|
| |
|
|
A253389
|
|
a(n) is the repeating digit pattern in penultimate digit of successive powers of n (omitting initial powers without at least two digits).
|
|
0
|
|
|
|
13625124998637487500, 28428684442602686000, 1652983470, 2, 31975, 4400, 61964512293803548770, 8264462800, 0, 1234567890, 14233809528576619047, 16969012743858543270, 9412305876, 27, 15937, 18125674943632987050, 2376, 1652983470, 0, 24680, 84530839221546916077, 22644848642280060680, 27, 2, 7, 22840808842260464660, 28556013027144398697, 2488420660, 0
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
2,1
|
|
|
COMMENTS
|
This is looking one step past the well-known rules for the last digit of successive powers: Powers of integers ending in digit 2 always repeat 2486 in the last digit pattern, powers of integers ending in digit 3 always repeat 3971, of integers ending in 4 repeat 46, of integers ending in 1, 5, and 6 repeat themselves, of integers ending in 7 repeat 7931, of integers ending in 8 repeat 8426, and of integers ending in 9 repeat 91.
Is there a pattern in the repeating patterns in the penultimate digits? Possibly 99 patterns, for x = 01 to 99?
All pattern lengths are a divisor of 20, as n^2 == n^22 (mod 100). - Walter Roscello, Jan 22 2023
Some terms have quasi-periodic patterns with first nonzero digit(s) not in the period, such as a(14), a(15) and a(18) ignoring first digit 1 while a(22) ignoring first digit 2. In these cases, the periodic patterns of a(n) could be rewritten by rotation. - Lerong Zhu, May 10 2024
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
Powers of 2: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096...
Second-to-the-last digits, skipping the one-digit powers: 1,3,6,2,5,1,2,4,9,9,8,6,3,7,4,8,7,5,0,0,...
Find repeating pattern and concatenate digits: 13625124998637487500
10 does not repeat its penultimate digit (1), so a(10)=0.
|
|
|
CROSSREFS
|
Cf. A160590 (penultimate digit of 2^n).
|
|
|
KEYWORD
|
nonn,base,easy
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
