Is there a reasonably standard symbol depending on a parameter, like $\delta_i$ or something, that takes the value $1$ when $i$ is even and $0$ when $i$ is odd? or the other way around?
$$ \frac{1 + (-1)^i}{2} $$ is far too cumbersome.
Edit: I really don't see why it was closed for lack of context. The context should be obvious -- the numerous formulas where such a symbol would be useful. If I asked, "is there a symbol depending on two parameters i and j which is 1 when i=j and 0 otherwise", I would have gotten the answer, "yes, the Kronecker delta symbol!" -- not a vote to put on hold.
But okay, if you want a particular context, I'll offer that this would give an alternative way to write coefficients of power series with alternating zero and nonzero terms.
There are other ways around that, so here's a different context. I'm writing a paper where I need to write a formula for the general term in the sequence of polynomials:
$x$, $-x+x^2$, $3x+x^2+x^3$, $-3x+4x^2-x^3+x^4$, $\ldots$
The coefficient of the $i$th degree term in the $m$th degree polynomial is, up to a sign,
$$\binom{\left\lfloor \frac{m+i-2}{2} \right\rfloor}{i-1}$$ then multiplied by $m/i$ when $m-i$ is even or $(m-1)/i$ when $m-i$ is odd. This latter factor would be easy to write using the symbol I asked about. I could also write this as a sum of two polynomials, one with odd degree terms and the other with even, but I would prefer one formula.
Consider the Iverson bracket convention.
If $p$ is a statement: $[p] = \begin{cases} 1; & p \\ 0; & \neg p\end{cases}$
If $P$ is a predicate: $[P(x)] = \begin{cases} 1; & P(x) \\ 0; & \neg P(x)\end{cases}$
So your desired notation would be $[2 | i]$ or $[i \equiv 0 \bmod 2]$
Or maybe (these two I made up):
$[\mathcal{E}(i)]$, or $[i \equiv 0]_2$
Or the characteristic function on evens:
$$\chi_{\mathcal E}(i) = \mathbf{1}_{\mathcal E}(i)$$
I use $\mathcal E$ and $\mathcal O$ for the set of even and odd numbers in my personal notes, but of course you can use some other symbol.
edit: I've seen $[p]_q$ to mean the remainder of $p / q$, I think, somewhere. So that would be $[i]_2$.