Can we generalize $\sum_{n=0}^\infty \binom{4n}{n}\frac{1}{(3n+1)\,2^{4n+1}}=\frac1{T}$ for tribonacci constant $T$? (Inspired by this post.) Given the tribonacci constant $\Phi_3$, the tetranacci constant $\Phi_4$, etc. How do prove that,
$$\sum_{n=0}^\infty \binom{4n}{n}\frac{1}{(3n+1)\,2^{4n+1}}=\frac{1}{2}{\;}_3F_2 \left(\frac14,\frac24,\frac34; \color{blue}{\frac23,\frac43};\frac{4^4}{\color{red}{2^4}\,3^3} \right)=\frac1{\Phi_3}$$
$$\sum_{n=0}^\infty \binom{5n}{n}\frac{1}{(4n+1)\,2^{5n+1}}=\frac{1}{2}{\;}_4F_3 \left(\frac15,\frac25,\frac35,\frac45; \color{blue}{\frac24,\frac34,\frac54};\frac{5^5}{\color{red}{2^5}\,4^4} \right)=\frac1{\Phi_4}$$
$$\sum_{n=0}^\infty \binom{6n}{n}\frac{1}{(5n+1)\,2^{6n+1}}=\frac{1}{2}{\;}_5F_4 \left(\frac16,\frac26,\frac36,\frac46,\frac56; \color{blue}{\frac25,\frac35,\frac45,\frac65};\frac{6^6}{\color{red}{2^6}\,5^5} \right)=\frac1{\Phi_5}$$
and so on? And what is the corresponding hypergeometric formula for $\displaystyle \frac1{\Phi_2}$ with golden ratio $\Phi_2$?

(Courtesy of Jack D'Aurizio) By the Lagrange inversion theorem, a solution to,
$$x^k-x+1=0$$
is given by,
$$x=\sum_{n=0}^\infty \binom{kn}{n}\frac{1}{(k-1)n+1}$$

Similarly, given the general polynomial of the $k$-nacci constants as,
  $$x^k(2-x)-1=0$$
  how do we show that,
  $$\frac1{x}=\frac1{\Phi_k}=\sum_{n=0}^\infty \binom{(k+1)n}{n}\frac{1}{(kn+1)\,2^{(k+1)n+1}}$$


Note that $\frac{1}{kn+1}\binom{(k+1)n}{n}$ is the number of $(k+1)$-ary trees on $n$ vertices. Let $T_{k+1}$ be the ordinary generating function for the class of $(k+1)$-ary trees. Then $T_{k+1}$ satisfies the equation
$$
T_{k+1}(z)=1+zT_{k+1}^{k+1}(z).
$$
Let $a=\frac{1}{2}T_{k+1}\left(\frac{1}{2^{k+1}}\right)$, then we need to prove that $\frac{1}{a}=\Phi_k$. From the functional equation, we have
$$
2a=1+\frac{1}{2^{k+1}}(2a)^{k+1}=1+a^{k+1},
$$
so $a$ satisfies the equation
$$
1-2x+x^{k+1}=0,
$$
and thus $\frac{1}{a}$ satisfies
$$
x^{k+1}-2x^k+1=(x-1)(x^k-x^{k-1}-\dots-x-1)=0.
$$
We will leave the argument that $0<a<1$ for $k\ge 2$ to the reader. Then $\frac{1}{a}$ is a root of the equation $\phi_k(x)=x^k-x^{k-1}-\dots-x-1=0$. From Proposition 1 on p. 6 of this paper, we see that $\phi_k(x)$ has a unique root greater than $1$, so $a$ is uniquely determined by this polynomial. However, $\Phi_k$ is also the unique root greater than $1$ of $\phi_k(x)$, so $\frac{1}{a}=\Phi_k$.
Perhaps, it would be interesting to interpret both sides probabilistically. A few years ago, I think I have actually heard of that problem for $k=1$, where the sum is $1$, but I'm not sure it's still open or not.
