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Zhi-Wei Sun
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Natural numbers in the form $\lfloor\frac{a^3+b^3}2+\frac{c^3+d^3}6\rfloor$

Let $\mathbb N=\{0,1,2,\ldots\}$. Several years ago I proved that $$\{aw^3+bx^3+cy^3+dz^3:\ w,x,y,z\in\mathbb N\}\not=\mathbb N$$ for any positive integers $a,b,c,d$ (cf. http://maths.nju.edu.cn/~zwsun/179b.pdf ).

I'm curious whether there are positive integers $m$ and $n$ such that $$\left\lfloor\frac{a^3+b^3}m+\frac{c^3+d^3}n\right\rfloor=\mathbb N.$$ My computation suggests that $(m,n)=(2,6)$ might meet my purpose. Moreover, I have formulated the following conjecture.

Conjecture. Each $n\in\mathbb N$ can be written as the integral part of $(a^3+b^3)/2+(c^3+d^3)/6$ with $a,b,c,d\in\mathbb N$, $a\ge\max\{b,1\}$ and $c\ge\max\{d,1\}$.

I have verified this for all $n=0,\ldots,60000$. For example, $219$ has a unique required representation: $$219=\left\lfloor \frac{4^3+0^3}2 +\frac{10^3+5^3}6\right\rfloor.$$ For the number of ways to write $n\in\mathbb N$ in the given form, one may consult http://oeis.org/A343326.

QUESTION. Is the above conjecture true? How to prove it?

Your comments are welcome!

Zhi-Wei Sun
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