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A241299
Initial digit of the decimal expansion of n^(n^n) or n^^3 (in Don Knuth's up-arrow notation).
16
0, 1, 1, 7, 1, 1, 2, 3, 6, 4, 1, 3, 4, 6, 1, 3, 1, 3, 1, 1, 3, 2, 3, 5, 5, 2, 2, 2, 8, 1, 1, 9, 1, 2, 3, 4, 8, 2, 4, 1, 1, 2, 8, 3, 2, 1, 4, 2, 5, 1, 6, 7, 2, 2, 2, 2, 2, 2, 8, 4, 1, 4, 8, 1, 5, 8, 4, 1, 4, 1, 2, 1, 9, 6, 6, 2, 1, 1, 7, 6, 1, 7, 7, 2, 4, 1, 8, 6, 1, 7, 1, 1, 3, 1, 2, 6, 3, 5, 1, 1, 1, 2, 2, 5, 4
OFFSET
0,4
COMMENTS
0^^3 = 0 since 0^^k = 1 for even k, 0 for odd k, k >= 0.
Conjecture: the distribution of the initial digits obey Zipf's law.
The distribution of the first 1000 terms beginning with 1: 302, 196, 124, 91, 72, 46, 71, 53, 45.
LINKS
Robert P. Munafo and Robert G. Wilson v, Table of n, a(n) for n = 0..1000
Cut the Knot.org, Benford's Law and Zipf's Law, A. Bogomolny, Zipf's Law, Benford's Law from Interactive Mathematics Miscellany and Puzzles.
Hans Havermann, Next 5 terms.
Eric Weisstein's World of Mathematics, Joyce Sequence.
Wikipedia, Zipf's law.
FORMULA
For n > 0, a(n) = floor(t/10^floor(log_10(t))) where t = n^(n^n).
a(n) = A000030(A002488(n)). - Omar E. Pol, Jul 04 2019
EXAMPLE
a(0) = 0, a(1) = 1, a(2) = 1 because 2^(2^2) = 16, a(3) = 7 because 3^(3^3) = 7625597484987 and its initial digit is 7, etc.
MATHEMATICA
g[n_] := Quotient[n^p, 10^(Floor[ p*Log10@ n] - (1004 + p))]; f[n_] := Block[{p = n}, Quotient[ Nest[ g@ # &, p, p], 10^(1004 + p)]]; Array[f, 105, 0]
KEYWORD
nonn,base,easy
AUTHOR
STATUS
approved