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?