Personal Investigation: Part Two

This post follows on from a previous one so it's best to read that first in order to understand what I'm on about. Here is the link:

• Personal Investigation: Part 1

After my previous post on Triads, it occurred to me that the same behaviour might occur with dyads. Starting with two numbers instead of three, I investigated the behaviour of the sequence:$$a_n=\text{ sum of digits }\big (a_{n-1}^k+a_{n-2}^k \big ),a_0=0,a_1=1, k \geq 2$$Not surprisingly, the behaviour was the same, with loops developing for \(2 \leq k \leq 50\).

For example, consider the case of$$a_n= \text{ sum of digits } \big ( a_{n-1}^2+a_{n-2}^2 \big ),a_0=0, a_1=1$$Here we the sequence begins 0, 1, 1, 2, 5, 11, 11, 8, 14, 8, 8, 11, 14, 11, 11, 8, ... but we see that we have the loop shown in bold. The total number of terms generated is seven and these are 0, 1, 2, 5, 8, 11 and 14. The number of terms generated by the different values of \(k\) is as follows (the first element in the ordered pair is the \(k\) value and the second element is the number of terms):

[(2, 7), (3, 13), (4, 8), (5, 18), (6, 7), (7, 30), (8, 13), (9, 29), (10, 12), (11, 27), (12, 11), (13, 37), (14, 19), (15, 22), (16, 15), (17, 37), (18, 13), (19, 41), (20, 22), (21, 37), (22, 13), (23, 28), (24, 12), (25, 34), (26, 27), (27, 48), (28, 14), (29, 23), (30, 11), (31, 39), (32, 35), (33, 19), (34, 10), (35, 48), (36, 12), (37, 44), (38, 26), (39, 40), (40, 28), (41, 35), (42, 22), (43, 59), (44, 35), (45, 32), (46, 24), (47, 47), (48, 15), (49, 61), (50, 46)

It can be seen that the largest number of terms (61) occurs when \(k=49\). As for the maximum values, the sequence that arises is as follows:

14, 28, 35, 45, 56, 91, 83, 110, 98, 135, 128, 151, 155, 181, 197, 218, 200, 241, 275, 271, 296, 286, 308, 319, 341, 351, 350, 376, 353, 410, 443, 433, 395, 495, 443, 501, 551, 521, 548, 565, 614, 620, 614, 604, 611, 646, 641, 716, 701

Figure 1 shows the graph of this sequence and highlights its basically linear behaviour:

Figure 1

Figure 2 shows the code (permalink) to generate the necessary details: 

Figure 2

A logical continuation of this investigation would be to look at tetrads, pentads, hexads, heptads, octads, enneads, decads and so on. Let's take the case of the tetrad with \(k=2\):$$a_n=\text{ sum of digits } \big (a_{n-1}^2+a_{n-2}^2 +a_{n-3}^2 +a_{n-4}^2 \big ),a_0=0,a_1=1,a_2=2,a_3=3$$Here the sequence of terms generated is:

0, 1, 2, 3, 5, 12, 11, 20, 15, 17, 9, 23, 8, 18, 26, 18, 20, 14, 21, 11, 15, 20, 17, 9, 23, 21, 8, 8, 18, 20, 15, 5, 20, 6, 20, 15, 8, 14, 21, 17, 18, 8, 11, 24, 14, 21, 11, 11, 24, 17, 9, 14, 8, 9, 8, 9, 11, 14, 12, 11, 15, 20, 17, ...

Once again, a loop arises, shown in bold typeface above.

The total number of terms generated is 19: 

0, 1, 2, 3, 5, 6, 8, 9, 11, 12, 14, 15, 17, 18, 20, 21, 23, 24 and 26.

For \(2 \leq k \leq 50 \), the sequence of terms for maximum values associated with each \(k\) is:

26, 28, 47, 56, 57, 82, 99, 118, 119, 147, 173, 183, 188, 206, 209, 214, 236, 282, 263, 271, 297, 311, 335, 338, 371, 357, 389, 409, 399, 442, 459, 468, 485, 457, 525, 541, 558, 567, 566, 633, 579, 651, 659, 666, 666, 699, 722, 714, 735, ...

Here is the permalink for the tetrad calculations. Figure 3 shows a graph of this sequence (up to the 50th term) and highlights once again the fact that the progression is basically linear. 

Figure 3