Sieve of Sundaram

Algorithm

Start with a list of the integers from 1 to n. From this list, remove all numbers of the form i + j + 2ij where:

The remaining numbers are doubled and incremented by one, giving a list of the odd prime numbers (i.e., all primes except 2) below 2n + 2.

The sieve of Sundaram sieves out the composite numbers just as sieve of Eratosthenes does, but even numbers are not considered; the work of "crossing out" the multiples of 2 is done by the final double-and-increment step. Whenever Eratosthenes' method would cross out k different multiples of a prime 2 i + 1 , Sundaram's method crosses out i + j ( 2 i + 1 ) for 1 ≤ j ≤ ⌊ k / 2 ⌋ .

Correctness

If we start with integers from 1 to n, the final list contains only odd integers from 3 to 2 n + 1 . From this final list, some odd integers have been excluded; we must show these are precisely the composite odd integers less than 2 n + 2 .

Let q be an odd integer of the form 2 k + 1 . Then, q is excluded if and only if k is of the form i + j + 2 i j , that is q = 2 ( i + j + 2 i j ) + 1 . Then we have:

So, an odd integer is excluded from the final list if and only if it has a factorization of the form ( 2 i + 1 ) ( 2 j + 1 ) — which is to say, if it has a non-trivial odd factor. Therefore the list must be composed of exactly the set of odd prime numbers less than or equal to 2 n + 2 .