A392143 - OEIS

%I #12 Jan 09 2026 17:30:24

%S 0,1,512,15752961,94606929690624,220903392825527587890625,

%T 311954920641940794545461153939587072,

%U 374142991911400415397080306521056997980469670707201,507528679944764261210790982765425081637339300928309681979392000000000,999999990000000044999999880000000209999999748000000209999999880000000044999999990000000001

%N a(n) = (n^(n - 1) - 1)^n.

%C For n > 3, a(n) = (n^(n - 1) - 1)^n is a perfect power of exact order n (by Mihailescu's theorem).

%C For the number of stable digits of the tetration a(n)^^b (b > 0 integer), in the radix-n numeral system, see Identity (5) in "Universal +-1 congruence speed invariant in any numeral system" (in Links).

%H Marco Ripà, <a href="https://doi.org/10.59400/jam1771">On the relation between perfect powers and tetration frozen digits</a>, Journal of AppliedMath, 2024, 2(5), 1771.

%H Marco Ripà, <a href="https://doi.org/10.5281/zenodo.17677512">Radix-r Congruence Speed Verification Tool (Python Script)</a>, Zenodo, 2025.

%H Marco Ripà, <a href="https://doi.org/10.5281/zenodo.17851020">Universal +-1 congruence speed invariant in any numeral system</a>, Zenodo, 2025.

%H Marco Ripà and Gabriele Di Pietro, <a href="https://doi.org/10.5281/zenodo.15276824">A Compact Notation for Peculiar Properties Characterizing Integer Tetration</a>, Zenodo, 2025.

%F a(n) = (n^(n-1)-1)^n = (n^n-n)^n/n^n.

%e For n = 3, a(3) = (3^(3 - 1) - 1)^2 = (9 - 1)^3 = 512.

%t Table[(n^(n-1)-1)^n,{n,1,9}]

%Y Cf. A253604, A373387, A390597.

%K nonn,easy

%O 1,3

%A _Marco Ripà_, Jan 01 2026