So we add the condition n>=6. Part of 'sufficiently large'. --Rich --- Quoting Michael Kleber <michael.kleber@gmail.com>:
Ah, sorry, right. I meant to subtract the smallest cube larger than half the original number, so that what was left is smaller than the term you subtracted.
--Michael On Feb 19, 2012 11:20 AM, "Dan Asimov" <dasimov@earthlink.net> wrote:
If I'm understanding: What ensures that n^3 - (n-1)^3 won't require another (n-1)^3 in its sum of cubes?
--Dan
<< Sure -- just let n^3 - (n-1)^3 be sufficiently large as well, no?
--Michael
On Feb 19, 2012 10:48 AM, "Dan Asimov" <dasimov@earthlink.net> wrote: In what follows, "cube" means the cube of a positive integer.
It is known that every sufficiently large integer is the sum of distinct cubes. See < http://oeis.org/A001476 >.
QUESTION: Is every sufficiently large integer the sum of *more than one* distinct cubes?
--Dan
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