A364837 - OEIS

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

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OFFSET

0,1

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.

LINKS

Table of n, a(n) for n=0..87.

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

FORMULA

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