Game of dots: winning strategy? The game begins with a row of $n$ numbers, in increasing order from $1$ to $n$. For example, if $n=7$, we have a row of numbers $(1,2,3,4,5,6,7)$.
On each turn, a player must either remove 1 number, or remove 2 consecutive numbers. For example, the first player to move can remove $2$ or remove 5 and 6 together.
The player who removes the last number wins. Is there a winning strategy for the player who goes first?
p.s. Sorry for the initial confusion. Here are some clarifications. (1) There are two players. (2) Let's say 4 is removed on the first turn. This does NOT make 3, 5 consecutive. So a player can never remove 3 and 5 together.
 A: Example strategy for $n=7$:
The first player takes $4$, and then until the last element:


*

*If the second player takes $x$, then the first player takes $8-x$.

*If the second player takes $(x,x+1)$, then the first player takes $(7-x,8-x)$.

General strategy for an odd $n$:
The first player takes $\dfrac{n+1}{2}$, and then until the last element:


*

*If the second player takes $x$, then the first player takes $n+1-x$.

*If the second player takes $(x,x+1)$, then the first player takes $(n-x,n+1-x)$.

General strategy for an even $n$:
The first player takes $(\dfrac{n}{2},\dfrac{n}{2}+1)$, and then until the last element:


*

*If the second player takes $x$, then the first player takes $n+1-x$.

*If the second player takes $(x,x+1)$, then the first player takes $(n-x,n+1-x)$.

Conceptual proof:
A pickable element is either a single number or a pair of consecutive numbers.
You can think of the element in the middle as a mirror.
It is the only pickable element that doesn't have a "reflecting counterpart".
So by picking this element first, you guarantee that for every element that your opponent picks, you can pick the corresponding element (located on "the other side of the mirror"), thus win the game...
