cp's OEIS Frontend

A364789 Initial digit of (n^n)^n (A002489(n)).

%I A364789 #58 Jan 31 2024 03:59:07
%S A364789 1,1,1,1,4,2,1,2,6,1,1,1,2,1,4,4,1,3,5,4,2,1,5,2,1,5,3,2,3,7,2,1,1,4,
%T A364789 2,3,9,7,1,1,1,1,2,1,5,5,2,4,3,1,2,2,1,3,4,3,2,6,1,2,2,1,8,3,1,3,8,1,
%U A364789 3,5,9,1,2,4,8,1,3,1,3,1,5,3,3,3,5,1,3
%N A364789 Initial digit of (n^n)^n (A002489(n)).
%C A364789 a(0) = 1 is from (0^0)^0 = 1 per A002489.
%C A364789 The author conjectures that this sequence obeys the well-known Benford's law.
%H A364789 Paolo Xausa, <a href="/A364789/b364789.txt">Table of n, a(n) for n = 0..1000</a>
%H A364789 Pointless Large numbers stuff by Cookiefonster, <a href="https://sites.google.com/site/pointlesslargenumberstuff/home/2/weakoperators">2.03 The Weak Hyper-Operators</a>.
%H A364789 Wikipedia, <a href="https://en.wikipedia.org/wiki/Benford%27s_law">Benford's law</a>.
%F A364789 a(n) = floor(((n^n)^n)/10^floor(log_10((n^n)^n))).
%F A364789 a(n) = A000030(A002489(n)).
%e A364789 a(4) = 4, since (4^4)^4 = 4^(4^2) = 4294967296.
%t A364789 A364789[n_] := If[n == 0, 1, First[IntegerDigits[(n^n)^n]]];
%t A364789 Array[A364789, 100, 0] (* _Paolo Xausa_, Jan 31 2024 *)
%o A364789 (Python)
%o A364789 def A364789(n): return int(str((n**n)**n)[0]) # _Chai Wah Wu_, Aug 10 2023
%Y A364789 Cf. A000030, A002489, A241299, A363746, A364837, A364855.
%Y A364789 Cf. A229522 (final digit).
%K A364789 base,easy,nonn
%O A364789 0,5
%A A364789 _Marco Ripà_, Aug 08 2023