A253389 a(n) is the repeating digit pattern in penultimate digit of successive powers of n (omitting initial powers without at least two digits).

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

Offset: 2

Author

Erik Maher, Dec 31 2014

Keywords

• nonn
• base
• easy

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

Examples

			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).

Extensions

Missing a(8) inserted by Walter Roscello, Jan 22 2023

a(12)-a(30) from Lerong Zhu, May 10 2024