteger a) {
if (BigInteger.ModPow(a, d, n) == 1) {
return false;
}
for (int i = 0; i < s; i++) {
var t = BigInteger.Pow(2, i);
if (BigInteger.ModPow(a, t * d, n) == n - 1) {
return false;
}
}
return true;
}

for (int i = 0; i < 8; i++) {
var a = GetRandom(2, n);
if (TrialComposite(a)) {
return false;
}
}
return true;
}
}

class Program {
static List<List<BigInteger>> Pierpont(int n) {
var p = new List<List<BigInteger>> {
new List<BigInteger>(),
new List<BigInteger>()
};
for (int i = 0; i < n; i++) {
}
p[0][0] = 2;

var count = 0;
var count1 = 1;
var count2 = 0;
List<BigInteger> s = new List<BigInteger> { 1 };
var i2 = 0;
var i3 = 0;
var k = 1;
BigInteger n2;
BigInteger n3;
BigInteger t;

while (count < n) {
n2 = s[i2] * 2;
n3 = s[i3] * 3;
if (n2 < n3) {
t = n2;
i2++;
} else {
t = n3;
i3++;
}
if (t > s[k - 1]) {
k++;
t += 1;
if (count1 < n && t.IsProbablePrime()) {
p[0][count1] = t;
count1++;
}
t -= 2;
if (count2 < n && t.IsProbablePrime()) {
p[1][count2] = t;
count2++;
}
count = Math.Min(count1, count2);
}
}

return p;
}

static void Main() {
var p = Pierpont(250);

Console.WriteLine("First 50 Pierpont primes of the first kind:");
for (int i = 0; i < 50; i++) {
Console.Write("{0,8} ", p[0][i]);
if ((i - 9) % 10 == 0) {
Console.WriteLine();
}
}
Console.WriteLine();

Console.WriteLine("First 50 Pierpont primes of the second kind:");
for (int i = 0; i < 50; i++) {
Console.Write("{0,8} ", p[1][i]);
if ((i - 9) % 10 == 0) {
Console.WriteLine();
}
}
Console.WriteLine();

Console.WriteLine("250th Pierpont prime of the first kind: {0}", p[0][249]);
Console.WriteLine("250th Pierpont prime of the second kind: {0}", p[1][249]);
Console.WriteLine();
}
}
}
```
Output:
```First 50 Pierpont primes of the first kind:
2        3        5        7       13       17       19       37       73       97
109      163      193      257      433      487      577      769     1153     1297
1459     2593     2917     3457     3889    10369    12289    17497    18433    39367
52489    65537   139969   147457   209953   331777   472393   629857   746497   786433
839809   995329  1179649  1492993  1769473  1990657  2654209  5038849  5308417  8503057

First 50 Pierpont primes of the second kind:
2        3        5        7       11       17       23       31       47       53
71      107      127      191      383      431      647      863      971     1151
2591     4373     6143     6911     8191     8747    13121    15551    23327    27647
62207    73727   131071   139967   165887   294911   314927   442367   472391   497663
524287   786431   995327  1062881  2519423 10616831 17915903 18874367 25509167 30233087

250th Pierpont prime of the first kind: 62518864539857068333550694039553
250th Pierpont prime of the second kind: 4111131172000956525894875083702271```

## C++

Library: GMP
```#include <cassert>
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <vector>
#include <gmpxx.h>

using big_int = mpz_class;

bool is_prime(const big_int& n) {
return mpz_probab_prime_p(n.get_mpz_t(), 25);
}

template <typename integer>
class n_smooth_generator {
public:
explicit n_smooth_generator(size_t n);
integer next();
size_t size() const {
return results_.size();
}
private:
std::vector<size_t> primes_;
std::vector<size_t> index_;
std::vector<integer> results_;
std::vector<integer> queue_;
};

template <typename integer>
n_smooth_generator<integer>::n_smooth_generator(size_t n) {
for (size_t p = 2; p <= n; ++p) {
if (is_prime(p)) {
primes_.push_back(p);
queue_.push_back(p);
}
}
index_.assign(primes_.size(), 0);
results_.push_back(1);
}

template <typename integer>
integer n_smooth_generator<integer>::next() {
integer last = results_.back();
for (size_t i = 0, n = primes_.size(); i < n; ++i) {
if (queue_[i] == last)
queue_[i] = results_[++index_[i]] * primes_[i];
}
results_.push_back(*min_element(queue_.begin(), queue_.end()));
return last;
}

void print_vector(const std::vector<big_int>& numbers) {
for (size_t i = 0, n = numbers.size(); i < n; ++i) {
std::cout << std::setw(9) << numbers[i];
if ((i + 1) % 10 == 0)
std::cout << '\n';
}
std::cout << '\n';
}

int main() {
const int max = 250;
std::vector<big_int> first, second;
int count1 = 0;
int count2 = 0;
n_smooth_generator<big_int> smooth_gen(3);
big_int p1, p2;

while (count1 < max || count2 < max) {
big_int n = smooth_gen.next();
if (count1 < max && is_prime(n + 1)) {
p1 = n + 1;
if (first.size() < 50)
first.push_back(p1);
++count1;
}
if (count2 < max && is_prime(n - 1)) {
p2 = n - 1;
if (second.size() < 50)
second.push_back(p2);
++count2;
}
}

std::cout << "First 50 Pierpont primes of the first kind:\n";
print_vector(first);
std::cout << "First 50 Pierpont primes of the second kind:\n";
print_vector(second);

std::cout << "250th Pierpont prime of the first kind: " << p1 << '\n';
std::cout << "250th Pierpont prime of the second kind: " << p2 << '\n';
return 0;
}
```
Output:
```First 50 Pierpont primes of the first kind:
2        3        5        7       13       17       19       37       73       97
109      163      193      257      433      487      577      769     1153     1297
1459     2593     2917     3457     3889    10369    12289    17497    18433    39367
52489    65537   139969   147457   209953   331777   472393   629857   746497   786433
839809   995329  1179649  1492993  1769473  1990657  2654209  5038849  5308417  8503057

First 50 Pierpont primes of the second kind:
2        3        5        7       11       17       23       31       47       53
71      107      127      191      383      431      647      863      971     1151
2591     4373     6143     6911     8191     8747    13121    15551    23327    27647
62207    73727   131071   139967   165887   294911   314927   442367   472391   497663
524287   786431   995327  1062881  2519423 10616831 17915903 18874367 25509167 30233087

250th Pierpont prime of the first kind: 62518864539857068333550694039553
250th Pierpont prime of the second kind: 4111131172000956525894875083702271
```

## D

Translation of: C#
```import std.algorithm;
import std.bigint;
import std.random;
import std.stdio;

immutable PRIMES = [
2,   3,   5,   7,  11,  13,  17,  19,  23,  29,  31,  37,  41,  43, 47,
53,  59,  61,  67,  71,  73,  79,  83,  89,  97, 101, 103, 107, 109, 113,
127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197,
199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281,
283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379,
383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463,
467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571,
577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659,
661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761,
769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863,
877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977,
];

BigInt getRandom(BigInt min, BigInt max) {
auto r = max - min;
auto hs = r.toHex;

BigInt result;
do {
string t = "0x";
for (int i = 0; i < hs.length; i++) {
t ~= "0123456789abcdef"[uniform(0, 16)];
}
result = BigInt(t) + min;
} while (result < min || max <= result);
return result;
}

//Modified from https://rosettacode.org/wiki/Miller-Rabin_primality_test#Python
bool isProbablePrime(BigInt n) {
if (n == 0 || n == 1) {
return false;
}

bool check(BigInt num) {
foreach (prime; PRIMES) {
if (num == prime) {
return true;
}
if (num % prime == 0) {
return false;
}
if (prime * prime > num) {
return true;
}
}
return true;
}

if (check(n)) {
auto large = PRIMES[\$ - 1];
if (n <= large) {
return true;
}
}

int s = 0;
auto d = n - 1;
while ((d & 1) == 0) {
d >>= 1;
s++;
}

bool trialComposite(BigInt a) {
if (powmod(a, d, n) == 1) {
return false;
}
for (int i = 0; i < s; i++) {
auto t = BigInt(2) ^^ i;
if (powmod(a, t * d, n) == n - 1) {
return false;
}
}
return true;
}

for (int i = 0; i < 8; i++) {
auto a = getRandom(BigInt(2), n);
if (trialComposite(a)) {
return false;
}
}
return true;
}

BigInt[][] pierpont(int n) {
BigInt[][] p = [[], []];
for (int i = 0; i < n; i++) {
p[0] ~= BigInt(0);
p[1] ~= BigInt(0);
}
p[0][0] = 2;

int count = 0;
int count1 = 1;
int count2 = 0;
BigInt[] s = [BigInt(1)];
int i2 = 0;
int i3 = 0;
int k = 1;
BigInt n2, n3, t;

while (count < n) {
n2 = s[i2] * 2;
n3 = s[i3] * 3;
if (n2 < n3) {
t = n2;
i2++;
} else {
t = n3;
i3++;
}
if (t > s[k - 1]) {
s ~= t;
k++;

t++;
if (count1 < n && t.isProbablePrime()) {
p[0][count1] = t;
count1++;
}

t -= 2;
if (count2 < n && t.isProbablePrime()) {
p[1][count2] = t;
count2++;
}

count = min(count1, count2);
}
}

return p;
}

void main() {
auto p = pierpont(250);

writeln("First 50 Pierpont primes of the first kind:");
for (int i = 0; i < 50; i++) {
writef("%8d ", p[0][i]);
if ((i - 9) % 10 == 0) {
writeln;
}
}
writeln;

writeln("First 50 Pierpont primes of the first kind:");
for (int i = 0; i < 50; i++) {
writef("%8d ", p[1][i]);
if ((i - 9) % 10 == 0) {
writeln;
}
}
writeln;

writefln("%dth Pierpont prime of the first kind: %d", p[0].length, p[0][\$ - 1]);
writefln("%dth Pierpont prime of the second kind: %d", p[1].length, p[1][\$ - 1]);
}
```
Output:
```First 50 Pierpont primes of the first kind:
2        3        5        7       13       17       19       37       73       97
109      163      193      257      433      487      577      769     1153     1297
1459     2593     2917     3457     3889    10369    12289    17497    18433    39367
52489    65537   139969   147457   209953   331777   472393   629857   746497   786433
839809   995329  1179649  1492993  1769473  1990657  2654209  5038849  5308417  8503057

First 50 Pierpont primes of the first kind:
2        3        5        7       11       17       23       31       47       53
71      107      127      191      383      431      647      863      971     1151
2591     4373     6143     6911     8191     8747    13121    15551    23327    27647
62207    73727   131071   139967   165887   294911   314927   442367   472391   497663
524287   786431   995327  1062881  2519423 10616831 17915903 18874367 25509167 30233087

250th Pierpont prime of the first kind: 62518864539857068333550694039553
250th Pierpont prime of the second kind: 4111131172000956525894875083702271```

## Delphi

Library: System.Math
Translation of: Go

Thanks Rudy Velthuis for the Velthuis.BigIntegers.Primes and Velthuis.BigIntegers library.

```program Pierpont_primes;

{\$APPTYPE CONSOLE}

uses
System.SysUtils,
System.Math,
System.StrUtils,
System.Generics.Collections,
System.Generics.Defaults,
Velthuis.BigIntegers,
Velthuis.BigIntegers.Primes;

function Pierpont(ulim, vlim: Integer; first: boolean): TArray<BigInteger>;
begin
var p: BigInteger := 0;
var p2: BigInteger := 1;
var p3: BigInteger := 1;

for var v := 0 to vlim - 1 do
begin
for var u := 0 to ulim - 1 do
begin
p := p2 * p3;
if first then
p := p + 1
else
p := p - 1;
if IsProbablePrime(p, 10) then
begin
SetLength(result, Length(result) + 1);
result[High(result)] := BigInteger(p);
end;
p2 := p2 * 2;
end;
p3 := p3 * 3;
p2 := 1;
end;

TArray.sort<BigInteger>(Result, TComparer<BigInteger>.Construct(
function(const Left, Right: BigInteger): Integer
begin
Result := BigInteger.Compare(Left, Right);
end));
end;

begin

writeln('First 50 Pierpont primes of the first kind:');
var pp := Pierpont(120, 80, True);
for var i := 0 to 49 do
begin
write(pp[i].ToString: 8, ' ');
if ((i - 9) mod 10) = 0 then
writeln;
end;

writeln('First 50 Pierpont primes of the second kind:');
var pp2 := Pierpont(120, 80, False);
for var i := 0 to 49 do
begin
write(pp2[i].ToString: 8, ' ');
if ((i - 9) mod 10) = 0 then
writeln;
end;

Writeln('250th Pierpont prime of the first kind:', pp[249].ToString);
Writeln('250th Pierpont prime of the second  kind:', pp2[249].ToString);

end.
```

## F#

This task uses Extensible Prime Generator (F#).

```// Pierpont primes . Nigel Galloway: March 19th., 2021
let fN g=let mutable g=g in ((fun()->g),fun()->g<-g+g;())
let fG y=let rec fG n g=seq{match g|>List.minBy(fun(n,_)->n()) with (f,s) when f()=n->yield f()+y; s(); yield! fG(n*3)(fN(n*3)::g)
|(f,s)           ->yield f()+y; s(); yield! fG n g}
seq{yield! fG 1 [fN 1]}|>Seq.filter isPrime
let pierpontT1,pierpontT2=fG 1,fG -1

pierpontT1|>Seq.take 50|>Seq.iter(printf "%d "); printfn ""
pierpontT2|>Seq.take 50|>Seq.iter(printf "%d "); printfn ""
```
Output:
```2 3 5 7 13 17 19 37 73 97 109 163 193 257 433 487 577 769 1153 1297 1459 2593 2917 3457 3889 10369 12289 17497 18433 39367 52489 65537 139969 147457 209953 331777 472393 629857 746497 786433 839809 995329 1179649 1492993 1769473 1990657 2654209 5038849 5308417 8503057
2 3 5 7 11 17 23 31 47 53 71 107 127 191 383 431 647 863 971 1151 2591 4373 6143 6911 8191 8747 13121 15551 23327 27647 62207 73727 131071 139967 165887 294911 314927 442367 472391 497663 524287 786431 995327 1062881 2519423 10616831 17915903 18874367 25509167 30233087
```

## Factor

```USING: fry grouping io kernel locals make math math.functions
math.primes prettyprint sequences sorting ;

: pierpont ( ulim vlim quot -- seq )
'[
_ <iota> _ <iota> [
[ 2 ] [ 3 ] bi* [ swap ^ ] 2bi@ * 1 @
dup prime? [ , ] [ drop ] if
] cartesian-each
] { } make natural-sort ; inline

: .fifty ( seq -- ) 50 head 10 group simple-table. nl ;

[let
[ + ] [ - ] [ [ 120 80 ] dip pierpont ] bi@
:> ( first second )

"First 50 Pierpont primes of the first kind:" print
first .fifty

"First 50 Pierpont primes of the second kind:" print
second .fifty

"250th Pierpont prime of the first kind: " write
249 first nth . nl

"250th Pierpont prime of the second kind: " write
249 second nth .
]
```
Output:
```First 50 Pierpont primes of the first kind:
2      3      5       7       13      17      19      37      73      97
109    163    193     257     433     487     577     769     1153    1297
1459   2593   2917    3457    3889    10369   12289   17497   18433   39367
52489  65537  139969  147457  209953  331777  472393  629857  746497  786433
839809 995329 1179649 1492993 1769473 1990657 2654209 5038849 5308417 8503057

First 50 Pierpont primes of the second kind:
2      3      5      7       11      17       23       31       47       53
71     107    127    191     383     431      647      863      971      1151
2591   4373   6143   6911    8191    8747     13121    15551    23327    27647
62207  73727  131071 139967  165887  294911   314927   442367   472391   497663
524287 786431 995327 1062881 2519423 10616831 17915903 18874367 25509167 30233087

250th Pierpont prime of the first kind: 62518864539857068333550694039553

250th Pierpont prime of the second kind: 4111131172000956525894875083702271
```

## FreeBASIC

```#define NPP 50

Function isPrime(Byval n As Ulongint) As Boolean
If n < 2 Then Return false
If n = 2 Then Return true
If n Mod 2 = 0 Then Return false
For i As Uinteger = 3 To Int(Sqr(n))+1 Step 2
If n Mod i = 0 Then Return false
Next i
Return true
End Function

Function is_23(Byval n As Uinteger) As Boolean
While n Mod 2 = 0
n /= 2
Wend
While n Mod 3 = 0
n /= 3
Wend
Return Iif(n=1, true, false)
End Function

Function isPierpont(n As Uinteger) As Uinteger
If Not isPrime(n) Then Return 0  'not prime
Dim As Uinteger p1 = is_23(n+1), p2 = is_23(n-1)
If p1 And p2 Then Return 3       'pierpont prime of both kinds
If p1 Then Return 1              'pierpont prime of the 1st kind
If p2 Then Return 2              'pierpont prime of the 2nd kind
Return 0                         'prime, but not pierpont
End Function

Dim As Uinteger pier(1 To 2, 1 To NPP), np(1 To 2) = {0, 0}
Dim As Uinteger x = 1, j
While np(1) <= NPP Or np(2) <= NPP
x += 1
j = isPierpont(x)
If j > 0 Then
If j Mod 2 = 1 Then
np(1) += 1
If np(1) <= NPP Then pier(1, np(1)) = x
End If
If j > 1 Then
np(2) += 1
If np(2) <= NPP Then pier(2, np(2)) = x
End If
End If
Wend

Print "First 50 Pierpoint primes of the first kind:"
For j = 1 To NPP
Print Using " ########"; pier(2, j);
If j Mod 10 = 0 Then Print
Next j
Print !"\nFirst 50 Pierpoint primes of the secod kind:"
For j = 1 To NPP
Print Using " ########"; pier(1, j);
If j Mod 10 = 0 Then Print
Next j
```
Output:
```First 50 Pierpont primes of the first kind:
2        3        5        7       13       17       19       37       73       97
109      163      193      257      433      487      577      769     1153     1297
1459     2593     2917     3457     3889    10369    12289    17497    18433    39367
52489    65537   139969   147457   209953   331777   472393   629857   746497   786433
839809   995329  1179649  1492993  1769473  1990657  2654209  5038849  5308417  8503057

First 50 Pierpont primes of the second kind:
2        3        5        7       11       17       23       31       47       53
71      107      127      191      383      431      647      863      971     1151
2591     4373     6143     6911     8191     8747    13121    15551    23327    27647
62207    73727   131071   139967   165887   294911   314927   442367   472391   497663
524287   786431   995327  1062881  2519423 10616831 17915903 18874367 25509167 30233087
```

## Go

### Brute force approach

Despite being inherently inefficient, this still works very quickly (less than 0.4 seconds on my machine).

A GMP wrapper, rather than Go's math/big package, has been used for extra speed (about 3.5 times quicker).

However, in order to be sure that the first 250 Pierpont primes will be generated, it is necessary to choose the loop sizes to produce somewhat more than this and then sort them into order.

```package main

import (
"fmt"
big "github.com/ncw/gmp"
"sort"
)

var (
one   = new(big.Int).SetUint64(1)
two   = new(big.Int).SetUint64(2)
three = new(big.Int).SetUint64(3)
)

func pierpont(ulim, vlim int, first bool) []*big.Int {
p := new(big.Int)
p2 := new(big.Int).Set(one)
p3 := new(big.Int).Set(one)
var pp []*big.Int
for v := 0; v < vlim; v++ {
for u := 0; u < ulim; u++ {
p.Mul(p2, p3)
if first {
} else {
p.Sub(p, one)
}
if p.ProbablyPrime(10) {
q := new(big.Int)
q.Set(p)
pp = append(pp, q)
}
p2.Mul(p2, two)
}
p3.Mul(p3, three)
p2.Set(one)
}
sort.Slice(pp, func(i, j int) bool {
return pp[i].Cmp(pp[j]) < 0
})
return pp
}
func main() {
fmt.Println("First 50 Pierpont primes of the first kind:")
pp := pierpont(120, 80, true)
for i := 0; i < 50; i++ {
fmt.Printf("%8d ", pp[i])
if (i-9)%10 == 0 {
fmt.Println()
}
}
fmt.Println("\nFirst 50 Pierpont primes of the second kind:")
pp2 := pierpont(120, 80, false)
for i := 0; i < 50; i++ {
fmt.Printf("%8d ", pp2[i])
if (i-9)%10 == 0 {
fmt.Println()
}
}
fmt.Println("\n250th Pierpont prime of the first kind:", pp[249])
fmt.Println("\n250th Pierpont prime of the second kind:", pp2[249])
}
```
Output:
```First 50 Pierpont primes of the first kind:
2        3        5        7       13       17       19       37       73       97
109      163      193      257      433      487      577      769     1153     1297
1459     2593     2917     3457     3889    10369    12289    17497    18433    39367
52489    65537   139969   147457   209953   331777   472393   629857   746497   786433
839809   995329  1179649  1492993  1769473  1990657  2654209  5038849  5308417  8503057

First 50 Pierpont primes of the second kind:
2        3        5        7       11       17       23       31       47       53
71      107      127      191      383      431      647      863      971     1151
2591     4373     6143     6911     8191     8747    13121    15551    23327    27647
62207    73727   131071   139967   165887   294911   314927   442367   472391   497663
524287   786431   995327  1062881  2519423 10616831 17915903 18874367 25509167 30233087

250th Pierpont prime of the first kind: 62518864539857068333550694039553

250th Pierpont prime of the second kind: 4111131172000956525894875083702271
```

### 3-Smooth approach

The strategy here is to generate successive 3-smooth numbers, add (or subtract) one, check if prime and, if so, append to a slice of 'big' integers until the required number is reached.

The Pierpoint primes of the first and second kind are generated at the same time so there is no real need for parallel processing.

This approach is considerably faster than the first version. Although not shown below, it produces the first 250 Pierpont primes (of both kinds) in under 0.2 seconds and the first 1000 in about 7.4 seconds. However, the first 2000 takes around 100 seconds.

These timings are for my Celeron @1.6GHz and should therefore be much faster on a more modern machine.

```package main

import (
"fmt"
big "github.com/ncw/gmp"
"time"
)

var (
one   = new(big.Int).SetUint64(1)
two   = new(big.Int).SetUint64(2)
three = new(big.Int).SetUint64(3)
)

func min(i, j int) int {
if i < j {
return i
}
return j
}

func pierpont(n int, first bool) (p [2][]*big.Int) {
p[0] = make([]*big.Int, n)
p[1] = make([]*big.Int, n)
for i := 0; i < n; i++ {
p[0][i] = new(big.Int)
p[1][i] = new(big.Int)
}
p[0][0].Set(two)
count, count1, count2 := 0, 1, 0
var s []*big.Int
s = append(s, new(big.Int).Set(one))
i2, i3, k := 0, 0, 1
n2, n3, t := new(big.Int), new(big.Int), new(big.Int)
for count < n {
n2.Mul(s[i2], two)
n3.Mul(s[i3], three)
if n2.Cmp(n3) < 0 {
t.Set(n2)
i2++
} else {
t.Set(n3)
i3++
}
if t.Cmp(s[k-1]) > 0 {
s = append(s, new(big.Int).Set(t))
k++
if count1 < n && t.ProbablyPrime(10) {
p[0][count1].Set(t)
count1++
}
t.Sub(t, two)
if count2 < n && t.ProbablyPrime(10) {
p[1][count2].Set(t)
count2++
}
count = min(count1, count2)
}
}
return p
}

func main() {
start := time.Now()
p := pierpont(2000, true)
fmt.Println("First 50 Pierpont primes of the first kind:")
for i := 0; i < 50; i++ {
fmt.Printf("%8d ", p[0][i])
if (i-9)%10 == 0 {
fmt.Println()
}
}
fmt.Println("\nFirst 50 Pierpont primes of the second kind:")
for i := 0; i < 50; i++ {
fmt.Printf("%8d ", p[1][i])
if (i-9)%10 == 0 {
fmt.Println()
}
}

fmt.Println("\n250th Pierpont prime of the first kind:", p[0][249])
fmt.Println("\n250th Pierpont prime of the second kind:", p[1][249])

fmt.Println("\n1000th Pierpont prime of the first kind:", p[0][999])
fmt.Println("\n1000th Pierpont prime of the second kind:", p[1][999])

fmt.Println("\n2000th Pierpont prime of the first kind:", p[0][1999])
fmt.Println("\n2000th Pierpont prime of the second kind:", p[1][1999])

elapsed := time.Now().Sub(start)
fmt.Printf("\nTook %s\n", elapsed)
}
```
Output:
```First 50 Pierpont primes of the first kind:
2        3        5        7       13       17       19       37       73       97
109      163      193      257      433      487      577      769     1153     1297
1459     2593     2917     3457     3889    10369    12289    17497    18433    39367
52489    65537   139969   147457   209953   331777   472393   629857   746497   786433
839809   995329  1179649  1492993  1769473  1990657  2654209  5038849  5308417  8503057

First 50 Pierpont primes of the second kind:
2        3        5        7       11       17       23       31       47       53
71      107      127      191      383      431      647      863      971     1151
2591     4373     6143     6911     8191     8747    13121    15551    23327    27647
62207    73727   131071   139967   165887   294911   314927   442367   472391   497663
524287   786431   995327  1062881  2519423 10616831 17915903 18874367 25509167 30233087

250th Pierpont prime of the first kind: 62518864539857068333550694039553

250th Pierpont prime of the second kind: 4111131172000956525894875083702271

1000th Pierpont prime of the first kind: 69269314716439690250482558089997110961545818230232043107188537422260188701607997086273960899938499201024414931399264696270849

1000th Pierpont prime of the second kind: 1308088756227965581249669045506775407896673213729433892383353027814827286537163695213418982500477392209371001259166465228280492460735463423

2000th Pierpont prime of the first kind: 23647056334818750458979408107288138983957799805326855934519920502493109431728722178351835778368596067773810122477389192659352731519830867553659739507195398662712180250483714053474639899675114018023738461139103130959712720686117399642823861502738433

2000th Pierpont prime of the second kind: 1702224134662426018061116932011222570937093650174807121918750428723338890211147039320296240754205680537318845776107057915956535566573559841027244444877454493022783449689509569107393738917120492483994302725479684822283929715327187974256253064796234576415398735760543848603844607

Took 1m40.781726122s
```

Uses arithmoi Library: https://hackage.haskell.org/package/arithmoi-0.11.0.0 for prime generation and prime testing.
n-smooth generation function based on Hamming_numbers#Avoiding_generation_of_duplicates

```import Control.Monad (guard)
import Data.List (intercalate)
import Data.List.Split (chunksOf)
import Math.NumberTheory.Primes (Prime, unPrime, nextPrime)
import Math.NumberTheory.Primes.Testing (isPrime)
import Text.Printf (printf)

data PierPointKind = First | Second

merge :: Ord a => [a] -> [a] -> [a]
merge [] b = b
merge a@(x:xs) b@(y:ys) | x < y     = x : merge xs b
| otherwise = y : merge a ys

nSmooth :: Integer -> [Integer]
nSmooth p = 1 : foldr u [] factors
where
factors = takeWhile (<=p) primes
primes = map unPrime [nextPrime 1..]
u n s = r
where
r = merge s (map (n*) (1:r))

pierpoints :: PierPointKind -> [Integer]
pierpoints k = do
n <- nSmooth 3
let x = case k of First  -> succ n
Second -> pred n
guard (isPrime x) >> [x]

main :: IO ()
main = do
printf "\nFirst 50 Pierpont primes of the first kind:\n"
mapM_ (\row -> mapM_ (printf "%12s" . commas) row >> printf "\n") (rows \$ pierpoints First)
printf "\nFirst 50 Pierpont primes of the second kind:\n"
mapM_ (\row -> mapM_ (printf "%12s" . commas) row >> printf "\n") (rows \$ pierpoints Second)
printf "\n250th Pierpont prime of the first kind: %s\n" (commas \$ pierpoints First !! 249)
printf "\n250th Pierpont prime of the second kind: %s\n\n" (commas \$ pierpoints Second !! 249)
where
rows = chunksOf 10 . take 50
commas = reverse . intercalate "," . chunksOf 3 . reverse . show
```
Output:
```First 50 Pierpont primes of the first kind:
2           3           5           7          13          17          19          37          73          97
109         163         193         257         433         487         577         769       1,153       1,297
1,459       2,593       2,917       3,457       3,889      10,369      12,289      17,497      18,433      39,367
52,489      65,537     139,969     147,457     209,953     331,777     472,393     629,857     746,497     786,433
839,809     995,329   1,179,649   1,492,993   1,769,473   1,990,657   2,654,209   5,038,849   5,308,417   8,503,057

First 50 Pierpont primes of the second kind:
2           3           5           7          11          17          23          31          47          53
71         107         127         191         383         431         647         863         971       1,151
2,591       4,373       6,143       6,911       8,191       8,747      13,121      15,551      23,327      27,647
62,207      73,727     131,071     139,967     165,887     294,911     314,927     442,367     472,391     497,663
524,287     786,431     995,327   1,062,881   2,519,423  10,616,831  17,915,903  18,874,367  25,509,167  30,233,087

250th Pierpont prime of the first kind: 62,518,864,539,857,068,333,550,694,039,553

250th Pierpont prime of the second kind: 4,111,131,172,000,956,525,894,875,083,702,271

./pierpoints  0.04s user 0.01s system 20% cpu 0.215 total
```

## J

Implementation (first kind first):

```   5 10\$(#~ 1 p:])1+/:~,*//2 3x^/i.20
2      3       5       7      13      17      19      37      73      97
109    163     193     257     433     487     577     769    1153    1297
1459   2593    2917    3457    3889   10369   12289   17497   18433   39367
52489  65537  139969  147457  209953  331777  472393  629857  746497  786433
839809 995329 1179649 1492993 1769473 1990657 2654209 5038849 5308417 8503057
5 10\$(#~ 1 p:])_1+/:~,*//2 3x^/i.20
2      3      5       7      11       17       23       31       47       53
71    107    127     191     383      431      647      863      971     1151
2591   4373   6143    6911    8191     8747    13121    15551    23327    27647
62207  73727 131071  139967  165887   294911   314927   442367   472391   497663
524287 786431 995327 1062881 2519423 10616831 17915903 25509167 30233087 57395627
```

and

```   249{ (#~ 1 p:])1+/:~,*//2 3x^/i.112
62518864539857068333550694039553
249{ (#~ 1 p:])_1+/:~,*//2 3x^/i.112
4111131172000956525894875083702271
```

## Java

```import java.math.BigInteger;
import java.text.NumberFormat;
import java.util.ArrayList;
import java.util.List;

public class PierpontPrimes {

public static void main(String[] args) {
NumberFormat nf = NumberFormat.getNumberInstance();
display("First 50 Pierpont primes of the first kind:", pierpontPrimes(50, true));
display("First 50 Pierpont primes of the second kind:", pierpontPrimes(50, false));
System.out.printf("250th Pierpont prime of the first kind:     %s%n%n", nf.format(pierpontPrimes(250, true).get(249)));
System.out.printf("250th Pierpont prime of the second kind: %s%n%n", nf.format(pierpontPrimes(250, false).get(249)));
}

private static void display(String message, List<BigInteger> primes) {
NumberFormat nf = NumberFormat.getNumberInstance();
System.out.printf("%s%n", message);
for ( int i = 1 ; i <= primes.size() ; i++ ) {
System.out.printf("%10s  ", nf.format(primes.get(i-1)));
if ( i % 10 == 0 ) {
System.out.printf("%n");
}
}
System.out.printf("%n");
}

public static List<BigInteger> pierpontPrimes(int n, boolean first) {
List<BigInteger> primes = new ArrayList<BigInteger>();
if ( first ) {
n -= 1;
}

BigInteger two = BigInteger.valueOf(2);
BigInteger twoTest = two;
BigInteger three = BigInteger.valueOf(3);
BigInteger threeTest = three;
int twoIndex = 0, threeIndex = 0;
List<BigInteger> twoSmooth = new ArrayList<BigInteger>();

BigInteger one = BigInteger.ONE;
BigInteger mOne = BigInteger.valueOf(-1);
int count = 0;
while ( count < n ) {
BigInteger min = twoTest.min(threeTest);
if ( min.compareTo(twoTest) == 0 ) {
twoTest = two.multiply(twoSmooth.get(twoIndex));
twoIndex++;
}
if ( min.compareTo(threeTest) == 0 ) {
threeTest = three.multiply(twoSmooth.get(threeIndex));
threeIndex++;
}
BigInteger test = min.add(first ? one : mOne);
if ( test.isProbablePrime(10) ) {
count++;
}
}
return primes;
}

}
```
Output:
```First 50 Pierpont primes of the first kind:
2           3           5           7          13          17          19          37          73          97
109         163         193         257         433         487         577         769       1,153       1,297
1,459       2,593       2,917       3,457       3,889      10,369      12,289      17,497      18,433      39,367
52,489      65,537     139,969     147,457     209,953     331,777     472,393     629,857     746,497     786,433
839,809     995,329   1,179,649   1,492,993   1,769,473   1,990,657   2,654,209   5,038,849   5,308,417   8,503,057

First 50 Pierpont primes of the second kind:
2           3           5           7          11          17          23          31          47          53
71         107         127         191         383         431         647         863         971       1,151
2,591       4,373       6,143       6,911       8,191       8,747      13,121      15,551      23,327      27,647
62,207      73,727     131,071     139,967     165,887     294,911     314,927     442,367     472,391     497,663
524,287     786,431     995,327   1,062,881   2,519,423  10,616,831  17,915,903  18,874,367  25,509,167  30,233,087

250th Pierpont prime of the first kind:     62,518,864,539,857,068,333,550,694,039,553

250th Pierpont prime of the second kind: 4,111,131,172,000,956,525,894,875,083,702,271
```

## jq

Works with: jq

Works with gojq, the Go implementation of jq

Since we do not know how often 2^u * 3^v ± 1 is prime, we use an adaptive approach based on adjusting the upper bounds for u and v that we need consider. The idea is to avoid any hand-waving about the correctness of the solution. Some "debug" statements that are informative about the adaptive steps have been retained as comments.

The standard (C-based) implementation of jq does not support very large integers, and the Go implementation, gojq, requires too much memory to solve the stretch problem, so the code for the stretch task is shown, but not executed.

Preliminaries

```def is_prime:
. as \$n
| if (\$n < 2)         then false
elif (\$n % 2 == 0)  then \$n == 2
elif (\$n % 3 == 0)  then \$n == 3
elif (\$n % 5 == 0)  then \$n == 5
elif (\$n % 7 == 0)  then \$n == 7
elif (\$n % 11 == 0) then \$n == 11
elif (\$n % 13 == 0) then \$n == 13
elif (\$n % 17 == 0) then \$n == 17
elif (\$n % 19 == 0) then \$n == 19
else {i:23}
| until( (.i * .i) > \$n or (\$n % .i == 0); .i += 2)
| .i * .i > \$n
end;

# pretty-printing
def lpad(\$len): tostring | (\$len - length) as \$l | (" " * \$l)[:\$l] + .;

def nwise(\$n):
def n: if length <= \$n then . else .[0:\$n] , (.[\$n:] | n) end;
n;

def table(\$ncols; \$colwidth):
nwise(\$ncols) | map(lpad(\$colwidth)) | join(" ");```

Pierpont primes

```# Input: the target number of primes of the form p(u,v) == 2^u * 3^v ± 1.
# The main idea is to let u run from 0:m and v run from 0:n where n ~~ 0.63 * m.
# Initially we start with plausible values for m and n (\$maxm and \$maxn respectively),
# and then check whether these have been chosen conservatively enough.
#
def pierponts(\$firstkind):
. as \$N
| (if \$firstkind then 1 else -1 end) as \$incdec

# Input:  [\$maxm, \$maxn]
# Output: an array of objects of the form {p, m, n}
# where .p is prime and .m and .n are the corresponding powers of 2 and 3
| def pp:
. as [\$maxm, \$maxn]
| [ ({p2:1, m:0},
(foreach range(0; \$maxm) as \$m (1; . * 2; {p2: ., m: (\$m + 1)}))) as \$a
| ({p3:1, n:0},
(foreach range(0; \$maxn) as \$n (1; . * 3; {p3: ., n: (\$n + 1)}))) as \$b
| {p: (\$a.p2 * \$b.p3 + \$incdec), m: \$a.m, n: \$b.n}
| select(.p|is_prime) ]
| unique_by(.p)
# | (length|debug) as \$debug # informative
| .;

# input: output of pp
# check that length is sufficient, and that \$maxm and \$maxn are large enough
# ( ".[\$N-1].m is \(.[\$N-1].m) vs \$maxm=\(\$maxm)" | debug) as \$debug |
# ( ".[\$N-1].n is \(.[\$N-1].n) vs \$maxn=\(\$maxn)" | debug) as \$debug |
length > \$N
and .[\$N-1].m < \$maxm - 3   # -2 is not sufficient
and .[\$N-1].n < \$maxn - 3   # -2 is not sufficient
;

# If our search has not been `adequate` then increase \$maxm and \$maxn
# input: [maxm, maxn, ([maxm,maxn]|pp)]
# output: pp
. as [\$maxm, \$maxn, \$one]
| if (\$one|adequate(\$maxm; \$maxn)) then \$one
else [\$maxm + 2, \$maxn + 1.6] as \$maxplus
# | ("retrying with \(\$maxplus)" | debug) as \$debug
|  (\$maxplus|pp) as \$two
|  \$maxplus + [\$two] | adapt
end;

# We start by selecting m and n so that
# m*n >> \$N, i.e., 0.63 * m^2 >> \$N , so m >> sqrt(1.585 * \$N)
# Using 7 as the constant to start with is sufficient to avoid too much rework.
((9 * \$N) | sqrt) as \$maxm
| (0.63 * \$maxm + 1) as \$maxn
| [\$maxm, \$maxn] as \$max
| (\$max | pp) as \$pp
| (\$max + [\$pp]) | [adapt[:\$N][].p] ;

# The stretch goal:
def stretch:
250
| "\nThe \(.)th Pierpoint prime of the first kind is \(pierponts(true)[-1])",
"\nThe \(.)th Pierpoint prime of the second kind is \(pierponts(false)[-1])" ;

50
| "\nThe first \(.) Pierpoint primes of the first kind:", (pierponts(true) | table(10;8)),
"\nThe first \(.) Pierpoint primes of the second kind:", (pierponts(false) | table(10;8))```
Output:
```The first 50 Pierpoint primes of the first kind:
2        3        5        7       13       17       19       37       73       97
109      163      193      257      433      487      577      769     1153     1297
1459     2593     2917     3457     3889    10369    12289    17497    18433    39367
52489    65537   139969   147457   209953   331777   472393   629857   746497   786433
839809   995329  1179649  1492993  1769473  1990657  2654209  5038849  5308417  8503057

The first 50 Pierpoint primes of the second kind:
2        3        5        7       11       17       23       31       47       53
71      107      127      191      383      431      647      863      971     1151
2591     4373     6143     6911     8191     8747    13121    15551    23327    27647
62207    73727   131071   139967   165887   294911   314927   442367   472391   497663
524287   786431   995327  1062881  2519423 10616831 17915903 18874367 25509167 30233087
```

## Julia

The generator method is very fast but does not guarantee the primes are generated in order. Therefore we generate two times the primes needed and then sort and return the lower half.

```using Primes

function pierponts(N, firstkind = true)
ret, incdec = BigInt[], firstkind ? 1 : -1
for k2 in 0:10000, k3 in 0:k2, switch in false:true
i, j = switch ? (k3, k2) : (k2, k3)
n = BigInt(2)^i * BigInt(3)^j + incdec
if isprime(n) && !(n in ret)
push!(ret, n)
if length(ret) == N * 2
return sort(ret)[1:N]
end
end
end
throw("Failed to find \$(N * 2) primes")
end

println("The first 50 Pierpont primes (first kind) are: ", pierponts(50))

println("\nThe first 50 Pierpont primes (second kind) are: ", pierponts(50, false))

println("\nThe 250th Pierpont prime (first kind) is: ", pierponts(250)[250])

println("\nThe 250th Pierpont prime (second kind) is: ", pierponts(250, false)[250])

println("\nThe 1000th Pierpont prime (first kind) is: ", pierponts(1000)[1000])

println("\nThe 1000th Pierpont prime (second kind) is: ", pierponts(1000, false)[1000])

println("\nThe 2000th Pierpont prime (first kind) is: ", pierponts(2000)[2000])

println("\nThe 2000th Pierpont prime (second kind) is: ", pierponts(2000, false)[2000])
```
Output:
```The first 50 Pierpont primes (first kind) are: BigInt[2, 3, 5, 7, 13, 17, 19, 37, 73, 97, 109, 163, 193, 257, 433, 487, 577, 769, 1153, 1297, 1459, 2593, 2917, 3457, 3889, 10369, 12289, 17497, 18433, 39367, 52489, 65537, 139969, 147457, 209953, 331777, 472393, 629857, 746497, 786433, 839809, 995329, 1179649, 1492993, 1769473, 1990657, 2654209, 5038849, 5308417, 8503057]

The first 50 Pierpont primes (second kind) are: BigInt[2, 3, 5, 7, 11, 17, 23, 31, 47, 53, 71, 107, 127, 191, 383, 431, 647, 863, 971, 1151, 2591, 4373, 6143, 6911, 8191, 8747, 13121, 15551, 23327, 27647, 62207, 73727, 131071, 139967, 165887, 294911, 314927, 442367, 472391, 497663, 524287, 786431, 995327, 1062881, 2519423, 10616831, 17915903, 18874367, 25509167, 30233087]

The 250th Pierpont prime (first kind) is: 62518864539857068333550694039553

The 250th Pierpont prime (second kind) is: 4111131172000956525894875083702271

The 1000th Pierpont prime (first kind) is: 69269314716439690250482558089997110961545818230232043107188537422260188701607997086273960899938499201024414931399264696270849

The 1000th Pierpont prime (second kind) is: 1308088756227965581249669045506775407896673213729433892383353027814827286537163695213418982500477392209371001259166465228280492460735463423

The 2000th Pierpont prime (first kind) is: 23647056334818750458979408107288138983957799805326855934519920502493109431728722178351835778368596067773810122477389192659352731519830867553659739507195398662712180250483714053474639899675114018023738461139103130959712720686117399642823861502738433

The 2000th Pierpont prime (second kind) is: 1702224134662426018061116932011222570937093650174807121918750428723338890211147039320296240754205680537318845776107057915956535566573559841027244444877454493022783449689509569107393738917120492483994302725479684822283929715327187974256253064796234576415398735760543848603844607
```

## Kotlin

Translation of: Go
```import java.math.BigInteger
import kotlin.math.min

val one: BigInteger = BigInteger.ONE
val two: BigInteger = BigInteger.valueOf(2)
val three: BigInteger = BigInteger.valueOf(3)

fun pierpont(n: Int): List<List<BigInteger>> {
val p = List(2) { MutableList(n) { BigInteger.ZERO } }
p[0][0] = two
var count = 0
var count1 = 1
var count2 = 0
val s = mutableListOf<BigInteger>()
var i2 = 0
var i3 = 0
var k = 1
var n2: BigInteger
var n3: BigInteger
var t: BigInteger
while (count < n) {
n2 = s[i2] * two
n3 = s[i3] * three
if (n2 < n3) {
t = n2
i2++
} else {
t = n3
i3++
}
if (t > s[k - 1]) {
k++
t += one
if (count1 < n && t.isProbablePrime(10)) {
p[0][count1] = t
count1++
}
t -= two
if (count2 < n && t.isProbablePrime(10)) {
p[1][count2] = t
count2++
}
count = min(count1, count2)
}
}
return p
}

fun main() {
val p = pierpont(2000)

println("First 50 Pierpont primes of the first kind:")
for (i in 0 until 50) {
print("%8d ".format(p[0][i]))
if ((i - 9) % 10 == 0) {
println()
}
}

println("\nFirst 50 Pierpont primes of the second kind:")
for (i in 0 until 50) {
print("%8d ".format(p[1][i]))
if ((i - 9) % 10 == 0) {
println()
}
}

println("\n250th Pierpont prime of the first kind: \${p[0][249]}")
println("\n250th Pierpont prime of the first kind: \${p[1][249]}")

println("\n1000th Pierpont prime of the first kind: \${p[0][999]}")
println("\n1000th Pierpont prime of the first kind: \${p[1][999]}")

println("\n2000th Pierpont prime of the first kind: \${p[0][1999]}")
println("\n2000th Pierpont prime of the first kind: \${p[1][1999]}")
}
```
Output:
```First 50 Pierpont primes of the first kind:
2        3        5        7       13       17       19       37       73       97
109      163      193      257      433      487      577      769     1153     1297
1459     2593     2917     3457     3889    10369    12289    17497    18433    39367
52489    65537   139969   147457   209953   331777   472393   629857   746497   786433
839809   995329  1179649  1492993  1769473  1990657  2654209  5038849  5308417  8503057

First 50 Pierpont primes of the second kind:
2        3        5        7       11       17       23       31       47       53
71      107      127      191      383      431      647      863      971     1151
2591     4373     6143     6911     8191     8747    13121    15551    23327    27647
62207    73727   131071   139967   165887   294911   314927   442367   472391   497663
524287   786431   995327  1062881  2519423 10616831 17915903 18874367 25509167 30233087

250th Pierpont prime of the first kind: 62518864539857068333550694039553

250th Pierpont prime of the first kind: 4111131172000956525894875083702271

1000th Pierpont prime of the first kind: 69269314716439690250482558089997110961545818230232043107188537422260188701607997086273960899938499201024414931399264696270849

1000th Pierpont prime of the first kind: 1308088756227965581249669045506775407896673213729433892383353027814827286537163695213418982500477392209371001259166465228280492460735463423

2000th Pierpont prime of the first kind: 23647056334818750458979408107288138983957799805326855934519920502493109431728722178351835778368596067773810122477389192659352731519830867553659739507195398662712180250483714053474639899675114018023738461139103130959712720686117399642823861502738433

2000th Pierpont prime of the first kind: 1702224134662426018061116932011222570937093650174807121918750428723338890211147039320296240754205680537318845776107057915956535566573559841027244444877454493022783449689509569107393738917120492483994302725479684822283929715327187974256253064796234576415398735760543848603844607```

## Lua

```local function isprime(n)
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
local f, limit = 5, math.sqrt(n)
for f = 5, limit, 6 do
if n % f == 0 then return false end
if n % (f+2) == 0 then return false end
end
return true
end

local function s3iter()
local s, i2, i3 = {1}, 1, 1
return function()
local n2, n3, val = 2*s[i2], 3*s[i3], s[#s]
s[#s+1] = math.min(n2, n3)
i2, i3 = i2 + (n2<=n3 and 1 or 0), i3 + (n2>=n3 and 1 or 0)
return val
end
end

local function pierpont(n)
local list1, list2, s3next = {}, {}, s3iter()
while #list1 < n or #list2 < n do
local s3 = s3next()
if #list1 < n then
if isprime(s3+1) then list1[#list1+1] = s3+1 end
end
if #list2 < n then
if isprime(s3-1) then list2[#list2+1] = s3-1 end
end
end
return list1, list2
end

local N = 50
local p1, p2 = pierpont(N)

print("First 50 Pierpont primes of the first kind:")
for i, p in ipairs(p1) do
io.write(string.format("%8d%s", p, (i%10==0 and "\n" or " ")))
end

print()

print("First 50 Pierpont primes of the second kind:")
for i, p in ipairs(p2) do
io.write(string.format("%8d%s", p, (i%10==0 and "\n" or " ")))
end
```
Output:
```First 50 Pierpont primes of the first kind:
2        3        5        7       13       17       19       37       73       97
109      163      193      257      433      487      577      769     1153     1297
1459     2593     2917     3457     3889    10369    12289    17497    18433    39367
52489    65537   139969   147457   209953   331777   472393   629857   746497   786433
839809   995329  1179649  1492993  1769473  1990657  2654209  5038849  5308417  8503057

First 50 Pierpont primes of the second kind:
2        3        5        7       11       17       23       31       47       53
71      107      127      191      383      431      647      863      971     1151
2591     4373     6143     6911     8191     8747    13121    15551    23327    27647
62207    73727   131071   139967   165887   294911   314927   442367   472391   497663
524287   786431   995327  1062881  2519423 10616831 17915903 18874367 25509167 30233087```

## M2000 Interpreter

Translation of: freebasic
```Module Pierpoint_Primes {
Form 80
Set Fast !
const NPP=50
dim pier(1 to 2, 1 to NPP), np(1 to 2) = 0
def long x = 1, j
while np(1)<=NPP or np(2)<=NPP
x++
j = @is_pierpont(x)
if j>0 Else Continue
if j mod 2 = 1 then np(1)++ :if np(1) <= NPP then pier(1, np(1)) = x
if j > 1 then np(2)++ : if np(2) <= NPP then pier(2, np(2)) = x
end while

print "First ";NPP;" Pierpont primes of the first kind:"
for j = 1 to NPP
print pier(2, j),
next j
if pos>0 then print
print "First ";NPP;" Pierpont primes of the second kind:"
for j = 1 to NPP
print pier(1, j),
next j
if pos>0 then print
Set Fast
function is_prime(n as decimal)
if n < 2 then = false  : exit function
if n <4 then = true : exit function
if n mod 2 = 0 then = false : exit function
local i as long
for i = 3 to int(sqrt(n))+1 step 2
if n mod i = 0 then = false  : exit function
next i
= true
end function

function is_23(n as long)
while n mod 2 = 0
n = n div 2
end while
while n mod 3 = 0
n = n div 3
end while
if n = 1 then = true else = false
end function

function is_pierpont(n as long)
if not @is_prime(n) then = 0& : exit function 'not prime
Local p1 = @is_23(n+1), p2 = @is_23(n-1)
if p1 and p2 then = 3  : exit function      'pierpont prime of both kinds
if p1 then = 1 : exit function              'pierpont prime of the 1st kind
if p2 then = 2 : exit function              'pierpont prime of the 2nd kind
= 0                         'prime, but not pierpont
end function
}
Pierpoint_Primes```
Output:
```First 50 Pierpont primes of the first kind:
2        3        5        7       13       17       19       37       73       97
109      163      193      257      433      487      577      769     1153     1297
1459     2593     2917     3457     3889    10369    12289    17497    18433    39367
52489    65537   139969   147457   209953   331777   472393   629857   746497   786433
839809   995329  1179649  1492993  1769473  1990657  2654209  5038849  5308417  8503057

First 50 Pierpont primes of the second kind:
2        3        5        7       11       17       23       31       47       53
71      107      127      191      383      431      647      863      971     1151
2591     4373     6143     6911     8191     8747    13121    15551    23327    27647
62207    73727   131071   139967   165887   294911   314927   442367   472391   497663
524287   786431   995327  1062881  2519423 10616831 17915903 18874367 25509167 30233087
```

## Mathematica/Wolfram Language

```ClearAll[FindUpToMax]
FindUpToMax[max_Integer, b_Integer] := Module[{res, num},
res = {};
Do[
num = 2^u 3^v + b;
If[PrimeQ[num], AppendTo[res, num]]
,
{u, 0, Ceiling@Log[2, max]}
,
{v, 0, Ceiling@Log[3, max]}
];
res //= Select[LessEqualThan[max]];
res //= Sort;
res
]
Print["Piermont primes of the first kind:"]
Take[FindUpToMax[10^10, +1], UpTo[50]]
Print["Piermont primes of the second kind:"]
Take[FindUpToMax[10^10, -1], UpTo[50]]
Print["250th Piermont prime of the first kind:"]
Part[FindUpToMax[10^34, +1], 250]
Print["250th Piermont prime of the second kind:"]
Part[FindUpToMax[10^34, -1], 250]
Print["1000th Piermont prime of the first kind:"]
Part[FindUpToMax[10^130, +1], 1000]
Print["1000th Piermont prime of the second kind:"]
Part[FindUpToMax[10^150, -1], 1000]
```
Output:
```Piermont primes of the first kind:
{2,3,5,7,13,17,19,37,73,97,109,163,193,257,433,487,577,769,1153,1297,1459,2593,2917,3457,3889,10369,12289,17497,18433,39367,52489,65537,139969,147457,209953,331777,472393,629857,746497,786433,839809,995329,1179649,1492993,1769473,1990657,2654209,5038849,5308417,8503057}
Piermont primes of the second kind:
{2,3,5,7,11,17,23,31,47,53,71,107,127,191,383,431,647,863,971,1151,2591,4373,6143,6911,8191,8747,13121,15551,23327,27647,62207,73727,131071,139967,165887,294911,314927,442367,472391,497663,524287,786431,995327,1062881,2519423,10616831,17915903,18874367,25509167,30233087}
250th Piermont primes of the first kind:
62518864539857068333550694039553
250th Piermont primes of the second kind:
4111131172000956525894875083702271
1000th Piermont prime of the first kind:
69269314716439690250482558089997110961545818230232043107188537422260188701607997086273960899938499201024414931399264696270849
1000th Piermont prime of the second kind:
1308088756227965581249669045506775407896673213729433892383353027814827286537163695213418982500477392209371001259166465228280492460735463423```

## Nim

```import math, strutils

func isPrime(n: int): bool =
## Check if "n" is prime by trying successive divisions.
## "n" is supposed not to be a multiple of 2 or 3.
var d = 5
var delta = 2
while d <= int(sqrt(n.toFloat)):
if n mod d == 0: return false
inc d, delta
delta = 6 - delta
result = true

func isProduct23(n: int): bool =
## Check if "n" has only 2 and 3 for prime divisors
## (i.e. that "n = 2^u * 3^v").
var n = n
while (n and 1) == 0: n = n shr 1
while n mod 3 == 0: n = n div 3
result = n == 1

iterator pierpont(maxCount: Positive; k: int): int =
## Yield the Pierpoint primes of first or second kind according
## to the value of "k" (+1 for first kind, -1 for second kind).
yield 2
yield 3
var n = 5
var delta = 2   # 2 and 4 alternatively to skip the multiples of 2 and 3.
yield n
var count = 3
while count < maxCount:
inc n, delta
delta = 6 - delta
if isProduct23(n - k) and isPrime(n):
inc count
yield n

template pierpont1*(maxCount: Positive): int = pierpont(maxCount, +1)
template pierpont2*(maxCount: Positive): int = pierpont(maxCount, -1)

when isMainModule:

echo "First 50 Pierpont primes of the first kind:"
var count = 0
for n in pierpont1(50):
stdout.write (\$n).align(9)
inc count
if count mod 10 == 0: stdout.write '\n'

echo ""
echo "First 50 Pierpont primes of the second kind:"
count = 0
for n in pierpont2(50):
stdout.write (\$n).align(9)
inc count
if count mod 10 == 0: stdout.write '\n'
```
Output:
```First 50 Pierpont primes of the first kind:
2        3        5        7       13       17       19       37       73       97
109      163      193      257      433      487      577      769     1153     1297
1459     2593     2917     3457     3889    10369    12289    17497    18433    39367
52489    65537   139969   147457   209953   331777   472393   629857   746497   786433
839809   995329  1179649  1492993  1769473  1990657  2654209  5038849  5308417  8503057

First 50 Pierpont primes of the second kind:
2        3        5        7       11       17       23       31       47       53
71      107      127      191      383      431      647      863      971     1151
2591     4373     6143     6911     8191     8747    13121    15551    23327    27647
62207    73727   131071   139967   165887   294911   314927   442367   472391   497663
524287   786431   995327  1062881  2519423 10616831 17915903 18874367 25509167 30233087```

## Perl

Translation of: Raku
Library: ntheory
```use strict;
use warnings;
use feature 'say';
use bigint try=>"GMP";
use ntheory qw<is_prime>;

# index of mininum value in list
sub min_index { my \$b = \$_[my \$i = 0]; \$_[\$_] < \$b && (\$b = \$_[\$i = \$_]) for 0..\$#_; \$i }

sub iter1 { my \$m = shift; my \$e = 0; return sub { \$m ** \$e++;    } }
sub iter2 { my \$m = shift; my \$e = 1; return sub { \$m * (\$e *= 2) } }

sub pierpont {
my(\$max ) = shift || die 'Must specify count of primes to generate.';
my(\$kind) = @_ ? shift : 1;
die "Unknown type: \$kind. Must be one of 1 (default) or 2" unless \$kind == 1 || \$kind == 2;
\$kind = -1 if \$kind == 2;

my \$po3     = 3;
my @iterators;
push @iterators, iter1(2);
push @iterators, iter1(3); \$iterators[1]->();

my @pierpont;
do {
push @pierpont, \$min + \$kind if is_prime(\$min + \$kind);

push @iterators, iter2(\$po3);
\$po3 *= 3;
}
} until @pierpont == \$max;
@pierpont;
}

my @pierpont_1st = pierpont(250,1);
my @pierpont_2nd = pierpont(250,2);

say "First 50 Pierpont primes of the first kind:";
my \$fmt = "%9d"x10 . "\n";
for my \$row (0..4) { printf \$fmt, map { \$pierpont_1st[10*\$row + \$_] } 0..9 }
say "\nFirst 50 Pierpont primes of the second kind:";
for my \$row (0..4) { printf \$fmt, map { \$pierpont_2nd[10*\$row + \$_] } 0..9 }

say "\n250th Pierpont prime of the first kind:    " . \$pierpont_1st[249];
say "\n250th Pierpont prime of the second kind: "   . \$pierpont_2nd[249];
```
Output:
```First 50 Pierpont primes of the first kind:
2        3        5        7       13       17       19       37       73       97
109      163      193      257      433      487      577      769     1153     1297
1459     2593     2917     3457     3889    10369    12289    17497    18433    39367
52489    65537   139969   147457   209953   331777   472393   629857   746497   786433
839809   995329  1179649  1492993  1769473  1990657  2654209  5038849  5308417  8503057

First 50 Pierpont primes of the second kind:
2        3        5        7       11       17       23       31       47       53
71      107      127      191      383      431      647      863      971     1151
2591     4373     6143     6911     8191     8747    13121    15551    23327    27647
62207    73727   131071   139967   165887   294911   314927   442367   472391   497663
524287   786431   995327  1062881  2519423 10616831 17915903 18874367 25509167 30233087

250th Pierpont prime of the first kind:    62518864539857068333550694039553

250th Pierpont prime of the second kind: 4111131172000956525894875083702271```

## Phix

Library: Phix/mpfr
Translation of: Go

I also tried using a priority queue, popping the smallest (and any duplicates of it) and pushing *2 and *3 on every iteration, which worked pretty well but ended up about 20% slower, with about 1500 untested candidates left in the queue by the time it found the 2000th second kind.

```-- demo/rosetta/Pierpont_primes.exw
with javascript_semantics
include mpfr.e

function pierpont(integer n)
sequence p = {{mpz_init(2)},{}},
s = {mpz_init(1)}
integer i2 = 1, i3 = 1
mpz {n2, n3, t} = mpz_inits(3)
atom t1 = time()+1
while min(length(p[1]),length(p[2])) < n do
mpz_mul_si(n2,s[i2],2)
mpz_mul_si(n3,s[i3],3)
if mpz_cmp(n2,n3)<0 then
mpz_set(t,n2)
i2 += 1
else
mpz_set(t,n3)
i3 += 1
end if
if mpz_cmp(t,s[\$]) > 0 then
s = append(s, mpz_init_set(t))
if length(p[1]) < n and mpz_prime(t) then
p[1] = append(p[1],mpz_init_set(t))
end if
mpz_sub_ui(t,t,2)
if length(p[2]) < n and mpz_prime(t) then
p[2] = append(p[2],mpz_init_set(t))
end if
end if
if time()>t1 then
printf(1,"Found: 1st:%d/%d, 2nd:%d/%d\r",
{length(p[1]),n,length(p[2]),n})
t1 = time()+1
end if
end while
return p
end function

--constant limit = 2000         -- 2 mins
--constant limit = 1000         -- 8.1s
constant limit = 250            -- 0.1s
atom t0 = time()
sequence p = pierpont(limit)
constant fs = {"first","second"}

for i=1 to length(fs) do
printf(1,"First 50 Pierpont primes of the %s kind:\n",{fs[i]})
for j=1 to 50 do
printf(1,"%8s ", {mpz_get_str(p[i][j])})
if mod(j,10)=0 then printf(1,"\n") end if
end for
printf(1,"\n")
end for

constant t = {250,1000,2000}
for i=1 to length(t) do
integer ti = t[i]
if ti>limit then exit end if
for j=1 to length(fs) do
string zs = shorten(mpz_get_str(p[j][ti]))
printf(1,"%dth Pierpont prime of the %s kind: %s\n",{ti,fs[j],zs})
end for
printf(1,"\n")
end for
printf(1,"Took %s\n", elapsed(time()-t0))
```
Output:
```First 50 Pierpont primes of the first kind:
2        3        5        7       13       17       19       37       73       97
109      163      193      257      433      487      577      769     1153     1297
1459     2593     2917     3457     3889    10369    12289    17497    18433    39367
52489    65537   139969   147457   209953   331777   472393   629857   746497   786433
839809   995329  1179649  1492993  1769473  1990657  2654209  5038849  5308417  8503057

First 50 Pierpont primes of the second kind:
2        3        5        7       11       17       23       31       47       53
71      107      127      191      383      431      647      863      971     1151
2591     4373     6143     6911     8191     8747    13121    15551    23327    27647
62207    73727   131071   139967   165887   294911   314927   442367   472391   497663
524287   786431   995327  1062881  2519423 10616831 17915903 18874367 25509167 30233087

250th Pierpont prime of the first kind: 62518864539857068333550694039553
250th Pierpont prime of the second kind: 4111131172000956525894875083702271

1000th Pierpont prime of the first kind: 6926931471643969025...4931399264696270849 (125 digits)
1000th Pierpont prime of the second kind: 1308088756227965581...8280492460735463423 (139 digits)

2000th Pierpont prime of the first kind: 2364705633481875045...9642823861502738433 (248 digits)
2000th Pierpont prime of the second kind: 1702224134662426018...5760543848603844607 (277 digits)

Took 2 minutes and 01s
```

## Prolog

```?- use_module(library(heaps)).

three_smooth(Lz) :-
singleton_heap(H, 1, nothing),
lazy_list(next_3smooth, 0-H, Lz).

next_3smooth(Top-H0, N-H2, N) :-
min_of_heap(H0, Top, _), !,
get_from_heap(H0, Top, _, H1),
next_3smooth(Top-H1, N-H2, N).
next_3smooth(_-H0, N-H3, N) :-
get_from_heap(H0, N, _, H1),
N2 is N * 2,
N3 is N * 3,

first_kind(K) :-
three_smooth(Ns), member(N, Ns),
K is N + 1,
prime(K).

second_kind(K) :-
three_smooth(Ns), member(N, Ns),
K is N - 1,
prime(K).

show(Seq, N) :-
format("The first ~w values of ~s are: ", [N, Seq]),
once(findnsols(N, X, call(Seq, X), L)),
write(L), nl,
once(offset(249, call(Seq, TwoFifty))),
format("The 250th value of ~w is ~w~n", [Seq, TwoFifty]).

main :-
show(first_kind, 50), nl,
show(second_kind, 50), nl,
halt.

% primality checker -- Miller Rabin preceded with a round of trial divisions.

prime(N) :-
integer(N),
N > 1,
divcheck(
N,
[  2,   3,   5,   7,  11,  13,  17,  19,  23,  29,  31,
37,  41,  43,  47,  53,  59,  61,  67,  71,  73,  79,
83,  89,  97, 101, 103, 107, 109, 113, 127, 131, 137,
139, 149],
Result),
((Result = prime, !); miller_rabin_primality_test(N)).

divcheck(_, [],    unknown) :- !.
divcheck(N, [P|_], prime) :- P*P > N, !.
divcheck(N, [P|Ps], State) :- N mod P =\= 0, divcheck(N, Ps, State).

miller_rabin_primality_test(N) :-
bases(Bases, N),
forall(member(A, Bases), strong_fermat_pseudoprime(N, A)).

miller_rabin_precision(16).

bases([31, 73], N) :- N < 9_080_191, !.
bases([2, 7, 61], N) :- N < 4_759_123_141, !.
bases([2, 325, 9_375, 28_178, 450_775, 9_780_504, 1_795_265_022], N) :-
N < 18_446_744_073_709_551_616, !. % 2^64
bases(Bases, N) :-
miller_rabin_precision(T), RndLimit is N - 2,
length(Bases, T), maplist(random_between(2, RndLimit), Bases).

strong_fermat_pseudoprime(N, A) :-  % miller-rabin strong pseudoprime test with base A.
succ(Pn, N), factor_2s(Pn, S, D),
X is powm(A, D, N),
((X =:= 1, !); \+ composite_witness(N, S, X)).

composite_witness(_, 0, _) :- !.
composite_witness(N, K, X) :-
X =\= N-1,
succ(Pk, K), X2 is (X*X) mod N, composite_witness(N, Pk, X2).

factor_2s(N, S, D) :- factor_2s(0, N, S, D).
factor_2s(S, D, S, D) :- D /\ 1 =\= 0, !.
factor_2s(S0, D0, S, D) :-
succ(S0, S1), D1 is D0 >> 1,
factor_2s(S1, D1, S, D).

?- main.
```
Output:
```The first 50 values of first_kind are: [2,3,5,7,13,17,19,37,73,97,109,163,193,257,433,487,577,769,1153,1297,1459,2593,2917,3457,3889,10369,12289,17497,18433,39367,52489,65537,139969,147457,209953,331777,472393,629857,746497,786433,839809,995329,1179649,1492993,1769473,1990657,2654209,5038849,5308417,8503057]
The 250th value of first_kind is 62518864539857068333550694039553

The first 50 values of second_kind are: [2,3,5,7,11,17,23,31,47,53,71,107,127,191,383,431,647,863,971,1151,2591,4373,6143,6911,8191,8747,13121,15551,23327,27647,62207,73727,131071,139967,165887,294911,314927,442367,472391,497663,524287,786431,995327,1062881,2519423,10616831,17915903,18874367,25509167,30233087]
The 250th value of second_kind is 4111131172000956525894875083702271
```

## Python

Translation of: Go
```import random

# Copied from https://rosettacode.org/wiki/Miller-Rabin_primality_test#Python
def is_Prime(n):
"""
Miller-Rabin primality test.

A return value of False means n is certainly not prime. A return value of
True means n is very likely a prime.
"""
if n!=int(n):
return False
n=int(n)
#Miller-Rabin test for prime
if n==0 or n==1 or n==4 or n==6 or n==8 or n==9:
return False

if n==2 or n==3 or n==5 or n==7:
return True
s = 0
d = n-1
while d%2==0:
d>>=1
s+=1
assert(2**s * d == n-1)

def trial_composite(a):
if pow(a, d, n) == 1:
return False
for i in range(s):
if pow(a, 2**i * d, n) == n-1:
return False
return True

for i in range(8):#number of trials
a = random.randrange(2, n)
if trial_composite(a):
return False

return True

def pierpont(ulim, vlim, first):
p = 0
p2 = 1
p3 = 1
pp = []
for v in xrange(vlim):
for u in xrange(ulim):
p = p2 * p3
if first:
p = p + 1
else:
p = p - 1
if is_Prime(p):
pp.append(p)
p2 = p2 * 2
p3 = p3 * 3
p2 = 1
pp.sort()
return pp

def main():
print "First 50 Pierpont primes of the first kind:"
pp = pierpont(120, 80, True)
for i in xrange(50):
print "%8d " % pp[i],
if (i - 9) % 10 == 0:
print
print "First 50 Pierpont primes of the second kind:"
pp2 = pierpont(120, 80, False)
for i in xrange(50):
print "%8d " % pp2[i],
if (i - 9) % 10 == 0:
print
print "250th Pierpont prime of the first kind:", pp[249]
print "250th Pierpont prime of the second kind:", pp2[249]

main()
```
Output:
```First 50 Pierpont primes of the first kind:
2         3         5         7        13        17        19        37        73        97
109       163       193       257       433       487       577       769      1153      1297
1459      2593      2917      3457      3889     10369     12289     17497     18433     39367
52489     65537    139969    147457    209953    331777    472393    629857    746497    786433
839809    995329   1179649   1492993   1769473   1990657   2654209   5038849   5308417   8503057
First 50 Pierpont primes of the second kind:
2         3         5         7        11        17        23        31        47        53
71       107       127       191       383       431       647       863       971      1151
2591      4373      6143      6911      8191      8747     13121     15551     23327     27647
62207     73727    131071    139967    165887    294911    314927    442367    472391    497663
524287    786431    995327   1062881   2519423  10616831  17915903  18874367  25509167  30233087
250th Pierpont prime of the first kind: 62518864539857068333550694039553
250th Pierpont prime of the second kind: 4111131172000956525894875083702271```

## Quackery

Uses `smoothwith` from N-smooth numbers#Quackery and `isprime` from Primality by trial division#Quackery.

Using trial division to determine primality makes finding the 250th Pierpont primes impractical. This should be updated when a coding of a more suitable test becomes available.

```  [ stack ] is pierponts
[ stack ] is kind
[ stack ] is quantity

[ 1 - -2 * 1+ kind put
1+ quantity put
' [ -1 ] pierponts put
' [ 2 3 ] smoothwith
[ -1 peek
kind share +
dup isprime not iff
[ drop false ] done
pierponts share -1 peek
over = iff
[ drop false ] done
pierponts take
swap join
dup size swap
pierponts put
quantity share = ]
drop
quantity release
kind release
pierponts take

say "Pierpont primes of the first kind." cr
50 1 pierpontprimes echo
cr cr
say "Pierpont primes of the second kind." cr
50 2 pierpontprimes echo```
Output:
```Pierpont primes of the first kind.
[ 2 3 5 7 13 17 19 37 73 97 109 163 193 257 433 487 577 769 1153 1297 1459 2593 2917 3457 3889 10369 12289 17497 18433 39367 52489 65537 139969 147457 209953 331777 472393 629857 746497 786433 839809 995329 1179649 1492993 1769473 1990657 2654209 5038849 5308417 8503057 ]

Pierpont primes of the second kind.
[ 2 3 5 7 11 17 23 31 47 53 71 107 127 191 383 431 647 863 971 1151 2591 4373 6143 6911 8191 8747 13121 15551 23327 27647 62207 73727 131071 139967 165887 294911 314927 442367 472391 497663 524287 786431 995327 1062881 2519423 10616831 17915903 18874367 25509167 30233087 ]```

## Raku

(formerly Perl 6)

Works with: Rakudo version 2019.07.1

### Finesse version

This finesse version never produces more Pierpont numbers than it needs to fulfill the requested number of primes. It uses a series of parallel iterators with additional iterators added as required. No need to speculatively generate an overabundance. No need to rely on magic numbers. No need to sort them. It produces exactly what is needed, in order, on demand.

Library: ntheory
```use ntheory:from<Perl5> <is_prime>;

sub pierpont (\$kind is copy = 1) {
fail "Unknown type: \$kind. Must be one of 1 (default) or 2" if \$kind !== 1|2;
\$kind = -1 if \$kind == 2;
my \$po3 = 3;
my @iterators = [1,2,4,8 … *].iterator, [3,9,27 … *].iterator;

gather {
loop {
my \$key = @head.pairs.min( *.value ).key;

take \$min + \$kind if "{\$min + \$kind}".&is_prime;

@iterators.push: ([|((2,4,8).map: * * \$po3) … *]).iterator;
\$po3 *= 3;
}
}
}
}

say "First 50 Pierpont primes of the first kind:\n" ~ pierpont[^50].rotor(10)».fmt('%8d').join: "\n";

say "\nFirst 50 Pierpont primes of the second kind:\n" ~ pierpont(2)[^50].rotor(10)».fmt('%8d').join: "\n";

say "\n250th Pierpont prime of the first kind: " ~ pierpont[249];

say "\n250th Pierpont prime of the second kind: " ~ pierpont(2)[249];
```
Output:
```First 50 Pierpont primes of the first kind:
2        3        5        7       13       17       19       37       73       97
109      163      193      257      433      487      577      769     1153     1297
1459     2593     2917     3457     3889    10369    12289    17497    18433    39367
52489    65537   139969   147457   209953   331777   472393   629857   746497   786433
839809   995329  1179649  1492993  1769473  1990657  2654209  5038849  5308417  8503057

First 50 Pierpont primes of the second kind:
2        3        5        7       11       17       23       31       47       53
71      107      127      191      383      431      647      863      971     1151
2591     4373     6143     6911     8191     8747    13121    15551    23327    27647
62207    73727   131071   139967   165887   294911   314927   442367   472391   497663
524287   786431   995327  1062881  2519423 10616831 17915903 18874367 25509167 30233087

250th Pierpont prime of the first kind: 62518864539857068333550694039553

250th Pierpont prime of the second kind: 4111131172000956525894875083702271```

### Generalized Pierpont iterator

Alternately, a version that will generate generalized Pierpont numbers for any set of prime integers where at least one of the primes is 2.

(Cut down output as it is exactly the same as the first version for {2,3} +1 and {2,3} -1; leaves room to demo some other options.)

```sub smooth-numbers (*@list) {
cache my \Smooth := gather {
my %i = (flat @list) Z=> (Smooth.iterator for ^@list);
my %n = (flat @list) Z=> 1 xx *;

loop {
take my \$n := %n{*}.min;

for @list -> \k {
%n{k} = %i{k}.pull-one * k if %n{k} == \$n;
}
}
}
}

# Testing various smooth numbers

for   'OEIS: A092506 - 2 + Fermat primes:',        (2),        1,  6,
"\nOEIS: A000668 - Mersenne primes:",          (2),       -1, 10,
"\nOEIS: A005109 - Pierpont primes 1st:",      (2,3),      1, 20,
"\nOEIS: A005105 - Pierpont primes 2nd:",      (2,3),     -1, 20,
"\nOEIS: A077497:",                            (2,5),      1, 20,
"\nOEIS: A077313:",                            (2,5),     -1, 20,
"\nOEIS: A002200 - (\"Hamming\" primes 1st):", (2,3,5),    1, 20,
"\nOEIS: A293194 - (\"Hamming\" primes 2nd):", (2,3,5),   -1, 20,
"\nOEIS: A077498:",                            (2,7),      1, 20,
"\nOEIS: A077314:",                            (2,7),     -1, 20,
"\nOEIS: A174144 - (\"Humble\" primes 1st):",  (2,3,5,7),  1, 20,
"\nOEIS: A299171 - (\"Humble\" primes 2nd):",  (2,3,5,7), -1, 20,
"\nOEIS: A077499:",                            (2,11),     1, 20,
"\nOEIS: A077315:",                            (2,11),    -1, 20,
"\nOEIS: A173236:",                            (2,13),     1, 20,
"\nOEIS: A173062:",                            (2,13),    -1, 20

-> \$title, \$primes, \$add, \$count {

say "\$title smooth \{\$primes\} {\$add > 0 ?? '+' !! '-'} 1 ";
put smooth-numbers(|\$primes).map( * + \$add ).grep( *.is-prime )[^\$count]
}
```
Output:
```OEIS: A092506 - 2 + Fermat primes: smooth {2} + 1
2 3 5 17 257 65537

OEIS: A000668 - Mersenne primes: smooth {2} - 1
3 7 31 127 8191 131071 524287 2147483647 2305843009213693951 618970019642690137449562111

OEIS: A005109 - Pierpont primes 1st: smooth {2 3} + 1
2 3 5 7 13 17 19 37 73 97 109 163 193 257 433 487 577 769 1153 1297

OEIS: A005105 - Pierpont primes 2nd: smooth {2 3} - 1
2 3 5 7 11 17 23 31 47 53 71 107 127 191 383 431 647 863 971 1151

OEIS: A077497: smooth {2 5} + 1
2 3 5 11 17 41 101 251 257 401 641 1601 4001 16001 25601 40961 62501 65537 160001 163841

OEIS: A077313: smooth {2 5} - 1
3 7 19 31 79 127 199 499 1249 1279 1999 4999 5119 8191 12799 20479 31249 49999 51199 79999

OEIS: A002200 - ("Hamming" primes 1st): smooth {2 3 5} + 1
2 3 5 7 11 13 17 19 31 37 41 61 73 97 101 109 151 163 181 193

OEIS: A293194 - ("Hamming" primes 2nd): smooth {2 3 5} - 1
2 3 5 7 11 17 19 23 29 31 47 53 59 71 79 89 107 127 149 179

OEIS: A077498: smooth {2 7} + 1
2 3 5 17 29 113 197 257 449 1373 3137 50177 65537 114689 268913 470597 614657 1075649 3294173 7340033

OEIS: A077314: smooth {2 7} - 1
3 7 13 31 97 127 223 1567 3583 4801 6271 8191 19207 25087 33613 76831 131071 401407 524287 917503

OEIS: A174144 - ("Humble" primes 1st): smooth {2 3 5 7} + 1
2 3 5 7 11 13 17 19 29 31 37 41 43 61 71 73 97 101 109 113

OEIS: A299171 - ("Humble" primes 2nd): smooth {2 3 5 7} - 1
2 3 5 7 11 13 17 19 23 29 31 41 47 53 59 71 79 83 89 97

OEIS: A077499: smooth {2 11} + 1
2 3 5 17 23 89 257 353 1409 2663 30977 65537 170369 495617 5767169 23068673 59969537 82458113 453519617 3429742097

OEIS: A077315: smooth {2 11} - 1
3 7 31 43 127 241 967 5323 8191 117127 131071 524287 7496191 10307263 77948683 253755391 428717761 738197503 1714871047 2147483647

OEIS: A173236: smooth {2 13} + 1
2 3 5 17 53 257 677 3329 13313 35153 65537 2768897 13631489 2303721473 3489660929 4942652417 11341398017 10859007357953 1594691292233729 31403151600910337

OEIS: A173062: smooth {2 13} - 1
3 7 31 103 127 337 1663 5407 8191 131071 346111 524287 2970343 3655807 22151167 109051903 617831551 1631461441 2007952543 2147483647```

## REXX

The REXX language has a "big num" capability to handle almost any amount of decimal digits,   but
it lacks a robust   isPrime   function.   Without that, verifying very large primes is problematic.

```/*REXX program finds and displays  Pierpont primes  of the  first  and  second  kinds.  */
parse arg n .                                    /*obtain optional argument from the CL.*/
if n=='' | n==","  then n= 50                    /*Not specified?  Then use the default.*/
numeric digits n                                 /*ensure enough decimal digs (bit int).*/
big= copies(9, digits() )                        /*BIG:  used as a max number (a limit).*/
@.= '2nd';                      @.1= '1st'
do t=1  to -1  by -2;   usum= 0;   vsum= 0;    s= 0       /*T  is  1,  then  -1.*/
#= 0                                     /*number of Pierpont primes  (so far). */
\$=;    do j=0  until #>=n                /*\$:   the list  "  "      "      "    */
if usum<=s  then usum= get(2, 3);    if vsum<=s  then vsum= get(3, 2)
s= min(vsum, usum);  if \isPrime(s)  then iterate /*get min;  Not prime? */
#= # + 1;            \$= \$ s                       /*bump counter; append.*/
end   /*j*/
say
w= length(word(\$, #) )                                   /*biggest prime length.*/
say center(n   " Pierpont primes of the "   @.t ' kind',  max(10 *(w+1), 80), "═")

do p=1  by 10  to #;      _=;      do k=p  for 10;   _= _ right( word(\$, k),  w)
end   /*k*/
if _\==''  then say substr( strip(_, "T"),  2)
end   /*p*/
end     /*t*/
exit 0                                           /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
isPrime: procedure; parse arg x '' -1 _; if x<17  then return wordpos(x,"2 3 5 7 11 13")>0
if _==5  then return 0;           if x//2==0  then return 0       /*not prime. */
if x//3==0  then return 0;        if x//7==0  then return 0       /* "    "    */
do j=11  by 6  until j*j>x                                     /*skip ÷ 3's.*/
if x//j==0  then return 0;  if x//(j+2)==0  then return 0      /*not prime. */
end   /*j*/;                                     return 1      /*it's prime.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
get: parse arg c1,c2; m=big;  do   ju=0;  pu= c1**ju;  if pu+t>s  then return min(m, pu+t)
do jv=0;  pv= c2**jv;  if pv  >s  then iterate ju
_= pu*pv  +  t;        if _   >s  then m= min(_, m)
end   /*jv*/
end     /*ju*/           /*see the    RETURN    (above).  */
```
output   when using the default input:
```═════════════════════50  Pierpont primes of the  1st  kind══════════════════════
2       3       5       7      13      17      19      37      73      97
109     163     193     257     433     487     577     769    1153    1297
1459    2593    2917    3457    3889   10369   12289   17497   18433   39367
52489   65537  139969  147457  209953  331777  472393  629857  746497  786433
839809  995329 1179649 1492993 1769473 1990657 2654209 5038849 5308417 8503057

══════════════════════════50  Pierpont primes of the  2nd  kind═══════════════════════════
2        3        5        7       11       17       23       31       47       53
71      107      127      191      383      431      647      863      971     1151
2591     4373     6143     6911     8191     8747    13121    15551    23327    27647
62207    73727   131071   139967   165887   294911   314927   442367   472391   497663
524287   786431   995327  1062881  2519423 10616831 17915903 18874367 25509167 30233087
```

## Ring

```load "stdlib.ring"

see "working..." + nl

pierpont = []
limit1 = 18
limit2 = 8505000
limit3 = 50
limit4 = 21
limit5 = 30500000

for n = 0 to limit1
for m = 0 to limit1
num = pow(2,n)*pow(3,m) + 1
if isprime(num) and num < limit2
ok
if num > limit2
exit
ok
next
next

pierpont = sort(pierpont)

see "First 50 Pierpont primes of the first kind:" + nl + nl

for n = 1 to limit3
see "" + n + ". " + pierpont[n] + nl
next

see "done1..." + nl

pierpont = []

for n = 0 to limit4
for m = 0 to limit4
num = pow(2,n)*pow(3,m) - 1
if isprime(num) and num < limit5
ok
if num > limit5
exit
ok
next
next

pierpont = sort(pierpont)

see "First 50 Pierpont primes of the second kind:" + nl + nl

for n = 1 to limit3
see "" + n + ". " + pierpont[n] + nl
next

see "done2..." + nl```

Output:

```working...
First 50 Pierpont primes of the first kind:

1. 2
2. 3
3. 5
4. 7
5. 13
6. 17
7. 19
8. 37
9. 73
10. 97
11. 109
12. 163
13. 193
14. 257
15. 433
16. 487
17. 577
18. 769
19. 1153
20. 1297
21. 1459
22. 2593
23. 2917
24. 3457
25. 3889
26. 10369
27. 12289
28. 17497
29. 18433
30. 39367
31. 52489
32. 65537
33. 139969
34. 147457
35. 209953
36. 331777
37. 472393
38. 629857
39. 746497
40. 786433
41. 839809
42. 995329
43. 1179649
44. 1492993
45. 1769473
46. 1990657
47. 2654209
48. 5038849
49. 5308417
50. 8503057
done1...
First 50 Pierpont primes of the second kind:

1. 2
2. 3
3. 5
4. 7
5. 11
6. 17
7. 23
8. 31
9. 47
10. 53
11. 71
12. 107
13. 127
14. 191
15. 383
16. 431
17. 647
18. 863
19. 971
20. 1151
21. 2591
22. 4373
23. 6143
24. 6911
25. 8191
26. 8747
27. 13121
28. 15551
29. 23327
30. 27647
31. 62207
32. 73727
33. 131071
34. 139967
35. 165887
36. 294911
37. 314927
38. 442367
39. 472391
40. 497663
41. 524287
42. 786431
43. 995327
44. 1062881
45. 2519423
46. 10616831
47. 17915903
48. 18874367
49. 25509167
50. 30233087
done2...
```

## Ruby

```require 'gmp'

def  smooth_generator(ar)
next_smooth = 1
queues = ar.map{|num| [num, []] }
loop do
yield next_smooth
queues.each {|m, queue| queue << next_smooth * m}
next_smooth = queues.collect{|m, queue| queue.first}.min
queues.each{|m, queue| queue.shift if queue.first == next_smooth }
end
end

def pierpont(num = 1)
smooth_generator([2,3]).each{|smooth| yield smooth+num if GMP::Z(smooth + num).probab_prime? > 0}
end

def puts_cols(ar, n=10)
ar.each_slice(n).map{|slice|puts  slice.map{|n| n.to_s.rjust(10)}.join }
end

n, m = 50, 250
puts "First #{n} Pierpont primes of the first kind:"
puts_cols(pierpont.take(n))
puts "#{m}th Pierpont prime of the first kind: #{pierpont.take(250).last}",""
puts "First #{n} Pierpont primes of the second kind:"
puts_cols(pierpont(-1).take(n))
puts "#{m}th Pierpont prime of the second kind: #{pierpont(-1).take(250).last}"
```
Output:
```First 50 Pierpont primes of the first kind:
2         3         5         7        13        17        19        37        73        97
109       163       193       257       433       487       577       769      1153      1297
1459      2593      2917      3457      3889     10369     12289     17497     18433     39367
52489     65537    139969    147457    209953    331777    472393    629857    746497    786433
839809    995329   1179649   1492993   1769473   1990657   2654209   5038849   5308417   8503057
250th Pierpont prime of the first kind: 62518864539857068333550694039553

First 50 Pierpont primes of the second kind:
2         3         5         7        11        17        23        31        47        53
71       107       127       191       383       431       647       863       971      1151
2591      4373      6143      6911      8191      8747     13121     15551     23327     27647
62207     73727    131071    139967    165887    294911    314927    442367    472391    497663
524287    786431    995327   1062881   2519423  10616831  17915903  18874367  25509167  30233087
250th Pierpont prime of the second kind: 4111131172000956525894875083702271
```

## Sidef

```func smooth_generator(primes) {
var s = primes.len.of { [1] }
{
var n = s.map { .first }.min
{ |i|
s[i].shift if (s[i][0] == n)
s[i] << (n * primes[i])
} * primes.len
n
}
}

func pierpont_primes(n, k = 1) {
var g = smooth_generator([2,3])
1..Inf -> lazy.map { g.run + k }.grep { .is_prime }.first(n)
}

say "First 50 Pierpont primes of the 1st kind: "
say pierpont_primes(50, +1).join(' ')

say "\nFirst 50 Pierpont primes of the 2nd kind: "
say pierpont_primes(50, -1).join(' ')

for n in (250, 500, 1000) {
var p = pierpont_primes(n, +1).last
var q = pierpont_primes(n, -1).last
say "\n#{n}th Pierpont prime of the 1st kind: #{p}"
say "#{n}th Pierpont prime of the 2nd kind: #{q}"
}
```
Output:
```First 50 Pierpont primes of the 1st kind:
2 3 5 7 13 17 19 37 73 97 109 163 193 257 433 487 577 769 1153 1297 1459 2593 2917 3457 3889 10369 12289 17497 18433 39367 52489 65537 139969 147457 209953 331777 472393 629857 746497 786433 839809 995329 1179649 1492993 1769473 1990657 2654209 5038849 5308417 8503057

First 50 Pierpont primes of the 2nd kind:
2 3 5 7 11 17 23 31 47 53 71 107 127 191 383 431 647 863 971 1151 2591 4373 6143 6911 8191 8747 13121 15551 23327 27647 62207 73727 131071 139967 165887 294911 314927 442367 472391 497663 524287 786431 995327 1062881 2519423 10616831 17915903 18874367 25509167 30233087

250th Pierpont prime of the 1st kind: 62518864539857068333550694039553
250th Pierpont prime of the 2nd kind: 4111131172000956525894875083702271

500th Pierpont prime of the 1st kind: 2228588214163334773718162801501181906563609505773852212825423873
500th Pierpont prime of the 2nd kind: 1582451786724712011220565860035647126064921139018525742951994237124607

1000th Pierpont prime of the 1st kind: 69269314716439690250482558089997110961545818230232043107188537422260188701607997086273960899938499201024414931399264696270849
1000th Pierpont prime of the 2nd kind: 1308088756227965581249669045506775407896673213729433892383353027814827286537163695213418982500477392209371001259166465228280492460735463423
```

## Swift

3-smooth version.

Using AttaSwift's BigInt library.

```import BigInt
import Foundation

public func pierpoint(n: Int) -> (first: [BigInt], second: [BigInt]) {
var primes = (first: [BigInt](repeating: 0, count: n), second: [BigInt](repeating: 0, count: n))

primes.first[0] = 2

var count1 = 1, count2 = 0
var s = [BigInt(1)]
var i2 = 0, i3 = 0, k = 1
var n2 = BigInt(0), n3 = BigInt(0), t = BigInt(0)

while min(count1, count2) < n {
n2 = s[i2] * 2
n3 = s[i3] * 3

if n2 < n3 {
t = n2
i2 += 1
} else {
t = n3
i3 += 1
}

if t <= s[k - 1] {
continue
}

s.append(t)
k += 1
t += 1

if count1 < n && t.isPrime(rounds: 10) {
primes.first[count1] = t
count1 += 1
}

t -= 2

if count2 < n && t.isPrime(rounds: 10) {
primes.second[count2] = t
count2 += 1
}
}

return primes
}

let primes = pierpoint(n: 250)

print("First 50 Pierpoint primes of the first kind: \(Array(primes.first.prefix(50)))\n")
print("First 50 Pierpoint primes of the second kind: \(Array(primes.second.prefix(50)))")
print()
print("250th Pierpoint prime of the first kind: \(primes.first[249])")
print("250th Pierpoint prime of the second kind: \(primes.second[249])")
```
Output:
```First 50 Pierpoint primes of the first kind: [2, 3, 5, 7, 13, 17, 19, 37, 73, 97, 109, 163, 193, 257, 433, 487, 577, 769, 1153, 1297, 1459, 2593, 2917, 3457, 3889, 10369, 12289, 17497, 18433, 39367, 52489, 65537, 139969, 147457, 209953, 331777, 472393, 629857, 746497, 786433, 839809, 995329, 1179649, 1492993, 1769473, 1990657, 2654209, 5038849, 5308417, 8503057]

First 50 Pierpoint primes of the second kind: [2, 3, 5, 7, 11, 17, 23, 31, 47, 53, 71, 107, 127, 191, 383, 431, 647, 863, 971, 1151, 2591, 4373, 6143, 6911, 8191, 8747, 13121, 15551, 23327, 27647, 62207, 73727, 131071, 139967, 165887, 294911, 314927, 442367, 472391, 497663, 524287, 786431, 995327, 1062881, 2519423, 10616831, 17915903, 18874367, 25509167, 30233087]

250th Pierpoint prime of the first kind: 62518864539857068333550694039553
250th Pierpoint prime of the second kind: 4111131172000956525894875083702271```

## Wren

Translation of: Go
Library: Wren-big
Library: Wren-fmt

The 3-smooth version. Just the first 250 - a tolerable 14 seconds or so on my machine.

```import "/big" for BigInt
import "/fmt" for Fmt

var pierpont = Fn.new { |n, first|
var p = [ List.filled(n, null), List.filled(n, null) ]
for (i in 0...n) {
p[0][i] = BigInt.zero
p[1][i] = BigInt.zero
}
p[0][0] = BigInt.two
var count  = 0
var count1 = 1
var count2 = 0
var s = [BigInt.one]
var i2 = 0
var i3 = 0
var k  = 1
while (count < n) {
var n2 = s[i2] * 2
var n3 = s[i3] * 3
var t
if (n2 < n3) {
t = n2
i2 = i2 + 1
} else {
t = n3
i3 = i3 + 1
}
if (t > s[k-1]) {
k = k + 1
t = t.inc
if (count1 < n && t.isProbablePrime(5)) {
p[0][count1] = t.copy()
count1 = count1 + 1
}
t = t - 2
if (count2 < n && t.isProbablePrime(5)) {
p[1][count2] = t.copy()
count2 = count2 + 1
}
count = count1.min(count2)
}
}
return p
}

var p = pierpont.call(250, true)
System.print("First 50 Pierpont primes of the first kind:")
for (i in 0...50) {
Fmt.write("\$8i ", p[0][i])
if ((i-9)%10 == 0) System.print()
}
System.print("\nFirst 50 Pierpont primes of the second kind:")
for (i in 0...50) {
Fmt.write("\$8i ", p[1][i])
if ((i-9)%10 == 0) System.print()
}

System.print("\n250th Pierpont prime of the first kind: %(p[0][249])")
System.print("\n250th Pierpont prime of the second kind: %(p[1][249])")
```
Output:
```First 50 Pierpont primes of the first kind:
2        3        5        7       13       17       19       37       73       97
109      163      193      257      433      487      577      769     1153     1297
1459     2593     2917     3457     3889    10369    12289    17497    18433    39367
52489    65537   139969   147457   209953   331777   472393   629857   746497   786433
839809   995329  1179649  1492993  1769473  1990657  2654209  5038849  5308417  8503057

First 50 Pierpont primes of the second kind:
2        3        5        7       11       17       23       31       47       53
71      107      127      191      383      431      647      863      971     1151
2591     4373     6143     6911     8191     8747    13121    15551    23327    27647
62207    73727   131071   139967   165887   294911   314927   442367   472391   497663
524287   786431   995327  1062881  2519423 10616831 17915903 18874367 25509167 30233087

250th Pierpont prime of the first kind: 62518864539857068333550694039553

250th Pierpont prime of the second kind: 4111131172000956525894875083702271
```

## zkl

Translation of: Go
Library: GMP
GNU Multiple Precision Arithmetic Library

Using GMP's probabilistic primes makes it is easy and fast to test for primeness.

```var [const] BI=Import("zklBigNum");  // libGMP
var [const] one=BI(1), two=BI(2), three=BI(3);

fcn pierPonts(n){  //-->((bigInt first kind primes) (bigInt second))
pps1,pps2 := List(BI(2)), List();
count1, count2, s := 1, 0, List(BI(1));  // n==2_000, s-->266_379 elements
i2,i3,k := 0, 0, 1;
n2,n3,t := BI(0),BI(0),BI(0);
while(count1.min(count2) < n){
n2.set(s[i2]).mul(two);	// .mul, .add, .sub are in-place
n3.set(s[i3]).mul(three);
if(n2<n3){ t.set(n2); i2+=1; }
else     { t.set(n3); i3+=1; }
if(t > s[k-1]){
s.append(t.copy());
k+=1;
if(count1<n and t.probablyPrime()){
pps1.append(t.copy());
count1+=1;
}
if(count2<n and t.sub(two).probablyPrime()){
pps2.append(t.copy());
count2+=1;
}
}
}
return(pps1,pps2)
}```
```pps1,pps2 := pierPonts(2_000);

println("The first 50 Pierpont primes (first kind):");
foreach r in (5){ pps1[r*10,10].apply("%10d".fmt).concat().println() }

println("\nThe first 50 Pierpont primes (second kind):");
foreach r in (5){ pps2[r*10,10].apply("%10d".fmt).concat().println() }

foreach n in (T(250, 1_000, 2_000)){
println("\n%4dth Pierpont prime, first kind: ".fmt(n), pps1[n-1]);
println( "                      second kind: ",        pps2[n-1]);
}```
Output:
```The first 50 Pierpont primes (first kind):
2         3         5         7        13        17        19        37        73        97
109       163       193       257       433       487       577       769      1153      1297
1459      2593      2917      3457      3889     10369     12289     17497     18433     39367
52489     65537    139969    147457    209953    331777    472393    629857    746497    786433
839809    995329   1179649   1492993   1769473   1990657   2654209   5038849   5308417   8503057

The first 50 Pierpont primes (second kind):
2         3         5         7        11        17        23        31        47        53
71       107       127       191       383       431       647       863       971      1151
2591      4373      6143      6911      8191      8747     13121     15551     23327     27647
62207     73727    131071    139967    165887    294911    314927    442367    472391    497663
524287    786431    995327   1062881   2519423  10616831  17915903  18874367  25509167  30233087

250th Pierpont prime, first kind: 62518864539857068333550694039553
second kind: 4111131172000956525894875083702271

1000th Pierpont prime, first kind: 69269314716439690250482558089997110961545818230232043107188537422260188701607997086273960899938499201024414931399264696270849
second kind: 1308088756227965581249669045506775407896673213729433892383353027814827286537163695213418982500477392209371001259166465228280492460735463423

2000th Pierpont prime, first kind: 23647056334818750458979408107288138983957799805326855934519920502493109431728722178351835778368596067773810122477389192659352731519830867553659739507195398662712180250483714053474639899675114018023738461139103130959712720686117399642823861502738433
second kind: 1702224134662426018061116932011222570937093650174807121918750428723338890211147039320296240754205680537318845776107057915956535566573559841027244444877454493022783449689509569107393738917120492483994302725479684822283929715327187974256253064796234576415398735760543848603844607
```