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> 2017 can be written as a sum of cubes of five distinct integers. This gives no results in SageMath...

by ComodoHacker 10y ago

cperciva 10y ago
The greedy algorithm works here: 2017 = 12^3 + 6^3 + 4^3 + 2^3 + 1^3
- gus_massa 10y ago
  Technical note: 5 cubes is not enough for every number, so this property of 2017 is not trivial. From > Every positive integer can be written as the sum of nine (or fewer) positive cubes. This upper limit of nine cubes cannot be reduced because, for example, 23 cannot be written as the sum of fewer than nine positive cubes: > 23 = 2^3 + 2^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3. I couldn't find if it's common that 5 cubes is enough. [This looks like a nice exercise for the reader.]
  - Someone 10y ago
    http://oeis.org/A003328 http://oeis.org/A003328: Numbers that are the sum of 5 positive cubes 5, 12, 19, 26, 31, 33, 38, 40, 45, 52, 57, 59, 64, 68, 71, 75, 78, 82, 83, 89, 90, 94, 96, 97, 101, 108, 109, 115, 116, 120, 127, 129, 131, 134, 135, 136, 138, 143, 145, 146, 150, 152, 153, 155, 157, 162, 164, 169, 171, 172, 176, 181, 183, 188, 190, 192, 194 It seems this is fairly common (1757 is the 1000th such number), but of course that says nothing. Reading http://mathworld.wolfram.com/CubicNumber.html http://mathworld.wolfram.com/CubicNumber.html, it is true that every sufficiently large integer is a sum of no more than 7 positive cubes. It also states ”the only integers requiring nine positive cubes are 23 and 239. Wieferich proved that only 15 integers require eight cubes: 15, 22, 50, 114, 167, 175, 186, 212, 231, 238, 303, 364, 420, 428, and 454 (OEIS A018889).” Even stronger (same page): ”Deshouillers et al. (2000) conjectured that 7373170279850 is the largest integer that cannot be expressed as the sum of four nonnegative cubes” (nice title for a paper: ”7 373 170 279 850.”. See http://www.ams.org/journals/mcom/2000-69-229/S0025-5718-99-01116-3/S0025-5718-99-01116-3.pdf http://www.ams.org/journals/mcom/2000-69-229/S0025-5718-99-0...) If that is true, it is indeed common that 5 cubes is enough (since 4 almost always would be sufficient)
    - ahakki 10y ago
      yes, but 2017 is the sum of five destinct integer cubes
tjwei 10y ago
And 2017 = 2^3 + 4^3 + 6^3 + 9^3 + 10^3

> 2017 can be written as a sum of cubes of five distinct integers. This gives no results in SageMath...

cperciva 10y ago

The greedy algorithm works here: 2017 = 12^3 + 6^3 + 4^3 + 2^3 + 1^3

gus_massa 10y ago
Technical note: 5 cubes is not enough for every number, so this property of 2017 is not trivial. From > Every positive integer can be written as the sum of nine (or fewer) positive cubes. This upper limit of nine cubes cannot be reduced because, for example, 23 cannot be written as the sum of fewer than nine positive cubes: > 23 = 2^3 + 2^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3. I couldn't find if it's common that 5 cubes is enough. [This looks like a nice exercise for the reader.]
- Someone 10y ago
  http://oeis.org/A003328 http://oeis.org/A003328: Numbers that are the sum of 5 positive cubes 5, 12, 19, 26, 31, 33, 38, 40, 45, 52, 57, 59, 64, 68, 71, 75, 78, 82, 83, 89, 90, 94, 96, 97, 101, 108, 109, 115, 116, 120, 127, 129, 131, 134, 135, 136, 138, 143, 145, 146, 150, 152, 153, 155, 157, 162, 164, 169, 171, 172, 176, 181, 183, 188, 190, 192, 194 It seems this is fairly common (1757 is the 1000th such number), but of course that says nothing. Reading http://mathworld.wolfram.com/CubicNumber.html http://mathworld.wolfram.com/CubicNumber.html, it is true that every sufficiently large integer is a sum of no more than 7 positive cubes. It also states ”the only integers requiring nine positive cubes are 23 and 239. Wieferich proved that only 15 integers require eight cubes: 15, 22, 50, 114, 167, 175, 186, 212, 231, 238, 303, 364, 420, 428, and 454 (OEIS A018889).” Even stronger (same page): ”Deshouillers et al. (2000) conjectured that 7373170279850 is the largest integer that cannot be expressed as the sum of four nonnegative cubes” (nice title for a paper: ”7 373 170 279 850.”. See http://www.ams.org/journals/mcom/2000-69-229/S0025-5718-99-01116-3/S0025-5718-99-01116-3.pdf http://www.ams.org/journals/mcom/2000-69-229/S0025-5718-99-0...) If that is true, it is indeed common that 5 cubes is enough (since 4 almost always would be sufficient)
  - ahakki 10y ago
    yes, but 2017 is the sum of five destinct integer cubes

tjwei 10y ago

And 2017 = 2^3 + 4^3 + 6^3 + 9^3 + 10^3