Representations with Triangular Numbers A well known theorem of Gauss says that any natural number $n$ may
be written as the sum of three triangular numbers - 
$$
n={a_{1} \choose 2}+{a_{2} \choose 2}+{a_{3} \choose 2}
$$
The following question came up over the course of my work - Does there
exist a (slowly growing) function $\omega(n)$ which tends to infinity
with $n$ such that any natural number $n$ may be written in the
following form - 
$$
n=\sum_{i=1}^{k}{a_{i} \choose 2}\mbox{ subject to }\prod_{i=1}^{k}a_{i}\leq\frac{n}{\omega(n)}
$$
If we choose the largest possible $a_{1}$, then the largest possible
$a_{2}$ etc; we can guarantee $\prod_{i=1}^{k}a_{i}\leq Cn$ for some absolute constant $C$ (NOT QUITE - SEE BELOW). The question then is, can we do any better?
[EDIT 1: In my rush to post this, I did not check my calculations very well - if you actually do what is described, you only get $\prod_{i=1}^{k}a_{i} \leq 2^{\log_2\log_2 n}n^{1/2+1/4...} = O(n\log n)$ (as Erick Wong and Emil Jeřábek pointed out). The question of how small you can make $\prod_{i=1}^{k}a_i$ still remains.]
[EDIT 2: Will Jagy has computed what the best possible product for $n\leq10^7$ and based off this data, the existence of an $\omega$ as asked above seems unlikely. That for some infinite family of natural numbers $n$, the product must be $\Theta(n)$ is a plausible conjecture - though how one might prove something like that is not clear to me.]
 A: The sequence $a(n)$ of minimal products (starting, for convenience, at $n=0$), which Will Jagy found to begin 
$$1,2,4,3,6,12,4,8,16,12,5,10,\ldots$$
can be defined using the recursion
$$a(n) = \min\left( 2a(n-1),3a(n-3),4a(n-6),5a(n-10)\ldots\right) $$
for $n>0$ (with the understanding that $a(n)$ is undefined for $n<0$).  For example, 
$$a(11) = \min(2a(10),3a(8),4a(5),5a(1))=  \min(10,48,48,10)=10.$$
This should allow a reasonably rapid tabulation up to around $n=10000$.
A: If $t$ is a triangular number then $8t+1$ is an odd square, and vice versa. So writing $n$ as a sum of $k$ triangular numbers all at least $1$ becomes writing $8m+k$ as a sum of odd squares at least $9$ $$8n+k=\sum_1^k(2c_i+1)^2.$$ I wonder if any insight arises from expressing $N$ as a sum of odd squares so as to minimize $\frac{\prod (2c_i+1)}{N}$ and if the extremes occur at about $8n$ for $n$ extreme to this triangular numbers problem. If so then perhaps some insight is available. If we instead try to minimize $\frac{\prod (c_i)}{N}$ then we have an exact match.
For example the solutions, in increasing increasing order of merit, to $\binom{a}{2}+\binom{b}{2}+\binom{c}{2}=52$ are 
$[7,7,5][9,5,4],[8,7,3],[9,6,2],[10,4,2]$ with products $245,180,168,108,80.$
Since $8\cdot52+3=419$ we have that the same triples are the solutions to $(2a+1)^2+(2b+1)^+(2c+1)^2=419$
Alternately the "odd" solutions to $u^2+v^2+w^2=419$ are $[13,13,9],[17,9,7],[15,13,5],[17,11,3],[19,7,3]$ with products $1521, 1071, 975, 561, 399.$
Of course sometime we need to consider $8n+4$ as a sum of four odd squares. So maybe this helps or maybe not. However the extreme cases found so far seem to all come from $3,4$ or $5$ squares. I wonder if that could be proved. If so perhaps the squares perspective would be helpful.
I did not find any of the triangle problem sequences in the OEIS. For example $2, 5, 20, 119, 230, 7259, 26795$ is record breakers for a greedy triangular number partition problem 
Consider the sequence $2,2,3,6,21,231,\cdots$ defined by $a_1=a_2=2$ and $a_{k+1}=\binom{a_k+1}{2}$ for $k \ge 2$ Also let $b_i=\binom{a_i}{2}$ and $n_i=\sum_1^ib_j.$ So $n_i$ starts out $2,5,20,230,26795.$ The $n_i$ are half the terms of the sequence I mentioned above, I.E. half the record values of $\frac{\prod{a_i}}{n}$ alternating the lead with the similar series $2, 4, 14, 119, 7259,\cdots$ generated by $2,3,5,15,120.$  (I think this is correct after some initial irregularities, I might be wrong.) By my very rough calculations, $a_i \approx 2^{2^{i-2}}$ while $b_i \approx n_i \approx 2^{2^{i-1}}.$ The greedy algorithm expresses $n_i$ as $\sum_0^ib_j=\sum_0^i \binom{a_i}{2}$ with $\prod a_j \approx \log_2(n_i)n_i$. 
