How to prove that any natural number $n \geq 34$ can be written as the sum of distinct triangular numbers? Sloane's A053614 implies that $2, 5, 8, 12, 23$, and $33$ are the only natural numbers $n \geq 1$ which cannot be written as the sum of distinct triangular numbers (i.e., numbers of the form $\binom{k}{2}$, beginning $1,3,6,10,15,\ldots$).
Question: How to prove that any natural number $n \geq 34$ can be written as the sum of distinct triangular numbers?
We have


*

*$34=\binom{8}{2}+\binom{4}{2}$,

*$35=\binom{8}{2}+\binom{4}{2}+\binom{2}{2}$,

*$36=\binom{9}{2}$,

*$37=\binom{9}{2}+\binom{2}{2}$,

*$38=\binom{8}{2}+\binom{5}{2}$,


and so on.
Sloane's A061208 links to a math olympiad question (page 207) which asks to prove this for $n \leq 1997$, but the given proof is not in English, so I neither understand it, nor can I be sure if it works for all $n$.
 A: This follows from a theorem of Richert:

Theorem Suppose that $k \ge 2, N \ge 0, M \ge0$ satisfy


*

*Whenever $N<x \le N + M$, $x$ is a sum of distinct elements of some of the first $k$ elements of a set $S = \{s_1, s_2, \ldots\}$, where $s_1 < s_2 < \cdots$.

*$M \ge s_{k+1}$

*$2 s_i \ge s_{i+1}$
Then every integer greater than $N$ is a sum of distinct elements of $S$.

Take $k=8, N=33, M=45$ to obtain the desired result.

Proof of Theorem To prove the theorem, let $I_p = \{N+1, N+2, \ldots, N + s_{p+1}\}$. Then by assumption, all elements of $I_k$ are the sum of the first $k$ elements $s_1, \ldots, s_k$.
But now observe that if this is true for some general $p$, then $$I_p \cup \{m + s_{p+1} : m \in I_p\}$$ contains $I_{p+1}$, as a consequence of $s_{p+2} \le 2 s_{p+1}$. Hence all elements of $I_{p+1}$ are sums of the $s_1, \ldots, s_{p+1}$.
Hence inductively the result follows by considering $\bigcup_{p\ge k} I_k$, which contains all integers larger than $N$, and contains only elements which are distinct sums of $s_i$.
