How many factors of $2n^2$ are less than or equal to $n$?
I know that the number of factors of $n^2$ less than $n$ is half the number of factors of $n^2$ (each factor $< n$ corresponds with one greater than $n$), but $2n^2$ is a whole different case, it seems. Is there any way to find an expression for this? And if not, is there an algorithm for it? I have looked into both combinatorics and prime factorization, but have arrived at dead-ends.
Factors of $2n^2 \leq n$
number theoryproject euler
Best Answer
This is a very interesting question. Assume $n=2^{a}(2k+1)$ for some integer $a$ and $k$. Let $f(x)=$numbers of positive divisors of the integer $x$ . Since factors of $2n^2\leq n$ we need number of factors of $2^{2a+1}(2k+1)^2$. Thus we have $f(2^a(2k+1))+c_{a}$.Where $c_{a}$ is the error factor has a small factor of bound, which needs to be deterministic. Although it's a rough idea, I am not finding the bound but for hints, you can try small cases. However let $g(x)=$greatest integer lesser equal to x then, $$c_{a}\leq f(g(2^{\frac{2a+1}{2}}(2k+1)))-f(2^a(2k+1))$$.Where we know $f$ is famous divisor function or $\tau$ function and $g$ is the floor function. Use this link for bound, Lower bound for the sum of divisor function