[seqfan] Re: Is the set of numbers representable by sum of 5 positive cubes in exactly 3 ways finite?

[Sean A. Irvine]

Hi Hugo,

Thank you for this.  It would seem to suggest that a great many of these
sequences could be finite (at least for cubes).

Are you able from your results to get a count of the total for each case --
i.e., simply assuming they become finite and have no further terms after
the last non-zero value you observe for that number of representations.
I'm interested in the shape of the curve of number of representations (0,
1, 2, 3, ...) versus count of values achieving that.  Maybe that would
match up with some of theoretical considerations.

Sean.



On Sun, 16 May 2021 at 05:06, <hv at crypt.org> wrote:

> I've generated files that count the number of ways that n can be
> represented
> as a sum of (2, 3, 4, 5) cubes up to n=10^9, capped at 255 representations.
>
> For lack of a better way to slice up the results, I looked at them grouped
> by cubes, ie each group has k^3 < n <= (k+1)^3. Tabulating the number of
> representations for 5 cubes in this way starts:
> 1: 1       # 1..1, no representations
> 2: 6 1     # 2..8, 6 numbers with no representations, 1 with one
> 3: 16 3
> 4: 28 9
> 5: 44 17
> 6: 55 35 1
> 7: 79 34 14
> 8: 81 73 15
> 9: 91 102 24
> 10: 108 108 48 7
>
> Later rows show clearly that we're moving towards having lots of
> representations for each number. By eye, the last row with an example
> having 3 representations is 131 (ie the range 2197001-2248091), which
> starts:
>   131: 0 0 1 0 0 2 3 21 35 64 118 184 250
>
> By the time we get to row 400, we see no example having fewer than
> 84 representations, and the last row with an example having fewer
> than 255 representations is 658.
>
> I'm happy to send any of the resulting files to interested parties;
> they are (chosen to be) 1GB uncompressed, but compress down to about 35MB.
> I can also generate the same using 2 or 4 bytes per number to avoid the cap
> if that would be more useful: it took about 40 minutes to generate up to
> 1000^3, and generating up to k^3 seems to be cost time proportional to k^4.
>
> Hugo
>
> "D. S. McNeil" <dsm054 at gmail.com> wrote:
> :If we weaken positive cubes to nonnegative cubes, Deshouillers, Hennecart,
> :and Landreau (2000) give numerical and heuristic evidence that all numbers
> :past 7373170279850 are representable as the sum of 4 nonnegative cubes.
> :
> :So if they're right, then eventually we can just take some N and represent
> :each of (N-1^3, N-2^3, N-3^3, N-4^3) as the sum of four cubes and then
> take
> :1^3, 2^3, 3^3, 4^3 as our fifth cube, giving at least four 5-cube
> :representations for N.
> :
> :So I'd bet a fair amount that the set of numbers representable by the sum
> :of 5 positive cubes in exactly three ways is indeed finite.
> :
> :
> :Doug
> :
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