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Search: id:A135998
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| A135998 |
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Smallest error in trying to solve n^3 = x^3 + y^3. That is, for each n, find positive integers x <= y < n such that | n^3 - x^3 - y^3 | is minimal, and let a(n) := n^3 - x^3 - y^3. |
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+0
1
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| 6, 11, 10, -3, 27, 2, 44, 1, -24, -12, -1, -43, 16, -81, -8, -28, 8, 19, -29, 54, 56, 71, -8, 64, 69, 27, 72, -46, -133, 47, -64, 161, -8, 79, -27, -99, -57, -263, -133, 8, 254, -62, -155, 109, -15, -56, -64, 2, 259, 107, -17, 269, 216, -78, -20, 316, 164, -28, -27, 333, 181, 47, -70, 6, 704, 63, -64, 253, 343, -389, -216
(list; graph; listen)
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OFFSET
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2,1
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COMMENT
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a(n) is never zero, by Fermat's last theorem for cubes. There are infinitely many n for which a(n)=1,-1, and 2. It is not known if a(n) is ever 3, besides
a(5). By congruence considerations, a(n) is never +-4 mod 9. Presumably a(n) is roughly of order n.
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LINKS
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Daniel Bernstein, Representations using three cubes.
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EXAMPLE
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a(7) = 2 because 7^3 - 5^3 - 6^3 = 2, and this can't be improved,
a(12) = -1 because 12^3 - 9^3 - 10^3 = -1, and this can't be improved.
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CROSSREFS
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Sequence in context: A106540 A134012 A103704 this_sequence A015870 A004471 A046953
Adjacent sequences: A135995 A135996 A135997 this_sequence A135999 A136000 A136001
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KEYWORD
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sign
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AUTHOR
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Moshe Newman (mshnoiman(AT)hotmail.com), Mar 03 2008
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