0
$\begingroup$

In Pascal's triangle, each number is the sum of the two numbers directly above it. I experimented with a modified Pascal's triangle. First a n-tuple of natural numbers is put to the nth row of the triangle so that the first element of the n-tuple is put to the first item in the nth row of the triangle, second element of the tuple to the second item in the nth row of the triangle all the way to the nth item. Then the items of the n+1th row are calculated as the sum of the square-free part of the left number above and square-free part of the right number above (instead of a direct sum) the item. The next rows are calculated similarly ad infinitum.

I played with the modified triangle with the computer and numerical evidence suggests that for any n-tuple, there exists a natural m such that all entries in the modified triangle are less than m. In my experimentations, the slowest part was to compute the square-free part of a number and my tests were limited by this. I computed the square free part and primality by brute force.

The largest value in the triangle may grow quite fast as the number of elements and the elements themselves in the initial tuple grows. For instance with the tuple (7) the largest value seems to be 2352, with (2,3) 36576 and with (5,9,7) 127039544.

I wonder if there is something non-trivial going on. Is there a way to prove or disprove the conjecture or to improve the numerics from the brute force?

$$\begin{eqnarray}7\\7,7\\7,14,7\\ 7,21,21,7\\7,28,42,28,7\\7,14,49,49,14,7\\7,21,15,2,15,21,7\\7,28,36,17,17,36,28,7\end{eqnarray}$$

$\endgroup$
0
$\begingroup$

You can see a more detailed answer answer at the mathoverflow version. If I had more time, I would have made it shorter.

In brief, small numeric evidence strongly suggests unbounded growth of entries. Starting with $[7]$ there are much bigger entries than $2,352$ , It occurs in row $82$ and the largest entry there is $629,736.$ In row $93$ there is an entry $8,790,873.$

Having settled on this, here is a possible justification which is suggestive but far from rigorous: Suppose a certain row has $n$ entries which sum to $S.$ So the average entry is $\frac{S}n.$ That row is set but, in preparation for the next row, we can rewrite the row replacing each entry by its square-free part and getting a new sum of $S'$. Then the sum of the entries in the next row will be $2S'.$ Maybe nothing changes or maybe almost all of them do but the probability that any certain entry stays the same is roughly $\frac{6}{\pi^2} \gt 0.6$ so, ignoring the entries which change, we still estimate $S'\gt 0.6S$ so the next row has expected sum greater than $1.2S$ and expected average value greater than $\frac{1.2S}{n+1}.$

In $10000$ trials picking $100$ random positive integers less than $10000$ the sum of the next row ranged from $1.12S$ to $1.6S$ but was never less.

On the other hand, it can be shown that the entries in the kth position are bounded and, hence, eventually periodic. The entries in the first position are constant. In short, those in position $k$ grow at a rate bounded by a constant but with some regularity get reduced by $\frac14$ or more. That keeps them bounded.

In a little more detail: Suppose the entries in position $k-1$ are eventually periodic with period $t$ and values in a period adding to $T.$ Then , for position $k,$ there is a $p$ (the least prime not dividing T) so that at least once every $p^2t$ rows there is an entry which is reduced to a quarter of itself or less. This means that if $a$ is the $k$th entry in row $r$, then the $k$th entry in row $r+p^2t$ is less than $\frac{a}4+p^2T$ and this ,in turn, is less than $a$ when $a \gt \frac{4p^2T}3.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.