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A052427 Baxter-Hickerson numbers. 3
2, 64037, 6634003367, 666334000333667, 66663334000033336667, 6666633334000003333366667, 666666333334000000333333666667, 66666663333334000000033333336666667 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,1
COMMENTS
From Amiram Eldar, Nov 23 2020: (Start)
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)
REFERENCES
Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005. See p. 109.
LINKS
Lew Baxter, Cubes lacking zeros, sci.math newsgroup, April 22, 1999.
Ed Pegg, Jr., Cube conjecture, sci.math newsgroup, April 18, 1999.
Ed Pegg, Jr., Fun with Numbers, mathpuzzle websize.
Eric Weisstein's World of Mathematics, Baxter-Hickerson Function.
FORMULA
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
MAPLE
a(0) = 2, and 2^3 = 8 is zeroless.
a(1) = 64037, and 64037^3 = 262598918898653 is zeroless.
MATHEMATICA
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 *)
CROSSREFS
Subsequence of A052044.
Sequence in context: A059764 A285694 A306907 * A051833 A213619 A060895
KEYWORD
nonn
AUTHOR
EXTENSIONS
Offset changed to 0 by Amiram Eldar, Nov 23 2020
STATUS
approved

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Last modified April 25 07:07 EDT 2024. Contains 371964 sequences. (Running on oeis4.)