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

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

Author

Marco Ripà, Aug 10 2023

Keywords

• nonn
• base

Comments

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.

Examples

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

Links

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

Crossrefs

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

Programs

Mathematica

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

Python

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

Formula

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

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

Extensions

More terms from Jinyuan Wang, Aug 10 2023