I improved the segmented sieve of erastothenes , my algorithm doesnt repeat the multiples of primes using the equation $p^{2}_{n}p_{j}+2p_{n}p_{j} \times c =N$ wich shows when at least two multiples of primes are equal . the number of operations the algo does is the number of the odd composite numbers included in a given interval $I=[a,b]$ + the sum of primes from $ (p_{1}=3) to ((p_{n}\leq \sqrt{b} )-p) $.
For large $b$ , $b-a \approx N$ the total number of operations is then :
$(\frac{N}{2} -\frac{N}{ln(N)} )+( \frac{1}{2} (\frac{N}{ln(N)} +\frac{\sqrt{N} }{ln(\sqrt{N} } )-\sqrt{N} )$.
I dont know how to calculate the algo complexity . can you help please?
If you have an explicit expression for the number of operations, the computational complexity (in the usual big O notation), is the term that eventually dominates the others. If there are multiple positive maximal terms that do not diminish with respect to each other, choose any of them. So your expression suggests a computational complexity of O(N) (drop constant multiples as well, for the sake of conciseness).
The usual implementation of this sieve has complexity O(n ln(n) ln(n)), but it is known that it can be implemented with complexity O(n). However, such mild improvements in computational complexity are often not practical, as the constant can grow considerably, and really intense sieving is going to be done more efficiently with the sieve of Atkin anyway. I should also make the general note that memory usage and access patterns are not represented in the computational complexity, but these considerations are often even more important that instruction count (a main memory hit takes an amount of time on the order of hundreds of clock cycles).
It is not clear to me what modification you made, as it sounds like you are saying that you have found a away to avoid all redundant removals from the prime list. It would be the greatest discovery in the history of number theory if you found an efficient way to avoid redundant eliminations from the prime list, so if it seems you have done this, I suggest you re-examine your algorithm.