login
A369771
Number of the rightmost decimal digits of n^(n^n) that are the same as those of n^(n^(n^n)).
4
1, 0, 1, 1, 2, 2, 8, 4, 4, 2, 3, 10000000000, 4, 2, 3, 2, 13, 4, 3, 2, 3, 104857600000000000000000000, 4, 1, 2, 4, 12, 8, 2, 2, 3, 205891132094649000000000000000000000000000000, 4, 4, 3, 2, 7, 4, 3, 1, 3, 12089258196146291747061760000000000000000000000000000000000000000
OFFSET
-1,5
COMMENTS
The common digits might include leading 0's (such as at n = 5) and they are included in the total.
Let c be a positive integer and assume that k is a positive integer that is not a multiple of 10. If n = k*10^c, then a(n) = c*(n^n) which is all the rightmost 0's of n^(n^n).
LINKS
Marco Ripà and Luca Onnis, Number of stable digits of any integer tetration, Notes on Number Theory and Discrete Mathematics, 2022, 28(3), 441-457.
Wikipedia, Tetration.
FORMULA
For any n >=2, a(n) is such that n^(n^n) == n^(n^(n^n)) (mod 10^(a(n))) and n^(n^n) <> n^(n^(n^n)) (mod 10^(a(n)+1)).
EXAMPLE
a(-1) = 1 since (-1)^(-1) = -1 (which is a one-digit number);
a(0) = 0 since 0^0 = 1 so that 0^(0^0) = 0 and 0^(0^(0^0)) = 1 have no digits in common.
For n=5, a(n)=8 since 5^(5^5) == 908203125 (mod 10^9) while 5^(5^(5^5)) == 408203125 (mod 10^9).
CROSSREFS
KEYWORD
sign,base
AUTHOR
Marco Ripà, Jan 31 2024
STATUS
approved