sieve-of-eratosthenes

From Wikipedia: The complexity of the algorithm is O(n(logn)(loglogn)) bit operations. How do you arrive at that? That the complexity includes the loglogn term tells me that there is a sqrt(n) somewhere. Suppose I am running the sieve on the first 100 numbers ( n = 100 ), assuming that marking the numbers as composite takes constant time (array implementation), the number of times we use mark_composite() would be something like n/2 + n/3 + n/5 + n/7 + ... + n/97 = O(n^2) And to find the next prime number (for example to jump to 7 after crossing out all the numbers that are multiples of 5 ),

I wrote a prime number generator using Sieve of Eratosthenes and Python 3.1. The code runs correctly and gracefully at 0.32 seconds on ideone.com to generate prime numbers up to 1,000,000. # from bitstring import BitString def prime_numbers(limit=1000000): '''Prime number generator. Yields the series 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 ... using Sieve of Eratosthenes. ''' yield 2 sub_limit = int(limit**0.5) flags = [False, False] + [True] * (limit - 2) # flags = BitString(limit) # Step through all the odd numbers for i in range(3, limit, 2): if flags[i] is False: # if flags[i] is True: continue

How to write a Program to find n primes after a given number? e.g. first 10 primes after 100, or first 25 primes after 1000. Edited: below is what I tried. I am getting output that way, but can we do it without using any primality-testing function? #include<stdio.h> #include<conio.h> int isprime(int); main() { int count=0,i; for(i=100;1<2;i++) { if(isprime(i)) { printf("%d\n",i); count++; if(count==5) break; } } getch(); } int isprime(int i) { int c=0,n; for(n=1;n<=i/2;n++) { if(i%n==0) c++; } if(c==1) return 1; else return 0; } Sure. Read about the Sieve of Eratosthenes . Instead of checking