If we look at the digits in some base $d$, and divide the numbers made by the string of digits from high to low and low to high, the ratio appears to approach $(d-2)$ as $d$ gets larger.
Base 2: 1/1 = 1/1 = 1
Base 3: 21/12 = 7/5 = 1.4
Base 4: 321/123 = 57/27 = 2.11111
Base 10: 987654321/123456789 = 8.000000073
Base 16: FEDCBA987654321/... = 13.99999953
How would you go about proving the limit? The best I can do for a general form is,
$$\lim\limits_{d\rightarrow\infty}\frac{\sum\limits_{x=1}^{d-1}{xd^{x-1}}}{\sum\limits_{x=1}^{d-1}(d-x)d^{d-x-1}} - d \stackrel{?}{=} -2$$
But I have no idea how to proceed.
The thing you want to calculate is given by $$ R=\frac{\sum _{n=1}^{d-1} n d^{n-1}}{\sum _{n=1}^{d-1} n d^{d-n-1}} $$ The top sum is $$ \sum _{n=1}^{d-1} n d^{n-1}=\frac{d^{d+1}-2 d^d+1}{(d-1)^2} $$ While the bottom sum is $$ \sum _{n=1}^{d-1} n d^{d-n-1}=\frac{d^d-d^2+d-1}{(d-1)^2} $$ These sums can be computed by adapting the formulae here (https://en.wikipedia.org/wiki/List_of_mathematical_series#Low-order_polylogarithms). This gives: $$ R=\frac{d^{d+1}-2 d^d+1}{d^d-d^2+d-1} $$ When $d$ gets very large, all terms not to a power of $d$ are negligibly small. So we can write $$ R(d\gg1)\approx \frac{d^{d+1}-2d^d}{d^d}=\frac{dd^{d}-2d^d}{d^d}=d-2 $$