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

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

Offset: 0

Author

Marco Ripà, Aug 08 2023

Keywords

• base
• easy
• nonn

Comments

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

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

Examples

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

Links

• Paolo Xausa, Table of n, a(n) for n = 0..1000
• Pointless Large numbers stuff by Cookiefonster, 2.03 The Weak Hyper-Operators.
• Wikipedia, Benford's law.

Crossrefs

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

Cf. A229522 (final digit).

Programs

Mathematica

A364789[n_] := If[n == 0, 1, First[IntegerDigits[(n^n)^n]]];
Array[A364789, 100, 0] (* Paolo Xausa, Jan 31 2024 *)

Python

def A364789(n): return int(str((n**n)**n)[0]) # Chai Wah Wu, Aug 10 2023

Formula

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

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