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A344555
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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.
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2
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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
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OFFSET
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1,3
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COMMENTS
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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:
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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...
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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.
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LINKS
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EXAMPLE
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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.
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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