Why is it that certain powers in an expansion are missing? Let's consider the real roots of the following equation in x:
$$x^5+\epsilon x =1.$$
Here $\epsilon\ge0$ is a real parameter. When $\epsilon=0$ the only root is $x=1.$ For a general $\epsilon>0$ we know that there is no closed form solution, so this is the perfect candidate for a series expansion:
$$x(\epsilon)=1+a_1\epsilon+a_2\epsilon^2+a_3\epsilon^3+\dots$$
If we work out the coefficients 
$$\mathbf{a}=(-\frac{1}{5},-\frac{1}{25},-\frac{1}{125},0,-\frac{21}{15625},\dots)$$
we come to the conclusion that $$a_{4+5k}=0,\text{ for }k=0,1,2,\dots$$
My question is why does this happen every five indices? My guess is that it has to do with the order of the equation. However when I do the actual computation it's not at all obvious and it seems to happen by chance! Terrible!
All ideas/comments welcome
 A: Let us do the same for $$x^n+\epsilon x =1$$ and use $$x=1+\sum_{i=1}^{2n} a_i\epsilon^i$$ Now, using the binomial theorem or Taylor expansion around $\epsilon=0$, by identification, we can find that $$a_1=-\frac 1n$$ $$a_2=-\frac{(n-3)}{2 n^2}$$ $$a_3=-\frac{(n-4) (n-2)}{3 n^3}$$ $$a_4=-\frac{(n-5) (2 n-5) (3 n-5)}{24 n^4}$$ $$a_5=-\frac{(n-6) (n-3) (n-2) (2 n-3)}{10 n^5}$$ $$a_6=-\frac{(n-7) (2 n-7) (3 n-7) (4 n-7) (5 n-7)}{720 n^6}$$ $$a_7=-\frac{(n-8) (n-4) (n-2) (3 n-8) (3 n-4) (5 n-8)}{315 n^7}$$ $$a_8=-\frac{(n-9) (n-3) (2 n-9) (2 n-3) (4 n-9) (5 n-9) (7 n-9)}{4480 n^8}$$ $$a_9=-\frac{(n-10) (n-5) (n-2) (2 n-5) (3 n-10) (3 n-5) (4 n-5) (7 n-10)}{4536 n^9}$$In this, you can notice that $a_k$ has, as a factor, $(n-k-1)$ and this explains that.
A: The solution of $x^5+\varepsilon x=1$ can be computed through Bring radical. 
If we set $x=\varepsilon^{1/4} z$, by Lagrange inversion theorem we have:
$$ x(\varepsilon) = \varepsilon^{1/4}\,\text{BR}(\varepsilon^{-5/4})=\color{red}{-\sum_{k\geq 0}\binom{5k}{k}\frac{(-1)^k\varepsilon^{-(5k+1)}}{4k+1}} $$
