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Prove the Sieve of Eratosthenes algorithm has runtime complexity ≤ O(n log log n).
Sieve
Input: an integer n > 1.
Let A be an array of Boolean values, indexed by integers 2 to n, initially all set to true.
for i = 2,3,4,..., not exceeding √n:
if A[i] is true:
for j = i2,i2 + i,i2 +2i,i2 +3i,..., not exceeding n:
A[j] := false.
Output: all i such that A[i] is true.
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