Can we generalize $\sum_{n=0}^\infty \binom{4n}{n}\frac{1}{(3n+1)\,2^{4n+1}}=\frac1{T}$ for tribonacci constant $T$?

Question

(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 $\Phi_2$?

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(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, $$\frac1{x}=\frac1{\Phi_k}=\sum_{n=0}^\infty \binom{(k+1)n}{n}\frac{1}{(kn+1)\,2^{(k+1)n+1}}$$

Answer 1

What about this for golden ratio $\Phi_2$?

$$\sum_{n=0}^\infty \binom{3n}{n}\frac{1}{(2n+1)\,2^{3n+1}}=\frac{1}{2}{\;}_2F_1 \left(\frac13,\frac23; \color{blue}{\frac32};\frac{3^3}{\color{red}{2^3}\,2^2} \right)=\frac1{\Phi_2}$$

Further Observations added

The first constant in this sequence is $1$ which is its own inverse. $$\sum_{n=0}^\infty \binom{2n}{n}\frac{1}{(n+1)\,2^{2n+1}}=?=\frac1{\Phi_1}=1$$

If $\frac1{\Phi_1}$,$\frac1{\Phi_2}$,$\frac1{\Phi_3}$ and $\frac1{\Phi_4}$ are used as the roots of the successive polynomials instead of $\Phi_1$,$\Phi_2$,$\Phi_3$ and $\Phi_4$ then the polynomials are instead respectively

$$x-1=0$$ $$x^2+x-1=0$$ $$x^3+x^2+x-1=0$$ $$x^4+x^3+x^2+x-1=0$$

If you look at the relationship I found for finding roots Rational Series for Positive Integer Root of $x>1$ then it seems the number multiplying n to produce the binomial coefficients should be one less; that is for $\Phi_2$ it should be 2 instead of 3.

However if the above polynomials are each multiplied by $(x-1)$ then by rearrangement we can arrive at a remarkable sequence of self-similar formulae of the form

$$\frac{1}{\Phi_r}=\left(\frac{2}{\Phi_r}-1\right)^{1/(r+1)}$$

I am not sure this is the right way to proceed, but it is an interesting pattern nevertheless.

The Plot Thickens

Contrary to what I was saying above the sequence below seems to hold numerically as well

$$\sum_{n=0}^\infty \binom{rn}{n}\frac{1}{((r-1)n+1)\,2^{rn+1}(\Phi_r)^n}=\frac1{\Phi_r}$$

For the first formula in the sequence we have numerically

$$\sum_{n=0}^\infty \binom{2n}{n}\frac{1}{(n+1)\,2^{2n+1}x^n}=\frac{1}{1+\sqrt{\frac{x-1}{x}}}$$

Answer 2

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.