The roots of trinomial equations $x^p+xq=0$ ($p\in\mathbb{N}$) can be expressed in terms of the hypergeometric functions. I am wondering if at least one real root, for instance given by the following iterative solution $x_n=\frac{q  (1  p) x_{n1}^p}{ 1+p x_{n1}^{(p  1)}}$, $x_0=q^{1/p}$ can be expressed in a similar form for $p\in\mathbb{R}$, $p\ge2$.
An analogous formula does hold, although the corresponding functions are not hypergeometric if $p$ is irrational.
For given $p\in\mathbb{R}$, $p>1$, consider the power series $$h(z)=\sum_{k=0}^{\infty} \frac{(1)^k}{pkk+1}\binom{pk}{k}\, z^k$$ with radius of convergence $R=(p1)^{p1}/p^p.$
Then, for $0\le y\le R^{1/(p1)}$, the function $g(y):=yh(y^{p1})$ is the inverse function of $f(x):=x+x^p$. $$*$$ [edit] There is also an analogous inversion formula for three or more terms, to invert e.g. $f(x)=x+ax^p+bx^q$ with real exponents $p>1$ and $q>1$. If $H=H_{p,q}$ is the analytic function $$H(u,v)=\sum_{i\ge0,j\ge0}\frac{(1)^{i+j}}{ (p1)i+ (q1)j+1} {pi+qj \choose i,\, j}u^iv^j,$$ then $g(y):=yH(ay^{p1},by^{q1})$ is the local inverse of $f$ at $0$ (the multinomial coefficient in the double series is ${pi+qj \choose i,\, j}:=\frac{(pi+qj)(pi+qj1)\dots(pi+qjij+1)}{i!j!}$ .)

3$\begingroup$ Two elementary approaches: math.harvard.edu/~elkies/Misc/catalan.pdf math.harvard.edu/~elkies/Misc/catalan2.pdf $\endgroup$ – Noam D. Elkies Sep 5 '16 at 18:51

$\begingroup$ Great! Very interesting, exactly what I was looking for. It is clear approximately how one can verify that the power series solution solves the equation, but little bit more details would be of interest. Maybe more about the idea behind. $\endgroup$ – yarchik Sep 5 '16 at 19:12

1$\begingroup$ @Noam D. Elkies: very nice computations. (The way I found the above series is via a Lagrange Inversion Formula, that also works in the context of formal power series with real or complex exponents). $\endgroup$ – Pietro Majer Sep 5 '16 at 21:37