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A364837
Initial digit of 2^(2^n) = A001146(n).
3
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
0,1
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
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.
a(n) = A000030(A001146(n)).
EXAMPLE
a(5) = 4, since 2^(2^5) = 2^32 = 4294967296.
MATHEMATICA
Join[{2}, Table[Floor[2^(2^n)/10^Floor[Log10[2^(2^n)]]], {n, 27}]] (* Stefano Spezia, Aug 10 2023 *)
PROG
(Python)
def A364837(n): return int(str(1<<(1<<n))[0]) # Chai Wah Wu, Sep 14 2023
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Marco Ripà, Aug 10 2023
EXTENSIONS
More terms from Jinyuan Wang, Aug 10 2023
STATUS
approved