# Capere sponsione ranarum0518

Exultat rana numerum recta conantur capit. Sitque semper saltare capiendisque proferimus. Rana animi ad statum $$s \in \mathbb{Z}$$ et omnis moventur ea cerebrosus prosilit procul a $$z \in \mathbb{Z}$$ (si $$z>0$$ , ut cerebrosus prosilit vade ad dextram sive ad sinistram, si aliter). $$z$$ idem est, quia omne jump. Heres ex ratione status integer. Nec scit $$z$$ vel $$s$$ . Ostendit nobis quia non est ita de respondente Rana sed semper capit.

Primum omnium, $$a_1 = s$$ et $$a_{n+1} = a_n + z = s + n \cdot z$$ et $$s,z \in \mathbb{Z}$$ .

Nos iam elige

$$h:\mathbb{N} \to \mathbb{Z}^2: h(2^k r) = \left ( (-1)^{k+1} \left \lfloor \frac{k+1}{2} \right \rfloor, (-1)^{\frac{r+1}{2}} \left \lfloor \frac{r+1}{4} \right \rfloor \right )$$

functionis assignatis pro (sicut) Omne numero plurium numerorum integrorum tuple. Et elegit de hoc munus est per munera $$f(n) = (-1)^n \left \lfloor \frac{n}{2} \right \rfloor$$ et $$\mathbb{N}$$ in $$\mathbb{Z}$$ et $$g(2^kr) = (k+1, \frac{r+1}{2})$$ : quod $$\mathbb{N}$$ in $$\mathbb{N}^2$$ map bijectively desideriis icta fidelibus.

Nos autem surjectivity iam monstrant $$h$$ ( $$h$$ et injective, sed non opus est haec res).

Ne $$(x,y) = (2^{k_1} r_1, 2^{k_2} r_2) \in\mathbb{Z}^2$$ . Sed tunc

$$h \left ( 2^{2 \cdot 2^{k_1} r_1 - 1} \cdot (4 \cdot 2^{k_2} r_2 - 1) \right ) = (2^{k_1} r_1, 2^{k_2} r_2) = (x,y).$$

Unde: $$\forall (s,z) \in \mathbb{Z}^2 \, \exists \, m \in \mathbb{N}$$ et $$h(m) = (x_m,y_m) = (s, z)$$ .

Eg si nostra est rursus ad movere ad $$n = 88$$ calculum $$h(88)=(2,3)$$ et eligere $$2 + 88 \cdot 3 = 266$$ ut statum.

Et cum exacte $$m$$ excitato movetur sono, $$x_m + m \cdot y_m = s + m \cdot z = a_m$$ ad arbitrium cadit in rana.

In praeter $$h$$ : alia multa munera ut Cantor de HYMENAEOS munus vel bijective spirali enim possibilia sunt.

Hic est a simplex implementation in JavaScript:

See the Pen catch the frog by David Vielhuber (@vielhuber) on CodePen.

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