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From here I found a fast method to make prime list, the python version works well, but my Mathematica version does not. Obviously ,169 is not a prime number. What's wrong with my program?

'''Python code'''
n = 200
sieve = np.ones(n / 3 + (n % 6 == 2), dtype = np.bool)
sieve[0] = False
for i in xrange(int(n**0.5)/3+1):
    if sieve[i]:
        k=(3 * i + 1) | 1
        sieve[      ((k*k)/3)      :: 2 * k] = False
        sieve[(k * k + 4 * k - 2 * k * (i & 1))/3 :: 2 * k] = False    
print ((3 * np.nonzero(sieve)[0] + 1) | 1)

(*Mathematica code*)
Clear["`*"];
n = 200;
p = ConstantArray[1, Quotient[n, 3] + Boole[Mod[n, 6] == 2]];
p[[1]] = 0;
Do[
  If[p[[i]] != 0, 
    k = BitOr[3 (i - 1) + 1, 1];
    p[[Quotient[k^2, 3] + 1 ;; -1 ;; 2 k]] = 0;
    p[[Quotient[(k^2 + 4 k - 2 k BitAnd[i - 1, 1]), 3] + 1 ;; -1 ;; 2 k]] = 0;],
  {i, 1, Floor[n^0.5]/3}];
res = BitOr[3 (Flatten@SparseArray[p]["NonzeroPositions"] - 1) + 1, 1];   
Pick[res, PrimeQ @ res, False]
Improve this question
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7
  • $\begingroup$ The simple way: Prime[Range[50]] gives the first 50 primes. $\endgroup$
    – bill s
    Aug 26, 2013 at 17:32
  • $\begingroup$ Might I suggest you set your "Notebook's Default Context" to "Unique to this Notebook"? It eliminates some of the need to run Clear["`*"];. $\endgroup$
    – rcollyer
    Aug 26, 2013 at 17:32
  • 1
    $\begingroup$ @bills,Prime@Range@PrimePi[10^7] // Length on my laptop takes about 3s, that method only need 0.4s. $\endgroup$
    – expression
    Aug 26, 2013 at 17:41
  • $\begingroup$ {i, 1, Floor[n^0.5]/3+2} $\endgroup$
    – Tobias Hagge
    Aug 27, 2013 at 2:27
  • $\begingroup$ explorer, I believe "code-review" is best reserved for critique of working code, though it doesn't say that in the tag wiki. I have replaced the tag with "broken-code" which I think is more accurate. $\endgroup$
    – Mr.Wizard
    Aug 27, 2013 at 6:43

1 Answer 1

Reset to default
1
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n = 400;
p = ConstantArray[1, Quotient[n, 3] + Boole[Mod[n, 6] == 2]];
p[[1]] = 0;
Do[
  If[p[[i]] != 0, k = BitOr[3 (i - 1) + 1, 1];
   x = Quotient[k^2, 3] + 1;
   y = Quotient[(k^2 + 4 k - 2 k BitAnd[i - 1, 1]), 3] + 1;
   If[x <= Length@p, p[[x ;; -1 ;; 2 k]] = 0];
   If[y <= Length@p, p[[y ;; -1 ;; 2 k]] = 0];],
  {i, 1, Floor[Sqrt[n]]/3 + 2}
  ];
res = BitOr[3 (Flatten@SparseArray[p]["NonzeroPositions"] - 1) + 1, 1]
Pick[res, PrimeQ@res, False]

(* {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} *)
(* {} *)
Improve this answer
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