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A364837
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Initial digit of 2^(2^n) = A001146(n).
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3
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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
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0,1
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COMMENTS
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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.
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LINKS
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FORMULA
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a(n) = floor(2^(2^n)/10^floor(log_10(2^(2^n)))), for n > 0.
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EXAMPLE
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a(5) = 4, since 2^(2^5) = 2^32 = 4294967296.
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MATHEMATICA
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Join[{2}, Table[Floor[2^(2^n)/10^Floor[Log10[2^(2^n)]]], {n, 27}]] (* Stefano Spezia, Aug 10 2023 *)
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PROG
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(Python)
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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