A "Markov game" 
I call games similar to the one I describe below to be Markov games. I am selecting just that one or rather a 1-parameter series of games. The open challenge is to find out which of the players $\ 0\ $ or $\ 1\ $ has a winning strategy for each of the given parameter $\ W.$

NOTATION $\ n\%2=0\ $ for $\ n\ $ even, and $\ n\%2=1\ $ for $\ n\ $ odd.;
Let $ d(0)=J(0)=0.\ $ For arbitrary positive integer $\ n,\ $ player $\ n\%2\ $ selects a positive integer $\ d(n)\le d(n-1)+1;\ $ then $\ J(n)=J(n-1)+d(n).$
When players compete at game $\ M(W),\ $ where $\ W\ $ is an arbitrarily fixed positive integer, then the player that gets exactly $J(n)=W\ $ wins.
Let $\ \omega(W)=0\ $ if player $0$ has a winning strategy at $M(W);\ $
otherwise let $\ \omega(W)=1\ $ if player $1$ has a winning strategy at $M(W)$.
PROBLEM:   Compute function $\ \omega:\mathbb N\to\{0\ 1\}.$

For instance: $\ \omega(1)=1;\ \omega(2)=\omega(3)=0;\ \omega(4)=1, $
etc.   However, $\ \omega(120)\ $ or $\ \omega(5553)\ $ is a bit harder (and unknown to me).

 A: The answer doesn't change much with greater numbers.

The full answer is $$
\omega(W)=\left\{
\begin{array}{ll}
1, & W\%5=1,4\\
0, & W\%5=0,2,3
\end{array}\right.
$$

Let us say that position $(W-J,d)$ is winning if player $n\%2$ has a winning strategy for game $M(W)$ on his turn $n$ with $d(n)=d,J(n)=J$. It is losing otherwise. Clearly, it indeed depends only on the difference $W-J$.
$(i,d)$ is winning iff there exists losing $(j,f)$ s.t. $j+f=i,f\le d+1$. We are interested in whether $(W-1,1)$ is winning.
For this it suffices to consider only $(i,d)$ with $d\le 3$.
The pattern looks like this:
$
\begin{array}{ccc}
- & - & -\\
+ & + & +\\
+ & + & +\\
- & + & +\\
+ & + & +\\
- & - & -\\
\vdots & \vdots & \vdots
\end{array}
$
To see it, notice:

*

*$(0,d)$ is losing for all $d$, i.e. first row is filled with $-$'s

*if $(i,d)$ is winning, then $(i,d+1)$ is winning, i.e. to the right of $+$ is always another $+$
A: We already have a full answer from @JosephGordon who proved periodicity of $\ \omega,\ $ the length of the period being 5. Joseph -- many thanks!
Let me write a solution that at least to me is easier to follow.

Remember that player $\ \omega(n)\ $ is the one who has the winning strategy for Markov game $\ M(n).$

Theorem
For every positive integer $\ n\ $ the following two properties hold:

*

*Player $\ \omega(n)\ $ can win every game of $\ M(n)\ $ by selecting all of their own moves such that $\ d(k)\le 3;$


*$\ \omega(n+5) = \omega(5).\ $
Proof   Player $\ \omega(n)\ $, when playing game $\ M(n+5)\ $,
is able to arrive at position $\ n\ $ while utilizing moves such that
$\ d(k)\le 3\ $ each time. In particular, $\ d(t)\le 3\ $ when
$\ J(t)=n.\ $ Thus, now we have only four extensions of the game:

*

*$\ d(t+1)=4.\ $ Then player $\ \omega(n)\ $ plays $\ d(t+2)=1\ $
and wins (since $\ J(t+2)=n+5\ \text{and}\ t+2\equiv t\mod 2)$;


*$\ d(t+1)=3.\ $ Then player $\ \omega(n)\ $ plays $\ d(t+2)=2\ $
and wins;


*$\ d(t+1)=2.\ $ Then player $\ \omega(n)\ $ plays $\ d(t+2)=3\ $
and wins;


*$\ d(t+1)=1.\ $ Then player $\ \omega(n)\ $ plays $\ d(t+2)=1;\ $
then the other player plays $\ d(t+3)= 1$ or $2\ $, and player
$\ \omega(n)\ $ plays $\ 2$ or $1\ $ respectively, and wins.
The formal rest of the argument is totally routine and obvious.
End of PROOF.
The complete numerical description of $\ \omega\ $ is given by the above theorem and the initial $5$ values:
$$ \omega(1)=1;\,\ \omega(2)=\omega(3)=0;\,\ \omega(4)=1,
         \,\ \omega(5)=0 $$
For instance, $\ \omega(5)=0\ $ because J(1)=1, then player $0$ can play d(2)=1, i.e. J(2)=2, then -- after a move by player
$0$ -- player $\ 0\ $ will play $\ J(4)=5.$
