## The dynamics of a trick

### Proof

If $$D$$ is even, then the image of $$f_{D}$$ has $$\frac{19^{\frac{D}{2}}+1}{2}$$ elements.

If $$D$$ is odd, then the image of $$f_{D}$$ has $$\frac{19^{\frac{D-1}{2}}+1}{2}$$ elements.

Let us see two examples. If $$D=4$$ and $$n=abcd$$, with $$a\geq d$$ (it is sufficient to consider these cases because $$f_{D}$$ computes the difference between the biggest and the smallest number within the pair comprised by $$n$$ and $$i_{D}(n)$$), then $f_{4}(n)=\left|[10^{3} - 1](a-d) + [10^{2} - 10](b-c)\right|.$

Therefore:

• Either $$a=d$$, and thus $$f_{4}(n)= [10^{2} - 10]z_{2}$$ with $$z_{2}$$ in $$\{0,...,9\}$$, and we have $$10$$ possibilities for $$z_{2}$$, and hence the same number of possibilities for $$f_{4}(n)$$.
• Or $$a>d$$ (and there are $$9$$ possible values for $$a-d$$) and one has $f_{4}(n) = [10^{3} - 1](a-d) + [10^{2} - 10]z_{2}$ with $$z_{2}$$ being any element in the set $$\{-9,...,0,1,...,9\}$$, which gives $$9\times 19$$ possibilities for $$f_{4}(n)$$.

Hence, the image of $$f_{4}$$ contains $$10 + 9\times 19 = 1 + 9 + 9\times 19 = 181$$ elements. Note also that $$1 + 9 + 9\times 19 = \frac{19^{2} + 1}{2}$$.

If $$D=6$$ and $$n=abcdef$$ with $$a\geq f$$, then $f_{6}(n) = \left|[10^{5} - 1](a-f) + [10^{4} - 10](b-e) + [10^{3} - 10^{2} ](c-d)\right|$ and

• either $$a=f$$ and $$b=e$$, and one has $$f_{6}(n)= [10^{3} - 10^{2}]z_{3}$$, with $$z_{3}$$ in $$\{0,...,9\}$$, which gives $$10$$ possibilities for $$f_{6}(n)$$;
• or $$a=f$$ and $$b\neq e$$, and one has $$f_{6}(n)= [10^{4} - 10]z_{2} + [10^{3} - 10^{2}]z_{3}$$, with $$z_{2}$$ in $$\{0,...,9\}$$ and $$z_{3}$$ any element of $$\{-9,...,0,1,...,9\}$$, which gives $$9\times 19$$ elements in the image of $$f_{6}$$;
• or $$a>f$$ (there are $$9$$ possible values for $$a-f$$) and then $f_{6}(n) = [10^{5} - 1](a-f) + [10^{4} - 10]z_{2} + [10^{3} - 10^{2} ]z_{3}$ where $$z_{2}$$ and $$z_{3}$$ can be any element of $$\{-9,...,0,1,...,9\}$$, which gives $$9\times 19^{2}$$ elements in the image of $$f_{6}$$.

Consequently, the cardinal of the image of $$f_{D}$$ is $10 + 9\times 19 + 9\times 19\times 19 = 1 + 9 + 9\times 19+ 9\times 19^{2} = 3430.$ And one has $$1 + 9 + 9\times 19 + 9\times 19^{2} = 1 + 9\, (1 + 19 + 19^{2}) = \frac{19^{3} + 1}{2}$$.

Let us generalize these arguments to the set $$N_{D}$$. Since $f_{D}(x)=\left|x_{D-1}x_{D-2} ... x_{1}x_{0} - x_{0}x_{1} ... x_{D-2}x_{D-1}\right|,$ when $$D$$ is even we obtain $f_{D}(x)=\left|(10^{D-1} - 1)(x_{D-1} - x_{0})+ (10^{D-2} - 10)(x_{D-2} - x_{1})+ ... +\left (10^{\frac{D}{2}} - 10^{\frac{D}{2}-1}\right)\left(x_{\frac{D}{2}} - x_{\frac{D}{2}-1}\right)\right|$ and, when $$D$$ is odd, $f_{D}(x)=\left|(10^{D-1} - 1)(x_{D-1} - x_{0})+ (10^{D-2} - 10)(x_{D-2} - x_{1})+ ... + \left(10^{\frac{D+1}{2}} - 10^{\frac{D-3}{2}}\right)\left(x_{\frac{D+1}{2}} - x_{\frac{D-3}{2}}\right)\right|.$

Hence, in the image of $$f$$ we only find the nonnegative integers that can be written in the form $(10^{D-1} - 1)z_{1} + (10^{D-2} - 10)z_{2} +...+\left (10^{\frac{D}{2}} - 10^{\frac{D}{2}-1}\right)z_{\frac{D}{2}} \mbox{ if }D \mbox{ is even}$ $(10^{D-1} - 1)z_{1} + (10^{D-2} - 10)z_{2} +...+ \left(10^{\frac{D+1}{2}} - 10^{\frac{D-3}{2}}\right)z_{\frac{D-1}{2}} \mbox{ if }D \mbox{ is odd,}$ where $$0 \leq z_{1} \leq 9$$ and $$-9 \leq z_{j} \leq 9$$ for $$j>1$$.

Subject to these conditions only, we would obtain $$10\times 19^{\frac{D}{2}-1}$$ numbers when $$D$$ is even. Actually, the $$z_{i}´s$$ are not independent from each other. Let us see the case where $$D$$ is even (the other case is analogous). If there is some $$z_{j} \neq 0$$ and $$k = \mbox{minimum }\{0 \leq i \leq \frac{D}{2}: z_{i} \neq 0\}$$, then $$1 \leq z_{k} \leq 9$$ and $$-9 \leq z_{i} \leq 9$$ for $$i>k$$. Analyzing the possibilities for each value of $$k$$, we conclude that among the $$10^{D}$$ natural numbers with $$D$$ digits, only $10+9\times 19+9\times 19^{2}+...+9\times 10^{\frac{D}{2}-1}=\frac{19^{\frac{D}{2}}+1}{2}$ of them are in the image of $$f_{D}$$.

By a similar argument, we deduce that, for $$D$$ odd, the cardinal of that image is $$\frac{19^{\frac{D-1}{2}} + 1}{2}$$.

Notice that the frequency of numbers in the image of $$f_{D}$$ has limit $$0$$ when $$D$$ goes to infinity, since $\frac{19^\frac{D}{2} + 1}{ 2\times 10^{D}}< \left(\frac{1}{5}\right)^{\frac{D}{2}}.$