

A135998


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.


1



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
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OFFSET

2,1


COMMENTS

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.


LINKS

Table of n, a(n) for n=2..72.
Daniel Bernstein, Representations using three cubes.


EXAMPLE

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.


MATHEMATICA

a[n_] := SortBy[n^3Flatten[Table[x^3+y^3, {x, n1}, {y, x}]], Abs][[1]];
Table[a[n], {n, 2, 72}] (* JeanFrançois Alcover, Jul 05 2019, after Giovanni Resta in A308834 *)


CROSSREFS

Sequence in context: A334280 A134012 A103704 * A242825 A358068 A276136
Adjacent sequences: A135995 A135996 A135997 * A135999 A136000 A136001


KEYWORD

sign


AUTHOR

Moshe Shmuel Newman, Mar 03 2008


STATUS

approved



