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
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
KEYWORD
nonn,base,easy
AUTHOR
Erik Maher, Dec 31 2014
EXTENSIONS
Missing a(8) inserted by Walter Roscello, Jan 22 2023
a(12)-a(30) from Lerong Zhu, May 10 2024
STATUS
approved