Personal Investigation: Part One

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:

$$0^2+1^2+2^2=0+1+4 =5$$ The result is 5 and, because it only contains one digit, the sum of its digits is 5. Now let's make 1, 2, 5 our new triad and carry out the same procedure: $$1^2+2^2+5^2=1+4+25=29 \text{ with sum of digits }=11$$ If we continue this process, we get the following sequence of terms: $$0, 1, 2, 5, 3, 11, 11, 8, 9, 14, 8, 8, 9, 11, \\14, 20, 15, 11, 17, 14, 12, 17, 17, 11, 24, 23, 11, 11,\\ 15, 17,14, 8, 18, 17, 20, 5, 12, 20, 20, 17, 18, 5, \\17, 17, 9, 20, 14, 20, 24, 11,17, 23, 21, 17, 17, 11, ...$$ 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: $$0^3+1^3+2^3=0+1+8=9$$ Once again, because 9 has only one digit, its sum of digits in also 9. Let's use 1, 2 and 9 as our new triad: $$1^3+2^3+9^3=1+8+729=738 \text{ with sum of digits }=18$$ Continuing this process, we get the following sequence of terms: $$0, 1, 2, 9, 18, 26, 17, 16, 26, 26, 26, 24, \\ 34, 18, 28, 29, 18, 18, 17, 26, 16, 26, 26, ...$$ 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: $$a_n=\text{ sum of digits } \big (a_{n-1}^k+a_{n-2}^k+a_{n-3}^k \big )$$

where $a_0=0, a_1=1 \text{ and }a_3=2, 2 \leq k \leq 50$

For every value of $k$, I've taken the largest value in the set of numbers generated and formed a sequence of these largest values for $2 \leq k \leq 50$. Let's denote this term by $a_{max, \, k}$ where $k$ ranges from 2 to 50. The result is the following sequence:

$$[24, 34, 35, 58, 66, 93, 98, 99, 117, 140, \\147, 165, 191, 181, 197, 218, 236, 248, 266, 297,\\ 297, 311, 309, 347, 344, 362, 360, 408, 416, 436, \\425, 449, 479, 482, 479, 544, 546, 546, 566, 570,\\ 597, 624, 632, 702, 665, 664, 696, 710, 725]$$ We see that $a_{max,\,4}=35$ and $a_{max,\,49}=710$. The sequence more or less increases but not strictly so. I supposing that these loops (and consequent maximum values) continue for $k >50$ but I haven't tested so far. The sequence is probably worthy of submission to the OEIS.