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A052427
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Baxter-Hickerson numbers.
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3
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OFFSET
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0,1
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COMMENTS
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Named after Lew Baxter and Dean Hickerson.
Pegg (1999) conjectured that the sequence of zeroless cubes (A052045) is finite. On April 19, 1999, Hickerson gave the counterexample: if n == 2 (mod 3) and n >= 5, then the cube of (2*10^(5*n) - 10^(4*n) + 17*10^(3*n-1) + 10^(2*n) + 10^n - 2)/3 is zeroless. Three days later, Baxter gave a simpler variation which is valid for all n>=0 and is given in the Formula section. (End)
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REFERENCES
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Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005. See p. 109.
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LINKS
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FORMULA
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a(n) = (2*10^(5*n) - 10^(4*n) + 2*10^(3*n) + 10^(2*n) + 10^n + 1)/3 (Baxter, 1999). - Amiram Eldar, Nov 23 2020
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MAPLE
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a(0) = 2, and 2^3 = 8 is zeroless.
a(1) = 64037, and 64037^3 = 262598918898653 is zeroless.
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MATHEMATICA
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a[n_] := (2*10^(5*n) - 10^(4*n) + 2*10^(3*n) + 10^(2*n) + 10^n + 1)/3; Array[a, 10, 0] (* Amiram Eldar, Nov 23 2020 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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