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Autodesk Inventor 2010 x64 (64bit) + product key & Xforce keygen Autodesk Inventor 2010 x64 (64bit) + product key & Xforce keygen.Q:

Sieve of Eratosthenes — why should we expect interval $(2,k+1)$ to cover all k-smooth numbers?

In the Sieve of Eratosthenes, when we determine the sieve function, we use the logic that odd numbers can be divided by any odd number and even numbers are divided by 2, and so there will be an interval of $ (2,k+1)$ sifted from the number.
This is because any number $x$ can be written as $2
u +1$ and $
u$ is an integer. $
u \in (1,k+1)$ because of the constraints of $k$. $
u$ thus will be in this interval, which justifies using this sieve function.
My question is if we can’t have $k$ even, we get a smaller interval of $
u \in (1,k)$. I was wondering if there is any logic as to why we should expect only odd numbers to be k-smooth numbers.

A:

This is a trick that has been used many times. In your case,
$$
x \equiv 2
u +1 \text{ where }
u \in \Bbb Z
$$
So, you’re thinking of $x$ as a kind of compressed integer.
In some sense, it is. However, if I’m given a number with a particular prime factorisation $p^{2n
c6a93da74d

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