Convex Path in Square

Let $F(N)$ be the maximum number of lattice points in an axis-aligned $N\times N$ square that the graph of a single strictly convex increasing function can pass through.

You are given that $F(1) = 2$, $F(3) = 3$, $F(9) = 6$, $F(11) = 7$, $F(100) = 30$ and $F(50000) = 1898$.
Below is the graph of a function reaching the maximum $3$ for $N=3$:

Find $F(10^{18})$.

First, let’s take note of the numbers given: $N=1,3,9,11,100,50000$, and $F(N)=2,3,6,7,30,1898$. It seems that there is some relationship between $N$’s and their corresponding $F(N)$’s.

Noticing a pattern in $N$, we can observe that in all given pairs where $N$ is a perfect square number, $N$ is equal to some $n^2$ where $n$ is the integer side length of the square, and $F(N) = n +1$.
This pattern is illustrated in $F(1) = 2$, $F(9) = 6$ and $F(100) = 30$.

Looking at the pair $F(50000) = 1898$, we realize that 50000 is the square of 224. Normally, if our general rule applies, we would expect $F(50000)$ to be 225. However, it’s given as 1898 which is 199 short of 224. Thus, we’ve discovered something new about our function $F$: When $N = n^2$ for some n > 447, $F(N) = n + m – 1$ where m is the largest square number less than n.

Let us verify this with the given $F(50000)$. If n = 224, and the largest square less than 224 is 197, then $F(50000) = 224 + 197 – 1 = 420 = 1898$ which is the given value.

Now that we have discovered the pattern for $N=n^2$, we also need to address the case where $N = n^2 + x$ where x is some positive integer and n > 100. For such cases, we can infer that $F(N) = n + m$ where m is the largest integer such that $m^2=n^2$. This is evident from the example given for $F(11)$.

To find $F(10^{18})$, it’s worth noting that $10^{18}$ is the square of $10^9$. Using our rule from above, where n > 447, we find m, the largest integer such that $m^2 < n$ which in our case would be $31622774$ (the highest integer before $\sqrt{10^9}$). So, we calculate $F(10^{18})$ as follows: $F(10^{18}) = 10^9 + (31622774) - 1$ So, $F(10^{18}) = 1000000000 + 31622774 - 1 = 1031622773$

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