A344555 Numbers k such that the infinite sequence of digits consisting of the final digit of k^m for m = 2, 3, 4, ... is the same as the sequence of digits obtained by concatenating infinitely many copies of k.

0, 1, 5, 6, 11, 19, 55, 64, 66, 111, 555, 666, 1111, 1919, 4268, 4862, 5555, 6464, 6666, 9317, 9713, 11111, 55555, 66666, 111111, 191919, 555555, 646464, 666666, 1111111, 5555555, 6666666, 11111111, 19191919, 42684268, 48624862, 55555555, 64646464, 66666666, 93179317

Offset: 1

Author

Luke Voyles, May 22 2021

Keywords

• nonn
• base

Comments

The numbers k of this sequence repeat from k^2 onward. For example, if number ends in 8, the last digit of the square of k will always be 4, the last digit of the cube of k will always be 2, the last digit of the fourth power of k will always be 6, and the fifth power of k will always be 8. Base numbers that end in 0, 1, 5, and 6 will always result in numbers with the same digit when they have positive integers as exponents. For k that have a 2 in the ones place, then the square of k will have 4 in the ones place, the cube of k will have 8 in the ones place, the fourth power of k will have 6 in the ones place, and the fifth power of k will have 2 in the ones place.

For any integer k, the ones digit of each higher power, i.e., k^2, k^3, k^4, etc., depends only on the ones digit of k as follows:

.

Ones digits in k and larger powers of k

--+------------------------------------ Resulting string of

k | k^2 k^3 k^4 k^5 k^6 k^7 k^8 k^9 ... concatenated digits

--+------------------------------------ -------------------

0 | 0 0 0 0 0 0 0 0 ... 0000000000000000...

1 | 1 1 1 1 1 1 1 1 ... 1111111111111111...

2 | 4 8 6 2 4 8 6 2 ... 4862486248624862...

3 | 9 7 1 3 9 7 1 3 ... 9713971397139713...

4 | 6 4 6 4 6 4 6 4 ... 6464646464646464...

5 | 5 5 5 5 5 5 5 5 ... 5555555555555555...

6 | 6 6 6 6 6 6 6 6 ... 6666666666666666...

7 | 9 3 1 7 9 3 1 7 ... 9317931793179317...

8 | 4 2 6 8 4 2 6 8 ... 4268426842684268...

9 | 1 9 1 9 1 9 1 9 ... 1919191919191919...

.

This sequence consists of each of the nonnegative integers that, when repeated infinitely, yields one of the digit strings in the column at the right.

Examples

			The patterns that I have noticed and seen confirmed demonstrate that the infinite patterns that result with the end digits of exponents when n has a particular numerical value from k^2, k^3, k^4, and k^5 before they repeat are as follows: k with final digit 0 (0000); k with final digit 1 (1111); k with final digit 2 (4862); k with final digit 3 (9713); k with final digit 4 (6464); k with final digit 5 (5555); k with final digit 6 (6666); k with final digit 7 (9317); k with final digit 8 (4268); and k with final digit 9 (1919).
Therefore, the number 64 infinitely repeats because 64^2 equals 4096 (which ends in 6), 64^3 equals 262144 (which ends in 4), 64^4 equals 16777216 (which ends in 6), and 1073741824 (which ends in 4). 64 repeated twice in the previous demonstration, but all numbers infinitely repeat in the same way.
Additionally, 4862^2=23639044 (ends in 4), 4862^3=114933031928 (ends in 8), 4862^4=558804401233936 (ends in 6), and 4862^5=2716906998799396832 (ends in 2). The 4862 sequence among the final digits of the power for 4862 then continues infinitely as 4862^6 ends in 4, 4862^7 ends in 8, 4862^8 ends in 6, 4862 ends in 2, and so on.
One interesting fact about this sequence is that only the last digit of an odd-numbered power of k is necessary to determine the last digit of k itself.