## Let $f(n)$ be the number of divisors of $2n^2$ that are no greater than n. For example, $f(15)=8$ because there are 8 such divisors: 1,2,3,5,6,9,10,15. Note that 18 is also a divisor of $2\times 15^2$ but it is not counted because it is greater than 15.

Let $\displaystyle F(N) = \sum_{n=1}^N f(n)$. You are given $F(15)=63$, and $F(1000)=15066$.

Find $F(10^{12})$.

### Computing $F(10^{12})$ directly is not practical because $N=10^{12}$ is a huge value, and so it will take a lot of computing power and time to calculate it. However, here are some insights that can help in solving this problem more efficiently:

Note that for any natural number $N$, the number of positive divisors $d$ that are less than or equal to the square root of $N$ is precisely $f(n)$ in our problem. This is because any divisor $d$ of $N$ pairs up with a corresponding divisor $N/d$, one of these must be less than or equal to $\sqrt N$. Therefore, the number of distinct divisors of $N$ that are no greater than $\sqrt N$ equals the number of divisors of $N$ that are no greater than $\sqrt N$. Hence the problem becomes, “what is the total number of divisors for all the numbers from $1$ to $N^2$, where $N = 10^{12}$?”

Knowing that the number of divisors of a non-square integer $N = prod_{i = 1}^{k} p_i^{e_i}$ equals $(e_1+1) * (e_2+1) * … * (e_k+1)$, and the number of divisors for a perfect square equals $(2*e_1+1) * (2*e_2+1) * … * (2*e_k+1)$, we can use the formula $\prod_{i=1}^n(2*ceil(log_2(i)) – 1)$ as a starting point.

However, this formula is easy to compute only for relatively small values of N (a couple of thousands or so). For larger values, you’ll need more sophisticated algorithms, typically involving prime factorization and number theory. Usually, such problems are not solvable with pen and paper and may require using a computer programming language for efficient computation. Besides, this is a hard and unsolved problem closely related to the divisor function in number theory, and for which no general formula is known. A precise answer for N = 10^12 would almost certainly require running a complex computer program for an extended period, or even finding a new mathematical technique.

As such, providing a precise value for $F(10^{12})$ directly is not feasible without a suitable computational tool or a new approach to tackle this problem.

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