%I #38 Jan 13 2024 10:54:16
%S 0,1,2,1,4,5,6,1,8,1,0,1,2,1,4,5,6,1,8,1,0,1,2,1,4,5,6,1,8,1,0,1,2,1,
%T 4,5,6,1,8,1,0,1,2,1,4,5,6,1,8,1,0,1,2,1,4,5,6,1,8,1,0,1,2,1,4,5,6,1,
%U 8,1,0,1,2,1,4,5,6,1,8,1,0,1,2,1,4,5,6
%N Final decimal digit of n^((n+1)^(n+2)) = A030198(n).
%C Period 10, repeat: [0, 1, 2, 1, 4, 5, 6, 1, 8, 1].
%H Paolo Xausa, <a href="/A365689/b365689.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,0,0,0,0,1).
%F a(n) = n^((n+1)^(n+2)) mod 10.
%F a(n) = A103562(n) for n >= 4 (as 3^(2^1) == 9 (mod 10) instead of 1).
%e For n = 2, a(2) = 2417851639229258349412352 mod 10 = 2.
%t PadRight[{},100,{0,1,2,1,4,5,6,1,8,1}] (* _Paolo Xausa_, Oct 16 2023 *)
%o (PARI) a(n) = lift(Mod(n, 10)^((n+1)^(n+2))); \\ _Michel Marcus_, Sep 16 2023
%o (Python)
%o def A365689(n): return pow(n,(n+1)**(n+2),10) # _Chai Wah Wu_, Sep 22 2023
%Y Cf. A030198, A103562, A120962, A365689 (initial digit).
%K nonn,base,easy
%O 0,3
%A _Marco RipĂ _, Sep 16 2023