Repeat $$$C_0(27)$$$ for as long as possible until the remainder is short enough. More specifically, combine $$$\floor\left(\frac{n}{27}-1\right)$$$ copies of

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In the above, we used repeating units of size $$$27$$$. Here, we use repeating units of size $$$200$$$.

First, cover a region of size $$$50$$$, except for positions $$$9$$$ and $$$42$$$, using $$$9-16-25$$$ triplets: $$$1\ldots 8~X~1\ldots 16~1\ldots 16~X~9\ldots 16$$$. Then assemble $$$4$$$ of these regions, and pair up the gaps as follows: $$$9-109,42-142,59-159,92-192$$$. This uses $$$68$$$ colors.

Covering the array of length $$$n$$$ with such units of size $$$200$$$ produces the desired bound.

This concept of "stacking" regions of $$$50$$$ to give a region of $$$200$$$, for which we will fill up the gaps for, is important for later.

We refer to IMO 2001 C4 (link contains solution). The problem statement is as follows:

A triplet of nonnegative integers $$${x,y,z}$$$ (with $$$x<y<z$$$) is historic if $$${y-x,z-y}={1776,2001}$$$. Prove that the set of all nonnegative integers can be written as the union of pairwise disjoint historic sets.

The solution does not rely on properties of $$$1776$$$ and $$$2001$$$ other than the fact that they are distinct. Thus, replace $$$1776$$$ by $$$9$$$ and $$$2001$$$ by $$$16$$$, and "nonnegative" by "positive" by shifting by $$$1$$$, and we obtain a coloring of the infinite array using triplets.

We cut off the array at $$$n$$$, and almost all integers are covered by a triplet within the range $$$1$$$ to $$$n$$$. However, if $$$k>n-25$$$, it is possible that $$$k$$$ is covered by a triplet that "overflows". If colors were allowed to be used once only, this problem would have been solved, and we would have $$$F(n)\le \frac{n-25}{3}+25<\frac{n}{3}+17$$$.

However, the "gaps" must be filled, and we don't know where they are. Thus, similar to level $$$2$$$, we take $$$n=m^2$$$, and fill in the remaining gaps with pairs with distance $$$m^2$$$. Thus, this gives for $$$n=2m^2$$$, $$$F(n)\le 2\cdot\frac{m^2-25}{3}+25<\frac{n}{3}+9$$$.

For general $$$n$$$, let $$$m$$$ be the greatest integer such that $$$n-27\ge 2m^2$$$. Then we can cover the array of length $$$n$$$ with the above construction for $$$2m^2$$$, and an additional $$$C(n-2m^2)$$$. Note that $$$n-2m^2=O(\sqrt{n})$$$. Thus, the expression is $$$F(n)\le \frac{2m^2}{3}+9+O(\sqrt{n})=\frac{n}{3}+O(\sqrt{n})$$$.

We need to get rid of the $$$O(\sqrt{n})$$$ error; however, trying to simply cover the remaining region with iteratively smaller squares cannot yield a constant number of squares, and would yield a complexity of $$$O(\log\log n)$$$.

Instead, we use Lagrange's four-square theorem: any integer $$$k$$$ is the sum of four squares, $$$k=a^2+b^2+c^2+d^2$$$. Then let $$$x$$$ be $$$27$$$ if $$$n$$$ is odd, and $$$0$$$ if $$$n$$$ is even. Write $$$n=2k+x=2a^2+2b^2+2c^2+2d^2+x$$$. Appending $$$C_3(2a^2)$$$ to $$$C_3(2d^2)$$$ as well as $$$C_0(x)$$$, we can cover the array in no more than $$$\frac{2a^2}{3}+\frac{2b^2}{3}+\frac{2c^2}{3}+\frac{2d^2}{3}+36+13\le \frac{n}{3}+49$$$ colors. Thus, $$$C=50$$$ suffices.

It turns out that there can be quadruplets with the same color: let $$$(a,b,c)=(153^2,104^2,672^2)$$$, then $$$a^2+b^2=185^2$$$ and $$$b^2+c^2=680^2$$$, while $$$a^2+b^2+c^2=697^2<490000$$$. Let $$$C=490000$$$ and $$$n=CM=700^2M$$$. Thus, start with an array of length $$$C$$$. Fill in the elements $$$1,1+a^2,1+a^2+b^2,1+a^2+b^2+c^2$$$ with the same color, leaving $$$C-4$$$ "gaps". We arrange $$$M$$$ arrays in a row, considering each residue class mod $$$C$$$, except $$$1,1+a^2,1+a^2+b^2,1+a^2+b^2+c^2$$$, and fill in the other residue classes using $$$C_4(M)$$$ expanded by a factor of $$$C=700^2$$$.

This gives, for $$$n=CM$$$, $$$F(n)\le (C-4)F(M)+M<(C-4)\left(\frac{n/C}{3}+50\right)+\frac{n}{C}<\frac{(C-1)n}{3C}+50C$$$. For general $$$n$$$, add an extra $$$F(C+27)<C$$$ to give $$$F(n)<\frac{(C-1)n}{3C}+51C$$$.

We use the construction above but add on multiple "layers". For some $$$k$$$, let $$$n=C^kM$$$, and start with an array of length $$$C^k$$$, and for each $$$x$$$ consider the base $$$C$$$ expansion of $$$x$$$: $$$x=c_1c_2\ldots c_k$$$, where $$$c_i$$$ are allowed to be $$$0$$$. Then if $$$i$$$ is the least index such that $$$c_i\in{0,a^2,a^2+b^2,a^2+b^2+c^2}$$$, then $$$x$$$ is in the quadruplet with the three integers resulting from replacing $$$c_i$$$ with the other $$$3$$$ elements of the set $$${0,a^2,a^2+b^2,a^2+b^2+c^2}$$$. This leaves $$$(C-4)^k$$$ gaps, which are filled in similar to Level $$$5$$$. Then we have $F(C^kM)\le (C-4)^kF(M)+\frac{C^k-(C-4)^k}{4}\cdot M<\left(\frac{C^k}{4}+ \frac{(C-4)^k}{12}\right)M+50C^k$. In other words, for general $$$n$$$, we have $$$F(n)<\left(\frac{1}{4}+\frac{1}{12}\left(1-\frac{4}{C}\right)^k\right)n+51C^k$$$.

For $$$n>C^4$$$, choose $$$k$$$ such that $$$C^{2k}\le n\le C^{3k}$$$. Let $$$1-\frac{4}{C}=C^{-3\epsilon}$$$. Then we get $$$F(n)<\frac{n}{4}+\frac{1}{12}C^{-3\epsilon k}n+51C^k\le\frac{n}{4}+\frac{1}{12}(C^{3k})^{-\epsilon}n+O(\sqrt{n})<\frac{n}{4}+O(n^{1-\epsilon})$$$, thus we are done.