The thinking behind "4 times 5 is 12, and 4 times 6 is 13, and 4 times 7 is-oh dear! I shall never get to 20 at that rate!" from Alice in Wonderland?

This blog has a suggestion

The idea is that she is doing calculations in base 10 but the answers are coming out in different bases ...

$4 \times 5 = 12 $ ( in base 18 )
$4 \times 6 = 13 $ ( in base 21 )
$4 \times 7 = 14 $ ( in base 24 )

So she is expressing $4n$ in base $3+3n$

indeed she can't get to 20 that way - if she could, $n$ would be a solution to $4n=6+6n$ which has solution $n=-3$

In theory there are other possibilities...

Imagine that

$$x\circ y=5x-8^{6-y}7^{y-5}$$

and it reads as $x$ times $y$.

So,

$$4\circ y=20-8^{6-y}7^{y-5}.$$

In concreto $$4\circ 5=20-8\cdot7^{0}=12 \text{ and } 4\circ 6=20-8^{0}7^{1}=13$$

(and $4\circ 7=20-8^{-1}7^{2}=20-\frac{49}8...)$

PS: It can be shown easily that $x\circ y$ $(y=5,6,7,...)$ will never reach $4x$.

We have $4\cdot 5=20,\ldots ,4\cdot 10=40$. Assuming that one "has to stop" there (because one does stop there in elementary school), it would correspond to $12,13,14,15,16$ and $17$ in the "wrong" counting. So we will never reach $20$. Of course, there are many other possibilities.