The sieve of Sundaram: Difference between revisions

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As a check, the 1,000,000 Sundaram prime would again have been 15,485,867
As a check, the 1,000,000 Sundaram prime would again have been 15,485,867
</pre>
</pre>

=={{header|Haskell}}==
<lang haskell>import Data.List (intercalate, transpose)
import Data.List.Split (chunksOf)
import qualified Data.Set as S
import Text.Printf (printf)

--------------------- SUNDARAM PRIMES --------------------

sundaram :: Integral a => a -> [a]
sundaram n =
[ succ (2 * x)
| x <- [1 .. m],
x `S.notMember` excluded
]
where
m = div (pred n) 2
excluded =
S.fromList
[ 2 * i * j + i + j
| let fm = fromIntegral m,
i <- [1 .. floor (sqrt (fm / 2))],
let fi = fromIntegral i,
j <- [i .. floor ((fm - fi) / succ (2 * fi))]
]

nSundaramPrimes ::
(Integral a1, RealFrac a2, Floating a2) => a2 -> [a1]
nSundaramPrimes n =
sundaram $ floor $ (2.4 * n * log n) / 2



--------------------------- TEST -------------------------
main :: IO ()
main = do
putStrLn "First 100 Sundaram primes (starting at 3):\n"
(putStrLn . table " " . chunksOf 10) $
show <$> nSundaramPrimes 100

table :: String -> [[String]] -> String
table gap rows =
let ws = maximum . fmap length <$> transpose rows
pw = printf . flip intercalate ["%", "s"] . show
in unlines $ intercalate gap . zipWith pw ws <$> rows</lang>
{{Out}}
<pre>First 100 Sundaram primes (starting at 3):

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</pre>


=={{header|Julia}}==
=={{header|Julia}}==

Revision as of 16:50, 9 August 2021

The sieve of Sundaram is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

The sieve of Eratosthenes: you've been there; done that; have the T-shirt. The sieve of Eratosthenes was ancient history when Euclid was a schoolboy. You are ready for something less than 3000 years old. You are ready for The sieve of Sundaram.

Starting with the ordered set of +ve integers, mark every third starting at 4 (4;7;10...).

Step through the set and if the value is not marked output 2*n+1. So from 1 to 4 output 3 5 7.

4 is marked so skip for 5 and 6 output 11 and 13.

7 is marked, so no output but now also mark every fifth starting at 12 (12;17;22...)

as per to 10 and now mark every seventh starting at 17 (17;24;31....)

as per for every further third element (13;16;19...) mark every (9th;11th;13th;...) element.

The output will be the ordered set of odd primes.

Using your function find and output the first 100 and the millionth Sundaram prime.

The faithless amongst you may compare the results with those generated by The sieve of Eratosthenes.

References

ALGOL 68

Translation of: Nim

To run this with Algol 68G, you will need to increase the heap size by specifying e.g. -heap=64M on the command line. <lang algol68>BEGIN # sieve of Sundaram # 

   INT n = 8 000 000;
   INT none = 0, mark1 = 1, mark2 = 2;
   [ 1 : n ]INT mark;
   FOR i FROM LWB mark TO UPB mark DO mark[ i ] := none  OD;
   FOR i FROM   4 BY 3 TO UPB mark DO mark[ i ] := mark1 OD;

   INT            count := 0; # Count of primes.          #
   [ 1 : 100 ]INT list100;    # First 100 primes.         #
   INT            last  := 0; # Millionth prime.          #
   INT            step  := 5; # Current step for marking. #

   FOR i TO n WHILE last = 0 DO
       IF mark[ i ] = none THEN # Add/count a new odd prime.  #
           count +:= 1;
           IF   count <= 100 THEN
               list100[ count ] := 2 * i + 1
           ELIF count = 1 000 000 THEN
               last := 2 * i + 1
           FI
       ELIF mark[ i ] = mark1 THEN # Mark new numbers using current step. #
           IF i > 4 THEN
               FOR k FROM i + step BY step TO n DO
                   IF mark[ k ] = none THEN mark[ k ] := mark2 FI
               OD;
               step +:= 2
           FI
     # ELSE must be mark2 - Ignore this number. #
       FI
   OD; 

   print( ( "First 100 Sundaram primes:", newline ) );
   FOR i FROM LWB list100 TO UPB list100 DO
       print( ( whole( list100[ i ], -3 ) ) );
       IF i MOD 10 = 0 THEN print( ( newline ) ) ELSE print( ( " " ) ) FI
   OD;
   print( ( newline ) );
   IF last = 0 THEN
       print( ( "Not enough values in sieve. Found only ", whole( count, 0 ), newline ) )
   ELSE
       print( ( "The millionth Sundaram prime is ", whole( last, 0 ), newline ) )
   FI

END</lang>

Output:
  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

The millionth Sundaram prime is 15485867

C#

Generating prime numbers during sieve creation gives a performance boost over completing the sieve and then scanning the sieve for output. There are, of course, a few at the end to scan out.
Heh, nope. It's faster to do the sieving first, then the generation afterwards. <lang csharp>using System; using System.Collections.Generic; using System.Linq; using static System.Console;

class Program { 

   static string fmt(int[] a)
   {
       var sb = new System.Text.StringBuilder();
       for (int i = 0; i < a.Length; i++)
           sb.Append(string.Format("{0,5}{1}",
             a[i], i % 10 == 9 ? "\n" : " "));
       return sb.ToString();
   }

   static void Main(string[] args)
   {
       var sw = System.Diagnostics.Stopwatch.StartNew();
       var pr = PG.Sundaram(15_500_000).Take(1_000_000).ToArray();
       sw.Stop();
       Write("The first 100 odd prime numbers:\n{0}\n",
         fmt(pr.Take(100).ToArray()));
       Write("The millionth odd prime number: {0}", pr.Last());
       Write("\n{0} ms", sw.Elapsed.TotalMilliseconds);
   }

}

class PG { 

   public static IEnumerable<int> Sundaram(int n)
   {
       // yield return 2;
       int i = 1, k = (n + 1) >> 1, t = 1, v = 1, d = 1, s = 1;
       var comps = new bool[k + 1];
       for (; t < k; t = ((++i + (s += d += 2)) << 1) - d - 2)
           while ((t += d + 2) < k)
               comps[t] = true;
       for (; v < k; v++)
           if (!comps[v])
               yield return (v << 1) + 1;
   }

}</lang>

Output:

Under 1/5 1/8 of a second @ Tio.run 

The first 100 odd prime numbers:
    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

The millionth odd prime number: 15485867
124.8262 ms

P.S. for those (possibly faithless) who wish to have a conventional prime number generator, one can uncomment the yield return 2 line at the top of the function.

F#

<lang fsharp> // The sieve of Sundaram. Nigel Galloway: August 7th., 2021 let sPrimes()= 

 let sSieve=System.Collections.Generic.Dictionary<int,(unit -> int) list>()
 let rec fN g=match g with h::t->(let n=h() in if sSieve.ContainsKey n then sSieve.[n]<-h::sSieve.[n] else sSieve.Add(n,[h])); fN t|_->()
 let     fI n=if sSieve.ContainsKey n then fN sSieve.[n]; sSieve.Remove n|>ignore; None else Some(2*n+1) 
 let     fG n g=let mutable n=n in (fun()->n<-n+g; n)
 let     fE n g=if not(sSieve.ContainsKey n) then sSieve.Add(n,[fG n g]) else sSieve.[n]<-(fG n g)::sSieve.[g] 
 let     fL    =let mutable n,g=4,3 in (fun()->n<-n+3; g<-g+2; fE (n+g) g; n)
 sSieve.Add(4,[fL]); Seq.initInfinite((+)1)|>Seq.choose fI

sPrimes()|>Seq.take 100|>Seq.iter(printf "%d "); printfn "" printfn "The millionth Sundaram prime is %d" (Seq.item 999999 (sPrimes())) </lang>

Output:
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
The millionth Sundaram prime is 15485867

Go

Translation of: Wren
Library: Go-rcu

<lang go>package main

import ( 

   "fmt"
   "math"
   "rcu"
   "time"

)

func sos(n int) []int { 

   if n < 3 {
       return []int{}
   }
   var primes []int
   k := (n-3)/2 + 1
   marked := make([]bool, k) // all false by default
   limit := (int(math.Sqrt(float64(n)))-3)/2 + 1
   for i := 0; i < limit; i++ {
       p := 2*i + 3
       s := (p*p - 3) / 2
       for j := s; j < k; j += p {
           marked[j] = true
       }
   }
   for i := 0; i < k; i++ {
       if !marked[i] {
           primes = append(primes, 2*i+3)
       }
   }
   return primes

}

// odds only func soe(n int) []int { 

   if n < 3 {
       return []int{}
   }
   var primes []int
   k := (n-3)/2 + 1
   marked := make([]bool, k) // all false by default
   limit := (int(math.Sqrt(float64(n)))-3)/2 + 1
   for i := 0; i < limit; i++ {
       if !marked[i] {
           p := 2*i + 3
           s := (p*p - 3) / 2
           for j := s; j < k; j += p {
               marked[j] = true
           }
       }
   }
   for i := 0; i < k; i++ {
       if !marked[i] {
           primes = append(primes, 2*i+3)
       }
   }
   return primes

}

func main() { 

   const limit = int(16e6) // say
   start := time.Now()
   primes := sos(limit)
   elapsed := int(time.Since(start).Milliseconds())
   climit := rcu.Commatize(limit)
   celapsed := rcu.Commatize(elapsed)
   million := rcu.Commatize(1e6)
   millionth := rcu.Commatize(primes[1e6-1])
   fmt.Printf("Using the Sieve of Sundaram generated primes up to %s in %s ms.\n\n", climit, celapsed)
   fmt.Println("First 100 odd primes generated by the Sieve of Sundaram:")
   for i, p := range primes[0:100] {
       fmt.Printf("%3d ", p)
       if (i+1)%10 == 0 {
           fmt.Println()
       }
   }
   fmt.Printf("\nThe %s Sundaram prime is %s\n", million, millionth)

   start = time.Now()
   primes = soe(limit)
   elapsed = int(time.Since(start).Milliseconds())
   celapsed = rcu.Commatize(elapsed)
   millionth = rcu.Commatize(primes[1e6-1])
   fmt.Printf("\nUsing the Sieve of Eratosthenes would have generated them in %s ms.\n", celapsed)
   fmt.Printf("\nAs a check, the %s Sundaram prime would again have been %s\n", million, millionth)

}</lang>

Output:
Using the Sieve of Sundaram generated primes up to 16,000,000 in 62 ms.

First 100 odd primes generated by the Sieve of Sundaram:
  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 

The 1,000,000 Sundaram prime is 15,485,867

Using the Sieve of Eratosthenes would have generated them in 33 ms.

As a check, the 1,000,000 Sundaram prime would again have been 15,485,867

Haskell

<lang haskell>import Data.List (intercalate, transpose) import Data.List.Split (chunksOf) import qualified Data.Set as S import Text.Printf (printf)

SUNDARAM PRIMES --------------------

sundaram :: Integral a => a -> [a] sundaram n = 

 [ succ (2 * x)
   | x <- [1 .. m],
     x `S.notMember` excluded
 ]
 where
   m = div (pred n) 2
   excluded =
     S.fromList
       [ 2 * i * j + i + j
         | let fm = fromIntegral m,
           i <- [1 .. floor (sqrt (fm / 2))],
           let fi = fromIntegral i,
           j <- [i .. floor ((fm - fi) / succ (2 * fi))]
       ]

nSundaramPrimes :: 

 (Integral a1, RealFrac a2, Floating a2) => a2 -> [a1]

nSundaramPrimes n = 

 sundaram $ floor $ (2.4 * n * log n) / 2


TEST -------------------------

main :: IO () main = do 

 putStrLn "First 100 Sundaram primes (starting at 3):\n"
 (putStrLn . table " " . chunksOf 10) $
   show <$> nSundaramPrimes 100

table :: String -> String -> String table gap rows = 

 let ws = maximum . fmap length <$> transpose rows
     pw = printf . flip intercalate ["%", "s"] . show
  in unlines $ intercalate gap . zipWith pw ws <$> rows</lang>
Output:
First 100 Sundaram primes (starting at 3):

  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

Julia

Translation of: Python

<lang julia> """ The sieve of Sundaram is a simple deterministic algorithm for finding all the prime numbers up to a specified integer. This function is modified from the Python example Wikipedia entry wiki/Sieve_of_Sundaram, to give primes to the nth prime rather than the Wikipedia function that gives primes less than n. """ function sieve_of_Sundaram(nth, print_all=true) 

   @assert nth > 0
   k = Int(round(1.2 * nth * log(nth)))  # nth prime is at about n * log(n)
   integers_list = trues(k)
   for i in 1:k
       j = i
       while i + j + 2 * i * j < k
           integers_list[i + j + 2 * i * j + 1] = false
           j += 1
       end
   end
   pcount = 0
   for i in 1:k + 1
       if integers_list[i + 1]
           pcount += 1
           if print_all
               print(lpad(2 * i + 1, 4), pcount % 10 == 0 ? "\n" : "")
           end
           if pcount == nth
               println("\nSundaram primes start with 3. The $(nth)th Sundaram prime is $(2 * i + 1).")
               break
           end
       end
   end

end

sieve_of_Sundaram(100) @time sieve_of_Sundaram(1000000, false)

println("\nChecking:") using Primes; @show count(primesmask(15485867)) @time count(primesmask(15485867))

</lang>

Output:
  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

Sundaram primes start with 3. The 100th Sundaram prime is 547.

Sundaram primes start with 3. The 1000000th Sundaram prime is 15485867.
  0.127168 seconds (19 allocations: 1.977 MiB)

Checking:
count(primesmask(15485867)) = 1000001
  0.022684 seconds (6 allocations: 5.785 MiB)

Nim

<lang Nim>import strutils

const N = 8_000_000

type Mark {.pure.} = enum None, Mark1, Mark2

var mark: array[1..N, Mark] for n in countup(4, N, 3): mark[n] = Mark1


var count = 0 # Count of primes. var list100: seq[int] # First 100 primes. var last = 0 # Millionth prime. var step = 5 # Current step for marking.

for n in 1..N: 

 case mark[n]
 of None:
   # Add/count a new odd prime.
   inc count
   if count <= 100:
     list100.add 2 * n + 1
   elif count == 1_000_000:
     last = 2 * n + 1
     break
 of Mark1:
   # Mark new numbers using current step.
   if n > 4:
     for k in countup(n + step, N, step):
       if mark[k] == None: mark[k] = Mark2
     inc step, 2
 of Mark2:
   # Ignore this number.
   discard


echo "First 100 Sundaram primes:" for i, n in list100: 

 stdout.write ($n).align(3), if (i + 1) mod 10 == 0: '\n' else: ' '

echo() if last == 0: 

 quit "Not enough values in sieve. Found only $#.".format(count), QuitFailure

echo "The millionth Sundaram prime is ", ($last).insertSep()</lang>

Output:
First 100 Sundaram primes:
  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

The millionth Sundaram prime is 15_485_867


Phix

with javascript_semantics
function sos(integer n)
    if n<3 then return {} end if
    integer k = floor((n-3)/2) + 1
    sequence primes = {},
             marked = repeat(false,k)
    for i=1 to floor((floor(sqrt(n))-3)/2) + 1 do
        integer p = 2*i + 1,
                s = (p*p - 1) / 2
        for j=s to k by p do
            marked[j] = true
        end for
    end for
    for i=1 to k do
        if not marked[i] then
            primes = append(primes, 2*i+1)
        end if
    end for
    return primes
end function

sequence s = sos(16_000_000)
printf(1,"The first 100 odd prime numbers:\n%s\n",{join_by(apply(true,sprintf,{{"%3d"},s[1..100]}),1,10)})
printf(1,"The millionth odd prime number: %,d\n",{s[1_000_000]})
Output:
The first 100 odd prime numbers:
  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

The millionth odd prime number: 15,485,867

Python

Python :: Procedural

<lang python>from numpy import log

def sieve_of_Sundaram(nth, print_all=True): 

   """
   The sieve of Sundaram is a simple deterministic algorithm for finding all the
   prime numbers up to a specified integer. This function is modified from the
   Wikipedia entry wiki/Sieve_of_Sundaram, to give primes to their nth rather
   than the Wikipedia function that gives primes less than n.
   """
   assert nth > 0, "nth must be a positive integer"
   k = int((2.4 * nth * log(nth)) // 2)  # nth prime is at about n * log(n)
   integers_list = [True] * k
   for i in range(1, k):
       j = i
       while i + j + 2 * i * j < k:
           integers_list[i + j + 2 * i * j] = False
           j += 1
   pcount = 0
   for i in range(1, k + 1):
       if integers_list[i]:
           pcount += 1
           if print_all:
               print(f"{2 * i + 1:4}", end=' ')
               if pcount % 10 == 0:
                   print()

           if pcount == nth:
               print(f"\nSundaram primes start with 3. The {nth}th Sundaram prime is {2 * i + 1}.\n")
               break


sieve_of_Sundaram(100, True)

sieve_of_Sundaram(1000000, False)

</lang>

Output:
   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 

Sundaram primes start with 3. The 100th Sundaram prime is 547.

Sundaram primes start with 3. The 1000000th Sundaram prime is 15485867.

Python :: Functional

Composing functionally, and obtaining slightly better performance by defining a set (rather than list) of exclusions. <lang python>Sieve of Sundaram

from math import floor, log, sqrt from itertools import islice


  1. sundaram :: Int -> [Int]

def sundaram(n): 

   Sundaram prime numbers up to n
   m = (n - 1) // 2
   exclusions = {
       2 * i * j + i + j
       for i in range(1, 1 + floor(sqrt(m / 2)))
       for j in range(
           i, 1 + floor((m - i) / (1 + (2 * i)))
       )
   }
   return [
       1 + (2 * x) for x in range(1, 1 + m)
       if not x in exclusions
   ]


  1. nPrimesBySundaram :: Int -> [Int]

def nPrimesBySundaram(n): 

   First n primes, by sieve of Sundaram.
   
   return list(islice(
       sundaram(
           # Probable limit
           int((2.4 * n * log(n)) // 2)
       ),
       int(n)
   ))


  1. ------------------------- TEST -------------------------
  2. main :: IO ()

def main(): 

   First 100 Sundaram primes, 
      and millionth Sundaram prime.
   
   print("First hundred Sundaram primes, starting at 3:\n")
   print(table(10)([
       str(s) for s in nPrimesBySundaram(100)
   ]))
   print("\n\nMillionth Sundaram prime, starting at 3:")
   print(
       f'\n\t{nPrimesBySundaram(1E6)[-1]}'
   )


  1. ----------------------- GENERIC ------------------------
  1. chunksOf :: Int -> [a] -> a

def chunksOf(n): 

   A series of lists of length n, subdividing the
      contents of xs. Where the length of xs is not evenly
      divisible, the final list will be shorter than n.
   
   def go(xs):
       return (
           xs[i:n + i] for i in range(0, len(xs), n)
       ) if 0 < n else None
   return go


  1. table :: Int -> [String] -> String

def table(n): 

   A list of strings formatted as
      right-justified rows of n columns.
   
   def go(xs):
       w = len(xs[-1])
       return '\n'.join(
           ' '.join(row) for row in chunksOf(n)([
               s.rjust(w, ' ') for s in xs
           ])
       )
   return go


  1. MAIN ---

if __name__ == '__main__': 

   main()</lang>
Output:
First hundred Sundaram primes, starting at 3:

  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


Millionth Sundaram prime, starting at 3:

    15485867

REXX

For the calculation of the 1,000,000th Sundaram prime,   it requires a 64-bit version of REXX. <lang rexx>/*REXX program finds & displays N Sundaram primes, or displays the Nth Sundaram prime.*/ parse arg n cols . /*get optional number of primes to find*/ if n== | n=="," then n= 100 /*Not specified? Then assume default.*/ if cols== | cols=="," then cols= 10 /* " " " " " */ @.= .; lim= 16 * n /*default value for array; filter limit*/ 

      do    j=1  for n;   do k=1  for n  until _>lim;  _= j + k + 2*j*k;  @._=
                          end   /*k*/
      end      /*j*/

w= 10 /*width of a number in any column. */ 

                                    title= 'a list of '   commas(N)   " Sundaram primes"

if cols>0 then say ' index │'center(title, 1 + cols*(w+1) ) if cols>0 then say '───────┼'center("" , 1 + cols*(w+1), '─')

  1. = 0; idx= 1 /*initialize # of Sundaram primes & IDX*/

$= /*a list of Sundaram primes (so far). */ 

      do j=1  until #==n                        /*display the output (if cols > 0).    */
      if @.j\==.  then iterate                  /*Is the number not prime?  Then skip. */
      #= # + 1                                  /*bump number of Sundaram primes found.*/
      a= j                                      /*save J for calculating the Nth prime.*/
      if cols<=0  then iterate                  /*Build the list  (to be shown later)? */
      c= commas(j + j + 1)                      /*maybe add commas to  Sundaram  prime.*/
      $= $  right(c, max(w, length(c) ) )       /*add Sundaram prime──►list, allow big#*/
      if #//cols\==0  then iterate              /*have we populated a line of output?  */
      say center(idx, 7)'│'  substr($, 2);  $=  /*display what we have so far  (cols). */
      idx= idx + cols                           /*bump the  index  count for the output*/
      end   /*j*/

if $\== then say center(idx, 7)"│" substr($, 2) /*possible display residual output.*/ if cols>0 then say '───────┴'center("" , 1 + cols*(w+1), '─') say say 'found ' commas(#) " Sundaram primes, and the last Sundaram prime is " commas(a+a+1) exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ?</lang>

output   when using the default inputs:
 index │                                        a list of  100  Sundaram primes
───────┼───────────────────────────────────────────────────────────────────────────────────────────────────────────────
   1   │          3          5          7         11         13         17         19         23         29         31
  11   │         37         41         43         47         53         59         61         67         71         73
  21   │         79         83         89         97        101        103        107        109        113        127
  31   │        131        137        139        149        151        157        163        167        173        179
  41   │        181        191        193        197        199        211        223        227        229        233
  51   │        239        241        251        257        263        269        271        277        281        283
  61   │        293        307        311        313        317        331        337        347        349        353
  71   │        359        367        373        379        383        389        397        401        409        419
  81   │        421        431        433        439        443        449        457        461        463        467
  91   │        479        487        491        499        503        509        521        523        541        547
───────┴───────────────────────────────────────────────────────────────────────────────────────────────────────────────

found  100  Sundaram primes, and the last Sundaram prime is  547
output   when using the inputs of:     1000000   0 
found  1,000,000  Sundaram primes, and the last Sundaram prime is  15,485,867

Wren

Library: Wren-fmt
Library: Wren-seq

I've worked here from the second (optimized) Python example in the Wikipedia article for SOS which allows an easy transition to an 'odds only' SOE for comparison. <lang ecmascript>import "/fmt" for Fmt import "/seq" for Lst

var sos = Fn.new { |n| 

   if (n < 3) return []
   var primes = []
   var k = ((n-3)/2).floor + 1
   var marked = List.filled(k, true)
   var limit = ((n.sqrt.floor - 3)/2).floor + 1
   limit = limit.max(0)
   for (i in 0...limit) {
       var p = 2*i + 3
       var s = ((p*p - 3)/2).floor
       var j = s
       while (j < k) {
           marked[j] = false
           j = j + p
       }
   }
   for (i in 0...k) {
       if (marked[i]) primes.add(2*i + 3)
   }
   return primes

}

// odds only var soe = Fn.new { |n| 

   if (n < 3) return []
   var primes = []
   var k = ((n-3)/2).floor + 1
   var marked = List.filled(k, true)
   var limit = ((n.sqrt.floor - 3)/2).floor + 1
   limit = limit.max(0)
   for (i in 0...limit) {
       if (marked[i]) {
           var p = 2*i + 3
           var s = ((p*p - 3)/2).floor
           var j = s
           while (j < k) {
               marked[j] = false
               j = j + p
           }
       }
   }
   for (i in 0...k) {
       if (marked[i]) primes.add(2*i + 3)
   }
   return primes

}

var limit = 16e6 // say var start = System.clock var primes = sos.call(limit) var elapsed = ((System.clock - start) * 1000).round Fmt.print("Using the Sieve of Sundaram generated primes up to $,d in $,d ms.\n", limit, elapsed) System.print("First 100 odd primes generated by the Sieve of Sundaram:") for (chunk in Lst.chunks(primes[0..99], 10)) Fmt.print("$3d", chunk) Fmt.print("\nThe $,d Sundaram prime is $,d", 1e6, primes[1e6-1])

start = System.clock primes = soe.call(limit) elapsed = ((System.clock - start) * 1000).round Fmt.print("\nUsing the Sieve of Eratosthenes would have generated them in $,d ms.", elapsed) Fmt.print("\nAs a check, the $,d Sundaram prime would again have been $,d", 1e6, primes[1e6-1])</lang>

Output:
Using the Sieve of Sundaram generated primes up to 16,000,000 in 1,232 ms.

First 100 odd primes generated by the Sieve of Sundaram:
  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

The 1,000,000 Sundaram prime is 15,485,867

Using the Sieve of Eratosthenes would have generated them in 797 ms.

As a check, the 1,000,000 Sundaram prime would again have been 15,485,867
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