Pierpont primes
You are encouraged to solve this task according to the task description, using any language you may know.
A Pierpont prime is a prime number of the form: 2u3v + 1 for some non-negative integers u and v .
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.
- Task
- 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.
- See also
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 #
PR read "primes.incl.a68" PR
[]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
#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].Add(0);
p[1].Add(0);
}
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]) {
s.Add(t);
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++
#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
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
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);
readln;
end.
EasyLang
fastfunc isprim num .
if num mod 2 = 0
if num = 2
return 1
.
return 0
.
i = 3
while i <= sqrt num
if num mod i = 0
return 0
.
i += 2
.
return 1
.
fastfunc mod2x3x n .
while n mod 2 = 0
n /= 2
.
while n mod 3 = 0
n /= 3
.
return n
.
i = 2
cnt = 1
write 2 & " "
while cnt < 50
if mod2x3x i = 1 and isprim (i + 1) = 1
cnt += 1
write i + 1 & " "
.
i += 2
.
print ""
print ""
i = 4
write 2 & " "
cnt = 1
while cnt < 50
if mod2x3x i = 1 and isprim (i - 1) = 1
cnt += 1
write i - 1 & " "
.
i += 2
.- 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
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 {
p.Add(p, one)
} 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++
t.Add(t, one)
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
Haskell
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 ) {
primes.add(BigInteger.valueOf(2));
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);
twoSmooth.add(min);
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) ) {
primes.add(test);
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 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 $maxn are large enough
def adequate($maxm; $maxn):
# ( ".[$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 $maxn
# input: [maxm, maxn, ([maxm,maxn]|pp)]
# output: pp
def adapt:
. as [$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])" ;
# The primary task:
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
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>()
s.add(one)
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]) {
s.add(t)
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
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
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 $add_one = 3;
my @iterators;
push @iterators, iter1(2);
push @iterators, iter1(3); $iterators[1]->();
my @head = ($iterators[0]->(), $iterators[1]->());
my @pierpont;
do {
my $key = min_index(@head);
my $min = $head[$key];
push @pierpont, $min + $kind if is_prime($min + $kind);
$head[$key] = $iterators[$key]->();
if ($min >= $add_one) {
push @iterators, iter2($po3);
$add_one = $head[$#iterators] = $iterators[$#iterators]->();
$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: " . $pierpont_2nd[249];
- Output:
First 50 Pierpont primes of the first kind:
2 3 5 7 13 17 19 37 73 97
109 163 193 257 433 487 577 769 1153 1297
1459 2593 2917 3457 3889 10369 12289 17497 18433 39367
52489 65537 139969 147457 209953 331777 472393 629857 746497 786433
839809 995329 1179649 1492993 1769473 1990657 2654209 5038849 5308417 8503057
First 50 Pierpont primes of the second kind:
2 3 5 7 11 17 23 31 47 53
71 107 127 191 383 431 647 863 971 1151
2591 4373 6143 6911 8191 8747 13121 15551 23327 27647
62207 73727 131071 139967 165887 294911 314927 442367 472391 497663
524287 786431 995327 1062881 2519423 10616831 17915903 18874367 25509167 30233087
250th Pierpont prime of the first kind: 62518864539857068333550694039553
250th Pierpont prime of the second kind: 4111131172000956525894875083702271
Phix
I also tried using a priority queue, popping the smallest (and any duplicates of it) and pushing *2 and *3 on every iteration, which worked pretty well but ended up about 20% slower, with about 1500 untested candidates left in the queue by the time it found the 2000th second kind.
-- demo/rosetta/Pierpont_primes.exw with javascript_semantics include mpfr.e function pierpont(integer n) sequence p = {{mpz_init(2)},{}}, s = {mpz_init(1)} integer i2 = 1, i3 = 1 mpz {n2, n3, t} = mpz_inits(3) atom t1 = time()+1 while min(length(p[1]),length(p[2])) < n do mpz_mul_si(n2,s[i2],2) mpz_mul_si(n3,s[i3],3) if mpz_cmp(n2,n3)<0 then mpz_set(t,n2) i2 += 1 else mpz_set(t,n3) i3 += 1 end if if mpz_cmp(t,s[$]) > 0 then s = append(s, mpz_init_set(t)) mpz_add_ui(t,t,1) 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,
add_to_heap(H1, N2, nothing, H2),
add_to_heap(H2, N3, nothing, H3).
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
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
behead drop ] is pierpontprimes
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)
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.
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 $add-one = 3;
my @iterators = [1,2,4,8 … *].iterator, [3,9,27 … *].iterator;
my @head = @iterators».pull-one;
gather {
loop {
my $key = @head.pairs.min( *.value ).key;
my $min = @head[$key];
@head[$key] = @iterators[$key].pull-one;
take $min + $kind if "{$min + $kind}".&is_prime;
if $min >= $add-one {
@iterators.push: ([|((2,4,8).map: * * $po3) … *]).iterator;
$add-one = @head[+@iterators - 1] = @iterators[+@iterators - 1].pull-one;
$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: A347977 - (\"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
add(pierpont,num)
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
add(pierpont,num)
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..." + nlOutput:
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)
return to_enum(__method__, ar) unless block_given?
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)
return to_enum(__method__, num) unless block_given?
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
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]) {
s.add(t.copy())
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
XPL0
func IsPrime(N); \Return 'true' if N is prime
int N, D;
[if N < 2 then return false;
if (N&1) = 0 then return N = 2;
if rem(N/3) = 0 then return N = 3;
D:= 5;
while D*D <= N do
[if rem(N/D) = 0 then return false;
D:= D+2;
if rem(N/D) = 0 then return false;
D:= D+4;
];
return true;
];
func Pierpont(N); \Return 'true' if N is a multiple of 2^U*3^V
int N;
[while (N&1) = 0 do N:= N>>1;
while N>1 do
[N:= N/3;
if rem(0) # 0 then return false;
];
return true;
];
int Kind, N, Count;
[Format(9, 0);
Kind:= 1;
repeat Count:= 0; N:= 2;
loop [if IsPrime(N) then
[if Pierpont(N-Kind) then
[Count:= Count+1;
RlOut(0, float(N));
if rem(Count/10) = 0 then CrLf(0);
if Count >= 50 then quit;
];
];
N:= N+1;
];
CrLf(0);
Kind:= -Kind;
until Kind = 1;
]- 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
zkl
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;
t.add(one);
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
Categories:
- Programming Tasks
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