You are perhaps confused because we are subtracting then then and so on. Instead, you can think of it like this. Take all the subsets of size from and take their sum. Then take the generating function for that and remove the largest power of that you can (which will be ).

For example, with and we have

$$ \begin{array}{c|c} \text{subset} & \text{sum} \\\hline 123 & 6 \\ 124 & 7 \\ 125 & 8 \\ 134 & 8 \\ 135 & 9 \\ 145 & 10 \\ 234 & 9 \\ 235 & 10 \\ 245 & 11 \\ 345 & 12 \end{array} $$

Subsets aren't ordered so etc. The sum also doesn't care about the order.

Now we take the generating function

which is

This simplifies to

The quantity

is the q-binomial coefficient .

It seems that an asymptotics could be $$\binom{2n}{n}_q \sim C_q\, q^{n^2}$$ Computing the coefficient for $n=1000$, the values are $$\left( \begin{array}{cc} q & C_q \\ 2 & 3.462746619 \\ 3 & 1.785312342 \\ 4 & 1.452353642 \\ 5 & 1.315213556 \\ 6 & 1.241175663 \\ 7 & 1.195035240 \\ 8 & 1.163594397 \\ 9 & 1.140822757 \\ 10 & 1.123582755 \\ 11 & 1.110084028 \\ 12 & 1.099231752 \\ 13 & 1.090319360 \\ 14 & 1.082870737 \\ 15 & 1.076553491 \\ 16 & 1.071128609 \\ 17 & 1.066419860 \\ 18 & 1.062294483 \\ 19 & 1.058650573 \\ 20 & 1.055408622 \end{array} \right)$$

Considering the last term in your lists, this would give $116190496$ and $267965804863721413$.