Ramanujan primes
As the integers get larger, the spacing between prime numbers slowly lengthens, but the spacing between primes increases at a slower rate than the numbers themselves increase. A consequence of this difference in rates of increase is the existence of special primes, called Ramanujan primes.
The `n`th Ramanujan prime is defined to be the least integer for which there are at least n primes between x and x/2 for all x greater or equal to n.
- Task
- Generate and show the first 100 Ramanujan prime numbers.
- Find and show the 1000th Ramanujan prime number.
- Stretch task
- Find and show the 10,000th Ramanujan prime number.
- See also
- The pi prime function, not to be confused with the transcendental number π
- The OEIS entry: OEIS entry
- The Wikipedia entry: Ramanujan_prime.
C++
<lang cpp>#include <chrono>
- include <cmath>
- include <iomanip>
- include <iostream>
- include <numeric>
- include <vector>
class prime_counter { public:
explicit prime_counter(int limit);
int prime_count(int n) const { return n < 1 ? 0 : count_.at(n); }
private:
std::vector<int> count_;
};
prime_counter::prime_counter(int limit) : count_(limit, 1) {
if (limit > 0)
count_[0] = 0;
if (limit > 1)
count_[1] = 0;
for (int i = 4; i < limit; i += 2)
count_[i] = 0;
for (int p = 3, sq = 9; sq < limit; p += 2) {
if (count_[p]) {
for (int q = sq; q < limit; q += p << 1)
count_[q] = 0;
}
sq += (p + 1) << 2;
}
std::partial_sum(count_.begin(), count_.end(), count_.begin());
}
int ramanujan_max(int n) {
return static_cast<int>(std::ceil(4 * n * std::log(4 * n)));
}
int ramanujan_prime(const prime_counter& pc, int n) {
int max = ramanujan_max(n);
for (int i = max; i >= 0; --i) {
if (pc.prime_count(i) - pc.prime_count(i / 2) < n)
return i + 1;
}
return 0;
}
int main() {
std::cout.imbue(std::locale(""));
auto start = std::chrono::high_resolution_clock::now();
prime_counter pc(1 + ramanujan_max(100000));
for (int i = 1; i <= 100; ++i) {
std::cout << std::setw(5) << ramanujan_prime(pc, i)
<< (i % 10 == 0 ? '\n' : ' ');
}
std::cout << '\n';
for (int n = 1000; n <= 100000; n *= 10) {
std::cout << "The " << n << "th Ramanujan prime is " << ramanujan_prime(pc, n)
<< ".\n";
}
auto end = std::chrono::high_resolution_clock::now();
std::cout << "\nElapsed time: "
<< std::chrono::duration<double>(end - start).count() * 1000
<< " milliseconds\n";
}</lang>
- Output:
2 11 17 29 41 47 59 67 71 97 101 107 127 149 151 167 179 181 227 229 233 239 241 263 269 281 307 311 347 349 367 373 401 409 419 431 433 439 461 487 491 503 569 571 587 593 599 601 607 641 643 647 653 659 677 719 727 739 751 769 809 821 823 827 853 857 881 937 941 947 967 983 1,009 1,019 1,021 1,031 1,049 1,051 1,061 1,063 1,087 1,091 1,097 1,103 1,151 1,163 1,187 1,217 1,229 1,249 1,277 1,289 1,297 1,301 1,367 1,373 1,423 1,427 1,429 1,439 The 1,000th Ramanujan prime is 19,403. The 10,000th Ramanujan prime is 242,057. The 100,000th Ramanujan prime is 2,916,539. Elapsed time: 46.0828 milliseconds
F#
This task uses Extensible Prime Generator (F#) <lang fsharp> // Ramanujan primes. Nigel Galloway: September 7th., 2021 let fN g=if isPrime g then 1 else if g%2=1 then 0 else if isPrime(g/2) then -1 else 0 let rP p=let N,G=Array.create p 0,(Seq.item(3*p-2)(primes32()))+1 in let rec fG n g=if g=G then N else(if n<p then N.[n]<-g); fG(n+(fN g))(g+1) in fG 0 1 let n=rP 100000 n.[0..99]|>Array.iter(printf "%d "); printfn "" [1000;10000;100000]|>List.iter(fun g->printf $"The %d{g}th Ramanujan prime is %d{n.[g-1]}\n" ) </lang>
- Output:
2 11 17 29 41 47 59 67 71 97 101 107 127 149 151 167 179 181 227 229 233 239 241 263 269 281 307 311 347 349 367 373 401 409 419 431 433 439 461 487 491 503 569 571 587 593 599 601 607 641 643 647 653 659 677 719 727 739 751 769 809 821 823 827 853 857 881 937 941 947 967 983 1009 1019 1021 1031 1049 1051 1061 1063 1087 1091 1097 1103 1151 1163 1187 1217 1229 1249 1277 1289 1297 1301 1367 1373 1423 1427 1429 1439 The 1000th Ramanujan prime is 19403 The 10000th Ramanujan prime is 242057 The 100000th Ramanujan prime is 2916539
Go
This takes about 40 ms to find the 100,000th Ramanujan prime on my machine. The millionth takes about 520 ms.
<lang go>package main
import (
"fmt" "math" "rcu" "time"
)
var count []int
func primeCounter(limit int) {
count = make([]int, limit)
for i := 0; i < limit; i++ {
count[i] = 1
}
if limit > 0 {
count[0] = 0
}
if limit > 1 {
count[1] = 0
}
for i := 4; i < limit; i += 2 {
count[i] = 0
}
for p, sq := 3, 9; sq < limit; p += 2 {
if count[p] != 0 {
for q := sq; q < limit; q += p << 1 {
count[q] = 0
}
}
sq += (p + 1) << 2
}
sum := 0
for i := 0; i < limit; i++ {
sum += count[i]
count[i] = sum
}
}
func primeCount(n int) int {
if n < 1 {
return 0
}
return count[n]
}
func ramanujanMax(n int) int {
fn := float64(n) return int(math.Ceil(4 * fn * math.Log(4*fn)))
}
func ramanujanPrime(n int) int {
if n == 1 {
return 2
}
for i := ramanujanMax(n); i >= 2*n; i-- {
if i%2 == 1 {
continue
}
if primeCount(i)-primeCount(i/2) < n {
return i + 1
}
}
return 0
}
func main() {
start := time.Now()
primeCounter(1 + ramanujanMax(1e6))
fmt.Println("The first 100 Ramanujan primes are:")
rams := make([]int, 100)
for n := 0; n < 100; n++ {
rams[n] = ramanujanPrime(n + 1)
}
for i, r := range rams {
fmt.Printf("%5s ", rcu.Commatize(r))
if (i+1)%10 == 0 {
fmt.Println()
}
}
fmt.Printf("\nThe 1,000th Ramanujan prime is %6s\n", rcu.Commatize(ramanujanPrime(1000)))
fmt.Printf("\nThe 10,000th Ramanujan prime is %7s\n", rcu.Commatize(ramanujanPrime(10000)))
fmt.Printf("\nThe 100,000th Ramanujan prime is %6s\n", rcu.Commatize(ramanujanPrime(100000)))
fmt.Printf("\nThe 1,000,000th Ramanujan prime is %7s\n", rcu.Commatize(ramanujanPrime(1000000)))
fmt.Println("\nTook", time.Since(start))
}</lang>
- Output:
The first 100 Ramanujan primes are:
2 11 17 29 41 47 59 67 71 97
101 107 127 149 151 167 179 181 227 229
233 239 241 263 269 281 307 311 347 349
367 373 401 409 419 431 433 439 461 487
491 503 569 571 587 593 599 601 607 641
643 647 653 659 677 719 727 739 751 769
809 821 823 827 853 857 881 937 941 947
967 983 1,009 1,019 1,021 1,031 1,049 1,051 1,061 1,063
1,087 1,091 1,097 1,103 1,151 1,163 1,187 1,217 1,229 1,249
1,277 1,289 1,297 1,301 1,367 1,373 1,423 1,427 1,429 1,439
The 1,000th Ramanujan prime is 19,403
The 10,000th Ramanujan prime is 242,057
The 100,000th Ramanujan prime is 2,916,539
The 1,000,000th Ramanujan prime is 34,072,993
Took 519.655163ms
Java
<lang java>import java.util.Arrays;
public class RamanujanPrimes {
public static void main(String[] args) {
long start = System.nanoTime();
System.out.println("First 100 Ramanujan primes:");
PrimeCounter pc = new PrimeCounter(1 + ramanujanMax(100000));
for (int i = 1; i <= 100; ++i) {
int p = ramanujanPrime(pc, i);
System.out.printf("%,5d%c", p, i % 10 == 0 ? '\n' : ' ');
}
System.out.println();
for (int i = 1000; i <= 100000; i *= 10) {
int p = ramanujanPrime(pc, i);
System.out.printf("The %,dth Ramanujan prime is %,d.\n", i, p);
}
long end = System.nanoTime();
System.out.printf("\nElapsed time: %.1f milliseconds\n", (end - start) / 1e6);
}
private static int ramanujanMax(int n) {
return (int)Math.ceil(4 * n * Math.log(4 * n));
}
private static int ramanujanPrime(PrimeCounter pc, int n) {
for (int i = ramanujanMax(n); i >= 0; --i) {
if (pc.primeCount(i) - pc.primeCount(i / 2) < n)
return i + 1;
}
return 0;
}
private static class PrimeCounter {
private PrimeCounter(int limit) {
count = new int[limit];
Arrays.fill(count, 1);
if (limit > 0)
count[0] = 0;
if (limit > 1)
count[1] = 0;
for (int i = 4; i < limit; i += 2)
count[i] = 0;
for (int p = 3, sq = 9; sq < limit; p += 2) {
if (count[p] != 0) {
for (int q = sq; q < limit; q += p << 1)
count[q] = 0;
}
sq += (p + 1) << 2;
}
Arrays.parallelPrefix(count, (x, y) -> x + y);
}
private int primeCount(int n) {
return n < 1 ? 0 : count[n];
}
private int[] count; }
}</lang>
- Output:
First 100 Ramanujan primes:
2 11 17 29 41 47 59 67 71 97
101 107 127 149 151 167 179 181 227 229
233 239 241 263 269 281 307 311 347 349
367 373 401 409 419 431 433 439 461 487
491 503 569 571 587 593 599 601 607 641
643 647 653 659 677 719 727 739 751 769
809 821 823 827 853 857 881 937 941 947
967 983 1,009 1,019 1,021 1,031 1,049 1,051 1,061 1,063
1,087 1,091 1,097 1,103 1,151 1,163 1,187 1,217 1,229 1,249
1,277 1,289 1,297 1,301 1,367 1,373 1,423 1,427 1,429 1,439
The 1,000th Ramanujan prime is 19,403.
The 10,000th Ramanujan prime is 242,057.
The 100,000th Ramanujan prime is 2,916,539.
Elapsed time: 187.2 milliseconds
Julia
<lang julia>using Primes
@time let
MASK = primesmask(625000)
PIVEC = accumulate(+, MASK)
PI(n) = n < 1 ? 0 : PIVEC[n]
function Ramanujan_prime(n)
maxposs = Int(ceil(4n * (log(4n) / log(2))))
for i in maxposs:-1:1
PI(i) - PI(i ÷ 2) < n && return i + 1
end
return 0
end
for i in 1:100
print(lpad(Ramanujan_prime(i), 5), i % 20 == 0 ? "\n" : "")
end
println("\nThe 1000th Ramanujan prime is ", Ramanujan_prime(1000))
println("\nThe 10,000th Ramanujan prime is ", Ramanujan_prime(10000))
end
</lang>
- Output:
2 11 17 29 41 47 59 67 71 97 101 107 127 149 151 167 179 181 227 229 233 239 241 263 269 281 307 311 347 349 367 373 401 409 419 431 433 439 461 487 491 503 569 571 587 593 599 601 607 641 643 647 653 659 677 719 727 739 751 769 809 821 823 827 853 857 881 937 941 947 967 983 1009 1019 1021 1031 1049 1051 1061 1063 1087 1091 1097 1103 1151 1163 1187 1217 1229 1249 1277 1289 1297 1301 1367 1373 1423 1427 1429 1439 The 1000th Ramanujan prime is 19403 The 10,000th Ramanujan prime is 242057 0.272471 seconds (625.44 k allocations: 38.734 MiB, 33.07% compilation time)
Mathematica/Wolfram Language
<lang Mathematica>l = PrimePi[Range[10^6]] - PrimePi[Range[10^6]/2]; Multicolumn[1 + Position[l, #]-1, 1 & /@ Range[0, 99], {Automatic, 10}, Appearance -> "Horizontal"] 1 + Position[l, 999]-1, 1 1 + Position[l, 9999]-1, 1</lang>
- Output:
2 11 17 29 41 47 59 67 71 97 101 107 127 149 151 167 179 181 227 229 233 239 241 263 269 281 307 311 347 349 367 373 401 409 419 431 433 439 461 487 491 503 569 571 587 593 599 601 607 641 643 647 653 659 677 719 727 739 751 769 809 821 823 827 853 857 881 937 941 947 967 983 1009 1019 1021 1031 1049 1051 1061 1063 1087 1091 1097 1103 1151 1163 1187 1217 1229 1249 1277 1289 1297 1301 1367 1373 1423 1427 1429 1439 19403 242057
Nim
I compiled using command nim c -d:release -d:lto --gc:arc ramanujan_primes.nim, i.e. with runtime checks on, link time optimization and using Arc garbage collector. To find the 100_000th Ramanujan prime, the program runs in about 100 ms on my laptop (i5-8250U CPU @ 1.60GHz, 8 GB Ram, Linux Manjaro).
<lang Nim>import math, sequtils, strutils, times
let t0 = now()
type PrimeCounter = seq[int]
proc initPrimeCounter(limit: Positive): PrimeCounter =
doAssert limit > 1
result = repeat(1, limit)
result[0] = 0
result[1] = 0
for i in countup(4, limit - 1, 2): result[i] = 0
var p = 3
var p2 = 9
while p2 < limit:
if result[p] != 0:
for q in countup(p2, limit - 1, p shl 1):
result[q] = 0
p2 += (p + 1) shl 2
if p2 >= limit: break
inc p, 2
# Compute partial sums in place.
var sum = 0
for item in result.mitems:
sum += item
item = sum
func ramanujanMax(n: int): int {.inline.} = int(ceil(4 * n.toFloat * ln(4 * n.toFloat)))
proc ramanujanPrime(pi: PrimeCounter; n: int): int =
if n == 1: return 2
var max = ramanujanMax(n)
if (max and 1) == 1: dec max
for i in countdown(max, 2, 2):
if pi[i] - pi[i div 2] < n:
return i + 1
let pi = initPrimeCounter(1 + ramanujanMax(100_000))
for n in 1..100:
stdout.write ($ramanujanPrime(pi, n)).align(4), if n mod 20 == 0: '\n' else: ' '
echo "\nThe 1000th Ramanujan prime is ", ramanujanPrime(pi, 1_000) echo "The 10_000th Ramanujan prime is ", ramanujanPrime(pi, 10_000) echo "The 100_000th Ramanujan prime is ", ramanujanPrime(pi, 100_000)
echo "\nElapsed time: ", (now() - t0).inMilliseconds, " ms"</lang>
- Output:
2 11 17 29 41 47 59 67 71 97 101 107 127 149 151 167 179 181 227 229 233 239 241 263 269 281 307 311 347 349 367 373 401 409 419 431 433 439 461 487 491 503 569 571 587 593 599 601 607 641 643 647 653 659 677 719 727 739 751 769 809 821 823 827 853 857 881 937 941 947 967 983 1009 1019 1021 1031 1049 1051 1061 1063 1087 1091 1097 1103 1151 1163 1187 1217 1229 1249 1277 1289 1297 1301 1367 1373 1423 1427 1429 1439 The 1000th Ramanujan prime is 19403 The 10_000th Ramanujan prime is 242057 The 100_000th Ramanujan prime is 2916539 Elapsed time: 99 ms
Perl
<lang perl>use strict; use warnings; use ntheory 'primes';
sub count {
my($n,$p) = @_;
my $c = -1;
do { $c++ } until $$p[$c] > $n;
return $c;
}
my(@rp,@mem); my $primes = primes( 100_000_000 );
sub r_prime {
my $n = shift;
for my $x ( reverse 1 .. int 4*$n * log(4*$n) / log 2 ) {
my $y = int $x / 2;
return 1 + $x if ($mem[$x] //= count($x,$primes)) - ($mem[$y] //= count($y,$primes)) < $n
}
}
push @rp, r_prime($_) for 1..100; print "First 100:\n" . (sprintf "@{['%5d' x 100]}", @rp) =~ s/(.{100})/$1\n/gr;
print "\n\n 1000th: " . r_prime( 1000) . "\n"; print "\n10000th: " . r_prime(10000) . "\n"; # faster with 'ntheory' function 'ramanujan_primes'</lang>
- Output:
First 100:
2 11 17 29 41 47 59 67 71 97 101 107 127 149 151 167 179 181 227 229
233 239 241 263 269 281 307 311 347 349 367 373 401 409 419 431 433 439 461 487
491 503 569 571 587 593 599 601 607 641 643 647 653 659 677 719 727 739 751 769
809 821 823 827 853 857 881 937 941 947 967 983 1009 1019 1021 1031 1049 1051 1061 1063
1087 1091 1097 1103 1151 1163 1187 1217 1229 1249 1277 1289 1297 1301 1367 1373 1423 1427 1429 1439
1000th: 19403
10000th: 242057
Phix
You can run this online here.
with javascript_semantics sequence pi = {} procedure primeCounter(integer limit) pi = repeat(1,limit) if limit > 1 then pi[1] = 0 for i=4 to limit by 2 do pi[i] = 0 end for integer p = 3, sq = 9 while sq<=limit do if pi[p]!=0 then for q=sq to limit by p*2 do pi[q] = 0 end for end if sq += (p + 1)*4 p += 2 end while atom total = 0 for i=2 to limit do total += pi[i] pi[i] = total end for end if end procedure function ramanujanMax(integer n) return floor(4*n*log(4*n)) end function function ramanujanPrime(integer n) if n=1 then return 2 end if integer maxposs = ramanujanMax(n) for i=maxposs-odd(maxposs) to 1 by -2 do if pi[i]-pi[floor(i/2)] < n then return i + 1 end if end for return 0 end function atom t0 = time() integer lim = iff(platform()=JS?5:6) primeCounter(ramanujanMax(power(10,lim))) sequence r = apply(tagset(100),ramanujanPrime) printf(1,"%s\n",join_by(apply(true,sprintf,{{"%5d"},r}),1,20,"")) for p=3 to lim do integer n = power(10,p) printf(1,"The %,dth Ramanujan prime is %,d\n", {n,ramanujanPrime(n)}) end for
- Output:
2 11 17 29 41 47 59 67 71 97 101 107 127 149 151 167 179 181 227 229 233 239 241 263 269 281 307 311 347 349 367 373 401 409 419 431 433 439 461 487 491 503 569 571 587 593 599 601 607 641 643 647 653 659 677 719 727 739 751 769 809 821 823 827 853 857 881 937 941 947 967 983 1009 1019 1021 1031 1049 1051 1061 1063 1087 1091 1097 1103 1151 1163 1187 1217 1229 1249 1277 1289 1297 1301 1367 1373 1423 1427 1429 1439 The 1,000th Ramanujan prime is 19,403 The 10,000th Ramanujan prime is 242,057 The 100,000th Ramanujan prime is 2,916,539 The 1,000,000th Ramanujan prime is 34,072,993 "2.7s"
The last line is omitted under pwa/p2js since the primeCounter array is too much for Javascript to handle.
Raku
All timings are purely informational. Will vary by system specs and load.
Pure Raku
<lang perl6>use Math::Primesieve; use Lingua::EN::Numbers;
my $primes = Math::Primesieve.new;
my @mem;
sub ramanujan-prime (\n) {
1 + (1..(4×n × (4×n).log / 2.log).floor).first: :end, -> \x {
my \y = x div 2;
((@mem[x] //= $primes.count(x)) - (@mem[y] //= $primes.count(y))) < n
}
}
say 'First 100:'; say (1..100).map( &ramanujan-prime ).batch(10)».&comma».fmt("%6s").join: "\n"; say "\n 1,000th: { comma 1000.&ramanujan-prime }"; say "10,000th: { comma 10000.&ramanujan-prime }"; say (now - INIT now).fmt('%.3f') ~ ' seconds';</lang>
- Output:
First 100:
2 11 17 29 41 47 59 67 71 97
101 107 127 149 151 167 179 181 227 229
233 239 241 263 269 281 307 311 347 349
367 373 401 409 419 431 433 439 461 487
491 503 569 571 587 593 599 601 607 641
643 647 653 659 677 719 727 739 751 769
809 821 823 827 853 857 881 937 941 947
967 983 1,009 1,019 1,021 1,031 1,049 1,051 1,061 1,063
1,087 1,091 1,097 1,103 1,151 1,163 1,187 1,217 1,229 1,249
1,277 1,289 1,297 1,301 1,367 1,373 1,423 1,427 1,429 1,439
1,000th: 19,403
10,000th: 242,057
18.405 seconds
ntheory library
<lang perl6>use ntheory:from<Perl5> <ramanujan_primes nth_ramanujan_prime>; use Lingua::EN::Numbers;
say 'First 100:'; say ramanujan_primes( nth_ramanujan_prime(100) ).batch(10)».&comma».fmt("%6s").join: "\n";
for (2..12).map: {exp $_, 10} -> $limit {
say "\n{tc ordinal $limit}: { comma nth_ramanujan_prime($limit) }";
}
say (now - INIT now).fmt('%.3f') ~ ' seconds';</lang>
- Output:
First 100:
2 11 17 29 41 47 59 67 71 97
101 107 127 149 151 167 179 181 227 229
233 239 241 263 269 281 307 311 347 349
367 373 401 409 419 431 433 439 461 487
491 503 569 571 587 593 599 601 607 641
643 647 653 659 677 719 727 739 751 769
809 821 823 827 853 857 881 937 941 947
967 983 1,009 1,019 1,021 1,031 1,049 1,051 1,061 1,063
1,087 1,091 1,097 1,103 1,151 1,163 1,187 1,217 1,229 1,249
1,277 1,289 1,297 1,301 1,367 1,373 1,423 1,427 1,429 1,439
One hundredth: 1,439
One thousandth: 19,403
Ten thousandth: 242,057
One hundred thousandth: 2,916,539
One millionth: 34,072,993
Ten millionth: 389,433,437
One hundred millionth: 4,378,259,731
One billionth: 48,597,112,639
Ten billionth: 533,902,884,973
One hundred billionth: 5,816,713,968,619
One trillionth: 62,929,891,461,461
15.572 seconds
Wren
This takes about 1.1 seconds to find the 100,000th Ramanujan prime on my machine. The millionth takes 13.2 seconds.
<lang ecmascript>import "/trait" for Stepped
import "/seq" for Lst
import "/fmt" for Fmt
var count
var primeCounter = Fn.new { |limit|
count = List.filled(limit, 1)
if (limit > 0) count[0] = 0
if (limit > 1) count[1] = 0
for (i in Stepped.new(4...limit, 2)) count[i] = 0
var p = 3
var sq = 9
while (sq < limit) {
if (count[p] != 0) {
var q = sq
while (q < limit) {
count[q] = 0
q = q + p * 2
}
}
sq = sq + (p + 1) * 4
p = p + 2
}
var sum = 0
for (i in 0...limit) {
sum = sum + count[i]
count[i] = sum
}
}
var primeCount = Fn.new { |n| (n < 1) ? 0 : count[n] }
var ramanujanMax = Fn.new { |n| (4 * n * (4*n).log).ceil }
var ramanujanPrime = Fn.new { |n|
if (n == 1) return 2
for (i in ramanujanMax.call(n)..2*n) {
if (i % 2 == 1) continue
if (primeCount.call(i) - primeCount.call((i/2).floor) < n) return i + 1
}
return 0
}
primeCounter.call(1 + ramanujanMax.call(1e6)) System.print("The first 100 Ramanujan primes are:") var rams = (1..100).map { |n| ramanujanPrime.call(n) }.toList for (chunk in Lst.chunks(rams, 10)) Fmt.print("$,5d", chunk)
Fmt.print("\nThe 1,000th Ramanujan prime is $,6d", ramanujanPrime.call(1000))
Fmt.print("\nThe 10,000th Ramanujan prime is $,7d", ramanujanPrime.call(10000))
Fmt.print("\nThe 100,000th Ramanujan prime is $,9d", ramanujanPrime.call(100000))
Fmt.print("\nThe 1,000,000th Ramanujan prime is $,10d", ramanujanPrime.call(1000000))</lang>
- Output:
The first 100 Ramanujan primes are:
2 11 17 29 41 47 59 67 71 97
101 107 127 149 151 167 179 181 227 229
233 239 241 263 269 281 307 311 347 349
367 373 401 409 419 431 433 439 461 487
491 503 569 571 587 593 599 601 607 641
643 647 653 659 677 719 727 739 751 769
809 821 823 827 853 857 881 937 941 947
967 983 1,009 1,019 1,021 1,031 1,049 1,051 1,061 1,063
1,087 1,091 1,097 1,103 1,151 1,163 1,187 1,217 1,229 1,249
1,277 1,289 1,297 1,301 1,367 1,373 1,423 1,427 1,429 1,439
The 1,000th Ramanujan prime is 19,403
The 10,000th Ramanujan prime is 242,057
The 100,000th Ramanujan prime is 2,916,539
The 1,000,000th Ramanujan prime is 34,072,993