Hitting the Infinite Frog

EDIT : The previous solution I had (shoot at multiples of 6) was correct, but only if Hunter Jim knew the frog's pace. Since Hunter Jim does not know this, a new solution applies. This is given below.

The formal declaration of this problem is as follows:
$(\exists g:\mathbb{N}\rightarrow\mathbb{N})(\forall a,b\in\mathbb{N})(\exists n\in\mathbb{N})(g(n)=an+b)$ This roughly translates to "There is some strategy Hunter Jim can employ such that for a frog any pace or starting position, at some point, Hunter Jim will shoot that frog."

We will try to prove this now.

Let us construct a set $S$ in the following way:
$S= \{2^a3^b\mid a\in\mathbb{N}, b\in\mathbb{N}\}$ In non-mathy notation, $S$ is the set of all numbers that can be written as some power of two times some power of three. By the fundamental theorem of arithmetic, each $n\in S$ has some unique $a,b\in\mathbb{N}$ such that $n = 2^a3^b$ . As well as this, we also know that each $a$ and $b$ has a unique member of $S$ that it corresponds to.

From here, we proceed to construct $g$ .
For any number $n$ , either $n\in S$ or $n\not\in S$ .
If $n\not\in S$ , then $g(n) = 1$ . This is completely arbitrary, and it represents that Hunter Jim can really shoot anywhere here.
However, if $n\in S$ , then we know there is some unique $a$ and $b$ such that $n=2^a3^b$ . Knowing this, we can construct $g(n)=an+b$ .

Now, we can prove the proposition.
Let $a$ and $b$ be arbitrary positive integers representing the pace and starting position of the frog, respectively. Then, at shot $2^a3^b$ , we know that Hunter Jim will shoot at $g(2^a3^b) = a(2^a3^b)+b$ . Because the frog will have hopped $2^a3^b$ times at a pace of $a$ starting at $b$ , we know that the frog will be at position $a(2^a3^b)+b$ , meaning at shot $2^a3^b$ , Hunter Jim will shoot the frog. Because $a$ and $b$ are arbitrary, we prove the proposition. $\square$

Hunter Jim shoots at odd number multiples of 5 until he gets the frog, starting with 5.

With each round, Hunter Jim aims at a spot that advances by $2(n+1)-1 -(2n+1)=2$ steps, while the frog advances by $1$ step. Hence, wherever the frog is, Hunter Jim's aim gets closer to the frog by $1$ step with each round.

(Examples added here)

Here's how it could go, frog vs hunter's aim

(Frog jumps from 25)
{30,35,40,45,50,55}
(Hunter aims at 5 first)
{5,15,25,35,45,55}

So, on the 6th try, the hunter will get the frog. Now let's try a different scenario.

(Frog jumps from 30)
{35,40,45,50,55,60,65}
(Hunter aims at 5 first)
{5,15,25,35,45,55,65}

On the 7th try, the hunter will get the frog.

By the way, how come the frog is showng jumping backwards?

I will make Agnishom’s game a little harder : the frog has a finite number of lives, but we don’t know how many. The hunter must shoot the frog enough times to win. Clearly, if the hunter has a winning strategy for this game, it has a winning strategy in the simpler game.

Let $s,T:\mathbb{N}\longrightarrow\mathbb{N}$ be functions such that $t\mapsto (s(t),T(t))$ is a surjection and such that every point in the image $\mathbb{N}\times\mathbb{N}$ occurs infinitely often. ${}^{\color{#D61F06}{\dagger}}$ Interpretation: $s,T$ denote possible starting position and possible periods respectively. The shooting strategy is as follows:

$\mathrm{sh}(t) := s(t)+t\cdot T(t)$

where $\mathrm{sh}(t)$ denotes the position, at which the hunter shoots at step $t$ in the game.

Claim: Let $a,b,L\in\mathbb{N}$ . Assume the frog has $L$ lives, starts at $a$ and springs according to period $b$ . Then the hunter, whose strategy is independent of this information, will eventually hit the frog enough times to kill it. That is, the hunter always wins and $\mathrm{sh}(\cdot)$ codes a winning strategy.

Proof: The frog’s trajectory is given by $(a+tb)_{t\in\mathbb{N}}$ --- although it may die after finitely many steps in this sequence. Let $A:=\{t\in\mathbb{N}\mid (s(t),T(t))=(a,b)\}$ . Per construction $A$ is infinite and $\mathrm{sh}(t)=s(t)+t\cdot T(t)=a+tb$ for all $t\in A$ . Thus the hunter hits the frog infinitely often, in particular, more than $L$ times. Thus the hunter will kill eventually the frog. QED.

${}^{\color{#D61F06}{\dagger}}$ This is easy to construct: fix first any old infinite partition of the naturals into countably infinitely many infinite sets $\mathbb{N}=\bigcup_{n\in\mathbb{N}}A_{n}$ . Now let $\pi:\mathbb{N}\longrightarrow\mathbb{N}$ code what partition each element belongs to, ie $\pi(t)=n\Leftrightarrow t\in A_{n}$ . Let $\sigma:\mathbb{N}\longrightarrow\mathbb{N}\times\mathbb{N}$ be any surjection. Set $(s,T)=\sigma\circ \pi$ .