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It is known that every number can be represented by a sum of $3$ triangular numbers. According to Gauss (see formula $35$ in mathworld article) $$ \text{num}=\Delta+\Delta+\Delta $$ I did some numerical experiments that suggest the above formula is correct when triangular numbers are replaced by tetrahedral numbers $$ \Delta=\frac{n(n+1)(n+2)}6 $$ if $n$ is allowed to be negative.

Is this conjecture correct?

I tried to google representation of integers by tetrahedral numbers but didn't find anything.

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    $\begingroup$ For positive tetrahedral numbers, this is a significant strengthening of Pollock's tetrahedral numbers conjecture: Every positive integer is the sum of at most five tetrahedral numbers. $\endgroup$ – Eric Towers Oct 14 '17 at 17:18
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    $\begingroup$ This conjecture is equivalent to Every element of $6\mathbb{Z}$ can be represented as $(x^3+y^3+z^3)-(x+y+z)$ and it probably is a very difficult problem. This is an instance of a similar problem: it is not known if $33$ can be represented as the sum of three integer cubes. $\endgroup$ – Jack D'Aurizio Oct 14 '17 at 17:25
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    $\begingroup$ The question has now been raised on mathoverflow, mathoverflow.net/questions/325659/… $\endgroup$ – Gerry Myerson Mar 19 at 0:53
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    $\begingroup$ @Jack, $33$ has been done very recently, by Andrew Booker. See, e.g., aperiodical.com/2019/03/… $\endgroup$ – Gerry Myerson Mar 19 at 0:56
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If $\Delta(n)$ for negative $n$ is allowed, then certainly the integers $t=0\dots 10000$ are all possible. The most awkward of these is $t=6398=\Delta(-1121877)+\Delta(1037512)+\Delta(665832)$. The size of the summands might give you an idea of the size of the task of seeking an explicit solution for each total $t$.

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  • $\begingroup$ Thanks for you answer. I did only verify it to $t=150$. $\endgroup$ – Tyrell Oct 15 '17 at 14:03

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