Let $\mathbb N=\{0,1,2,\ldots\}$. Recall that the triangular numbers are those natural numbers
$$T_x=\frac {x(x+1)}2\quad \text{with}\  x\in\mathbb N.$$
As $T_x=\binom{x+1}2$, Gauss' triangular number theorem (first claimed by Fermat) can be restated as follows:
$$\left\{\binom x2+\binom y2+\binom z2:\ x,y,z\in\mathbb N\right\}=\mathbb N.$$

In view of the above, here I pose a conjecture on representations of integers involving binomial coefficients.

**Conjecture.** Any positive integer $n$ can be written as 
$$\binom{w}2+\binom{x}4+\binom{y}6+\binom{z}8\quad \text{with}\ w,x,y,z\in\{2,3,\ldots\}.$$

Observe that
$$\frac12+\frac14+\frac16+\frac18=\frac{25}{24}\approx  1.0416667.$$
I have verified the above conjecture for all $n=1,\ldots,2\times10^7$. For example, 
$$4655=\binom{85}2+\binom{14}4+\binom 96+\binom 78=\binom{94}2+\binom 74+\binom 96+\binom{11}8$$
and
$$192080=\binom{7}2+\binom{26}4+\binom{25}6+\binom{9}8=\binom{414}2+\binom{39}4+\binom 86+\binom{17}8.$$
See http://oeis.org/A306477 for related data. Note that $1061619$ is the first positive integer not representable as $\binom w2+\binom x4+\binom y6+\binom z9$ with $w,x,y,z\in\mathbb N$.

I also have some other similar conjectures. See http://oeis.org/A306460, http://oeis.org/A306462 and http://oeis.org/A306471. For example, I conjecture that $$\left\{\binom{2x}2+\binom y2+\binom z3:\ x,y,z=1,2,3,\ldots\right\}=\{1,2,3,\ldots\}$$
and
$$\left\{2\binom w3+\binom x3+\binom y3+\binom z3:\ w,x,y,z\in\mathbb N\right\}=\mathbb N.$$
The last equality implies Pollock's conjecture which states that any positive integer is the sum of at most five tetrahedral numbers. 

Your comments are welcome!