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$\delta$ is a positive number. Is this Taylor expansion of some function?

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    $\begingroup$ What does $\binom{n \delta}{n}$ mean? Is it a beta function? $\endgroup$
    – Igor Rivin
    Nov 16, 2017 at 2:36
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    $\begingroup$ As long as $n$ is a whole number we have ${m \choose n} \in {\bf Q}[m]$ as a degree-$n$ polynomial. (It's also close to $1/B(n,m-n)$.) And the power series is close to this: math.harvard.edu/~elkies/Misc/catalan.pdf $\endgroup$ Nov 16, 2017 at 2:45
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    $\begingroup$ When $\delta$ is a natural integer ${n\delta\choose n}$ is equal to $(\delta-1)n+1$ times the number of $\delta$-ary trees with $n$ nodes. The Fuss-Catalan numbers are given by $$C(\ell,n)=\frac{1}{(\ell-1)n+1}{\ell n\choose n}$$ $\endgroup$ Nov 16, 2017 at 3:04
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    $\begingroup$ The series $h(z):=\sum_n C(\ell,n)(-z)^n$ is linked to the local inverse at $0$ of $x\mapsto x+x^\ell$, even for non-integer real $\ell>1$: mathoverflow.net/questions/249060/… $\endgroup$ Nov 16, 2017 at 9:07

2 Answers 2

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The Bürmann-Lagrange theorem gives that
$$\sum_{n\geq 0} {n\delta \choose n} t^n = \frac{1}{1-\delta t(1+z)^{\delta -1}}=\frac{1+z}{1+(1-\delta) z}$$ where $z=z(t)=\sum_{n\geq 1} \frac{1}{n}{n\delta \choose n-1}t^n$ is the solution of $z=t(1+z)^\delta$ $\big($i.e. $z(t)$ is the local inverse at $0$ of $z \mapsto \frac{z}{(1+z)^\delta}\big)$.

See e.g problem 216 in section 3 of Polya/Szegő, Problems and Theorems in Analysis I.

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Where there is abinomial coefficient in a summation, most of the time the result is a generalized hypergeometric function. $$S_\delta=\sum_{n=0}^\infty\binom{n\delta}{n}x^n=\sum_{n=0}^\infty \frac{\Gamma (n \delta +1)}{\Gamma (n+1)\,\, \Gamma (n(\delta-1)+1)}x^n$$

If $\delta$ is a positive whole number, this is the definition of

$$S_\delta=\, _{n-1}F_{n-2}\left(\frac{1}{n},\frac{2}{n},\cdots,\frac{n-1}{n};\frac{1}{n-1},\frac{2} {n-1},\cdots,\frac{n-2}{n-1};\frac{n^n }{(n-1)^{n-1}}x\right)$$

If $\delta$ is a rational number, $S_\delta$ is a linear combination of this kind of function weigheted by some powers of $x$.

In the past, I had one nice $$S_3=\frac{2}{\sqrt{4-27 x}}\,\cos \left(\frac{1}{3} \sin ^{-1}\left(\frac{3 \sqrt{3x} }{2}\right)\right)$$

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