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Initial digit of (n^n)^n (A002489(n)).
4

%I #58 Jan 31 2024 03:59:07

%S 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 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 3,5,9,1,2,4,8,1,3,1,3,1,5,3,3,3,5,1,3

%N Initial digit of (n^n)^n (A002489(n)).

%C a(0) = 1 is from (0^0)^0 = 1 per A002489.

%C The author conjectures that this sequence obeys the well-known Benford's law.

%H Paolo Xausa, <a href="/A364789/b364789.txt">Table of n, a(n) for n = 0..1000</a>

%H 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 Wikipedia, <a href="https://en.wikipedia.org/wiki/Benford%27s_law">Benford's law</a>.

%F a(n) = floor(((n^n)^n)/10^floor(log_10((n^n)^n))).

%F a(n) = A000030(A002489(n)).

%e a(4) = 4, since (4^4)^4 = 4^(4^2) = 4294967296.

%t A364789[n_] := If[n == 0, 1, First[IntegerDigits[(n^n)^n]]];

%t Array[A364789, 100, 0] (* _Paolo Xausa_, Jan 31 2024 *)

%o (Python)

%o def A364789(n): return int(str((n**n)**n)[0]) # _Chai Wah Wu_, Aug 10 2023

%Y Cf. A000030, A002489, A241299, A363746, A364837, A364855.

%Y Cf. A229522 (final digit).

%K base,easy,nonn

%O 0,5

%A _Marco RipĂ _, Aug 08 2023