## The dynamics of a trick

### Conjectures I

Conjecture 1

In base $$2$$, for every $$D$$, all cycles in $$N_{D}$$ have period $$1$$.

Conjecture 2

Let us analyze the following figures and tables, depicting the dynamics in base $$2$$.

• $$D=4$$

Cycles and precycles of $$f_{4,2}$$.

$$D$$ $$4$$
Number of cycles $$4$$
Number of preimages of each cycle $$2,4,4,6$$
$$\mbox{Total}=2\,(1+2+2+3)=16=2^{4}$$
Sum (in base 10) of the numbers of preimages of each cycle $$15,30,30,45$$
$$\mbox{Total}=15\,(1+2+2+3)$$ $$=15\times 8$$ $$=\left(2^{4}-1\right)2^{3}$$

• $$D=6$$

Cycles and precycles of $$f_{6,2}$$.

$$D$$ $$6$$
Number of cycles $$8$$
Number of preimages of each cycle $$2,$$ $$4,$$ $$4,$$ $$4,$$ $$10,$$ $$10,$$ $$14,$$ $$16$$
$$\mbox{Total}=2\,(1+2+2+2+5+5+7+8)$$ $$=64$$ $$=2^{6}$$
Sum (in base 10) of the numbers of preimages of each cycle $$63,$$ $$126,$$ $$126,$$ $$126,$$ $$315,$$ $$315,$$ $$441,$$ $$504$$
$$\mbox{Total}=63\,(1+2+2+2+5+5+7+8)$$ $$=63\times 32$$ $$=\left(2^{6}-1\right)2^{5}$$

• $$D=8$$

Cycles and precycles of $$f_{8,2}$$.

$$D$$ $$8$$
Number of cycles $$16$$
Number of preimages of each cycle $$2,$$ $$4,$$ $$4,$$ $$4,$$ $$4,$$ $$8,$$ $$10,$$ $$10,$$ $$12,$$ $$12,$$ $$16,$$ $$24,$$ $$24,$$ $$34,$$ $$34,$$ $$54$$
$$\mbox{Total}=2\,(1+2+2+2+2+4+5+5+6+6+8+12+12+17+17+27)$$ $$=256$$ $$=2^{8}$$
Sum (in base 10) of the numbers of preimages of each cycle $$255,$$ $$510,$$ $$510,$$ $$510,$$ $$510,$$ $$1020,$$ $$1275,$$ $$1275,$$ $$1530,$$ $$1530,$$ $$2040,$$ $$3060,$$ $$3060,$$ $$4335,$$ $$4335,$$ $$6885$$
$$\mbox{Total}=255\,(1+2+2+2+2+4+5+5+6+6+8+12+12+17+17+27)$$ $$=255\times 128$$ $$=\left(2^{8}-1\right)2^{7}$$

More generally, in base 2:

• there is one and only one (fixed) cycle with two pre-images, $$F_{1} = \{0101...01\};$$ notice that the sum (in base 10) of the numbers of the preimages of this cycle is $$2^{D} - 1;$$
• in the remaining $$2^{\frac{D}{2}} - 1$$ cycles, which we denote by $$F_{2},...,F_{m}$$, where $$m=2^{\frac{D}{2}}-1$$, each with $$2p_{i}$$ preimages (an even number because, with the exception of the natural numbers which are equal to their reverse and of which we know there is an even number, a number and its reverse belong to the preimage of the same cycle) and whose numbers have sum (in base $$10$$) equal to $$S_{i}$$, one has $S_{i} = \left(2^{D} - 1\right)p_{i}$ and $\sum_{i=1}^{m}p_{i}=2^{D-1}.$

In particular, we conclude that the sum of all preimages of the $$2^{\frac{D}{2 }}$$ fixed cycles is $\sum_{i=1}^{m}\left(2^{D}-1\right)p_{i}=\left(2^{D}-1\right)2^{D-1}=\frac{\left(2^{D}-1\right)2^{D}}{2}$ $= \mbox{sum of all naturals from }1 \mbox{ to }2^{D}-1.$

Hence, the proof of this conjecture ensures that the first conjecture is also true.

Note: An analogous property, about the number of preimages of the fixed cycles and their respective sum, seems to hold for other bases.