A364855 -id:A364855 - cp's OEIS frontend

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

Marco Ripà, Aug 08 2023

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

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

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

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.

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

Cf. A229522 (final digit).

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

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

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

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

A364837 Initial digit of 2^(2^n) = A001146(n).

Original entry on oeis.org

2, 4, 1, 2, 6, 4, 1, 3, 1, 1, 1, 3, 1, 1, 1, 1, 2, 4, 1, 2, 6, 4, 2, 4, 1, 3, 1, 1, 1, 2, 4, 1, 3, 9, 9, 8, 7, 5, 2, 8, 8, 6, 4, 1, 3, 9, 9, 9, 9, 9, 8, 7, 5, 2, 8, 7, 6, 3, 1, 2, 5, 3, 1, 1, 1, 3, 1, 1, 3, 9, 8, 7, 5, 3, 1, 1, 1, 3, 1, 2, 4, 2, 5, 2, 6, 4, 1, 2

Offset: 0

Marco Ripà, Aug 10 2023

The sequence corresponds to the initial digit of 2vvn (since 2^(2^n) = ((((2^2)^2)^...)^2) (n times)), where vv indicates weak tetration (see links).
Conjecture: this sequence obeys Benford's law.
For any n > 1, the final digit of 2^(2^n) is 6.

			a(5) = 4, since 2^(2^5) = 2^32 = 4294967296.

Pointless Large numbers stuff by Cookiefonster, 2.03 The Weak Hyper-Operators.

Cf. A000030, A001146, A362004, A364789, A364855.

Join[{2},Table[Floor[2^(2^n)/10^Floor[Log10[2^(2^n)]]],{n,27}]] (* Stefano Spezia, Aug 10 2023 *)

def A364837(n): return int(str(1<<(1<Chai Wah Wu, Sep 14 2023

a(n) = floor(2^(2^n)/10^floor(log_10(2^(2^n)))), for n > 0.

a(n) = A000030(A001146(n)).

More terms from Jinyuan Wang, Aug 10 2023

A365707 Initial digit of n^(n+1) (A007778(n)).

Original entry on oeis.org

0, 1, 8, 8, 1, 1, 2, 5, 1, 3, 1, 3, 1, 3, 1, 6, 2, 1, 7, 3, 2, 1, 7, 4, 3, 2, 1, 1, 9, 7, 6, 5, 4, 4, 3, 3, 3, 3, 4, 4, 4, 5, 6, 7, 9, 1, 1, 1, 2, 3, 4, 6, 8, 1, 1, 2, 4, 6, 1, 1, 2, 4, 8, 1, 2, 4, 8, 1, 2, 5, 1, 1, 3, 7, 1, 3, 6, 1, 2, 6, 1, 3, 7, 1, 3, 8, 2

Offset: 0

Marco Ripà, Sep 16 2023

			a(3) = 8, since 3^(3+1) = 3^4 = 81.

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A000030, A007778, A061505, A138029, A364855, A365935.

seq(convert(n^(n+1),base,10)[-1],n=0..100); # Robert Israel, Feb 16 2024

Join[{0}, Table[Floor[n^(n+1)/10^Floor[Log10[n^(n+1)]]], {n, 86}]]

a(n) = floor((n^(n+1))/10^floor(log_10(n^(n+1)))).

a(n) = A000030(A007778(n)).

cp's OEIS Frontend

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

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A364837 Initial digit of 2^(2^n) = A001146(n).

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A365707 Initial digit of n^(n+1) (A007778(n)).

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