The evidence seems much stronger for the entries to be unbounded. On the other hand, each diagonal parallel to an edge has bounded entries and hence is eventually periodic since the previous is.
If you start with $[7]$ then row $82$ is
$7, 28, 22, 2, \cdots 10620, 7640, {\Large \mathbf{2352}}, 2590, \cdots $$289800, {\Large \mathbf{629736}}, 436408, 103008, 11003 \cdots$
So it does have $2352$ in it but the largest thing in that row is much larger.
The largest entries of the remaining rows up to $100$ are
$266536, 106073, 201592, 205778, 398021, 685698, 1175313,$$ 2248918, 3564296, 4520942, {\large\mathbf {8790873}}
, 6505052, 6988090,$$ 1301717, 2558155, 3815406, 3311175, 749856$
However, the band made of the first $m$ entries (also last $m$ entries) of each row is bounded.
For example it turns out that if a particular row starts with $7$ then from some point on the rows might start, as above
$7,14 \\ 7,21 \\ 7,28 \\ 7,14 \\ 7,21 \\ 7,28 \\$
I will abbreviate that pattern of second entries as $[14,21,28]$. There are only two other possible pattern for $7$, namely $[8,9]$ and $[10,17,24,13,20,12]$
This can be explained as follows: To see the possible sequence of second entries we can make a directed graph of out-degree $1$ with an edge from $c$ to the square-free part of $7+c.$ The three patterns above can be seen to be cycles in this graph. We can check that any $c$ under $17$ ends up at one of those cycles after a few steps. We can finish by showing that any $c \geq 17$ arrives in $4$ steps or less at a number less than $c.$
If we have a row starting $7,c$ then the starts of the next rows are
- $7,d$ for $d=7+\frac{c}{t^2}$
- $7,e$ for $e=7+\frac{d}{t^2}$
- $7,f$ for $f=7+\frac{e}{t^2}$
- $7,g$ for $g=7+\frac{f}{t^2}$
Here the $t^2$ in each denominator is the largest square dividing the corresponding numerator. Now perhaps $c,d,e,f=c,c+7,c+14,c+21$ but then
$c,d,e,f \bmod 4=3,2,1,0$ so one of $d,e,f,g$ is $7+\frac{c+7j}{t^2}$ for $t \geq 2$ and $j \leq 3.$ For $c \gt \frac{49}3,$ we have $7+\frac{c+21}{4} \lt c.$
So any row $7,c$ with $c>16$ is followed soon by $7,c'$ with $c' \lt c.$ Once a second entry repeats, a period is confirmed.
In general, if a row begins with $a$ which is square-free and $p$ is the smallest prime not dividing $a$ then eventually all second entries are no larger than $p^2a$ and they are periodic with a period no larger than $p^2-1.$
A quick example: For $a=15$ eventually one has in the second entries one of
- a fixed point of $16$ or $20$
- the period two cycle $[17,32]$
- the period $4$ cycle $ [22, 28, 52, 37]$
- the period $5$ cycle $[26, 41, 56, 29, 44]$
What happens in later diagonals is a similar argument. I'll give an example and a sketch.
Suppose we have first entry $7$ and second entries $[10,17,24,13,20,12].$ It turns out that after a while the first three entries of rows follow one of these patterns
$\\7, 10, 22\\7, 17, 32\\7, 24, 19\\7, 13, 25\\7, 20, 14\\7, 12, 19\\7, 10, 22\ \ \ \ $ $7, 10, 34\\7, 17, 44\\7, 24, 28\\7, 13, 13\\7, 20, 26\\7, 12, 31\\7, 10, 34\ \ \ \ $$7, 10, 14\\7, 17, 24\\7, 24, 23\\7, 13, 29\\7, 20, 42\\7, 12, 47\\7, 10, 50\\7, 17, 12\\7, 24, 20\\7, 13, 11\\7, 20, 24\\7, 12, 11\\7, 10, 14\\$
To show that the behavior is eventually periodic we show that there must be some $N$ large enough that any row starting $7,10,c$ with $c \geq N$ leads eventually to one starting $7,10,c'$ with $c' \lt c.$ This implies infinitely many rows starting $7,10,b$ with $b \lt N.$ So, eventually one appears for a second time and the periodic behavior is established.
To show that the possibilities I listed are the only ones that can occur, it suffices to not only show there is such an $N$ but to find one and then check that each start $7,10,c$ with $c \lt N$ leads to one of these periods. I will argue that $N=1000$ is large enough. In this particular case it turns out that $7,10,c$ results, 6 rows further down, in $7,10,c'$ with $c' \lt c$ with the exceptions of $1 \leq c \leq 14$ and $16,19,21,25.$
Here is how we see that there is some $N$: There is a bounded possible jump between successive third entries. Here they can never jump by more than $17.$ If with some frequency we divide by $4$ (or a larger square) then any large enough third entry is later followed by a smaller one. This is enough to keep all later third entries bounded which, in turn, means that eventually we will have two rows with the same three first entries and periodicity is established.
Here are specifics for this case: I claim that if we have a third entry $c$ then no future third entry can exceed $c+48\cdot 54$ and some future entry will be smaller than $\frac{c+54\cdot 48}{4}+5.$ But if $c \geq 936$ then $\frac{c+54\cdot 48}{4}+54 \leq c.$
If at some point we have a row starting $7,10,c$ then maybe the next rows start
$\\7, 10, c\\7, 17, c+10\\7, 24, c+27\\7, 13, c+33\\7, 20, c+46\\7, 12, c+51\\7, 10, c+54 $
The rows which start $7,10,x$ starting at x=c might perhaps have third entries $c+54j$ for $0 \leq j \leq 48.$ But these are distinct $\bmod 49$ so one of those entries would have to be a multiple of $49$ and the next such row would start $7,10,y$ for $$y \leq \frac{c+54\cdot 48}{4}+54$$
As remarked, for $c \geq 936$ we have $y \leq \frac{c+54\cdot 48}{4}+54 \leq c.$
In this specific example we note that the entries $c,c+10,c+27,c+33$ are distinct $\bmod 4.$ However that is particular to this example.
The argument given did not use anything specific to this example except that the second entries have some period (namely $6$) and the third entries have some upper bound on the possible increase every $6$ rows (namely $54$.) And, since $49$ is a square prime to $54$, we can't go through more than $48$ groups of $6$ rows without getting a division by at least $4.$
