It is difficult to find a title for this post but as it certainly involves triads, or groups of three, I've settled on that. Let's start with the triad 0, 1, 2 and square each term so that we have:
0, 1, 2, 5, 3, 11, 11, 8, 9, 14, 8, 8, 9, 11, 14, 20, 15, 11, 17, 14, 12, 17, 17, 11, 24, 23, 11, 11,15, 17,14, 8, 18, 17, 20, 5, 12, 20, 20, 17, 18, 5,17, 17, 9, 20, 14, 20, 24, 11,17, 23, 21, 17, 17, 11, ...
With the last three terms listed here, we have entered a loop because 17, 17, 11 appeared earlier. This is the loop:
17, 17, 11, 24, 23, 11, 11, 15, 17, 14, 8, 18, 17, 20, 5, 12, 20, 20, 17, 18, 5, 17, 17, 9, 20, 14, 20, 24, 11, 17, 23, 21
We find that there are only 17 numbers that appear, including our starting values of 0, 1 and 2. Here are the numbers expressed in order:
0, 1, 2, 3, 5, 8, 9, 11, 12, 14, 15, 17, 18, 20, 21, 23, 24
What happens if we raise our triad of starting numbers to the third power. Here we have:
0, 1, 2, 9, 18, 26, 17, 16, 26, 26, 26, 24, 34, 18, 28, 29, 18, 18, 17, 26, 16, 26, 26, ...
With the last three terms listed, we have again entered a loop because 16, 16 and 26 occurred earlier. Here is our loop:
16, 26, 26, 26, 24, 34, 18, 28, 29, 18, 18, 17, 26
We find that only twelve numbers appearing, including our starting values of 0, 1 and 2:
0, 1, 2, 9, 16, 17, 18, 24, 26, 28, 29, 34
If we continue raising our initial triads to higher and higher powers, it seems that we always get a finite set of numbers. In other words, sooner or later, a loop develops but must ensure to take enough terms. When raising to the 50th power for example, we must take at least 561 iterations and there are total of 106 terms. So what are we doing in general terms:
where
For every value of
[24, 34, 35, 58, 66, 93, 98, 99, 117, 140, 147, 165, 191, 181, 197, 218, 236, 248, 266, 297, 297, 311, 309, 347, 344, 362, 360, 408, 416, 436, 425, 449, 479, 482, 479, 544, 546, 546, 566, 570, 597, 624, 632, 702, 665, 664, 696, 710, 725]
We see that
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