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Pluviophile
Pluviophile
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Many well-known combinatorial identities has an analytic version. For example, the following identities $$ 2^n = \sum_{k=0}^n \binom{n}{k} $$ $$ \binom{2n}{n} = \sum_{k=1}^n \binom{n}{k}^2 $$ can be generalized to $$ 2^z = \sum_{n=0}^{\infty} \binom{z}{n} $$ and $$ \frac{\Gamma(1+2z)}{\Gamma(1+z)^2} = \sum_{n=0}^{\infty} \binom{z}{n}^2 $$ where $z$ is in a small neighborhood of zero in $\mathbb C$ and $\binom{z}{n}=\frac{z(z-1)\dots (z-n+1)}{n!}$. For the Dixon's identity, $$ \sum_{k=0}^n (-1)^k \binom{2n}{k}^3 = (-1)^n \frac{(3n)!}{(n!)^3}, $$ it is less obvious to guess the analytic version of it. After some trial and error, I found $$ \sum_{n=0}^{\infty} (-1)^n \binom{z}{n}^3 = \cos \left(\frac{\pi z}{2}\right) \frac{\Gamma(1+3z/2)}{\Gamma(1+z/2)^3} $$ To prove this, my idea is to find a functional equation satisfied by both sides of the formula. However, before delving into it, I want to ask if this formula exists somewhere in the literature and in which context.

Remarks. Although my question has been completely answered in the comment section, I want to add here some motivations and remarks. My goal is to derive formulas by extrapolating simple combinatorial identities. For example, with the analytic version of the formula involving $\binom{2n}{n}$, by taking $z=1/2$, we obtain $$ \sum_{k \geq 0} \frac{(C_n)^2}{16^n} = \frac{16}{\pi}-4 $$ where $C_n$ is the $n$-th Catalan number. Taking the derivative at $z=1/2$, we get $$ \sum_{n=1}^\infty \frac{(C_n)^2}{16^n} \left( 1+\frac{1}{3}+\dots+\frac{1}{2n-1}\right) = \frac{24-16 \ln 2}{\pi} - 4 $$ Taking the third derivative at $z=0$, we have $$ \sum_{n \geq 1} \frac{1}{n^2} \left(1+\frac{1}{2}+\dots+\frac{1}{n}\right)= 2\zeta(3) $$ Taking $z=-1/3$ and $z=-1/6$, we obtain infinite series with values depending on the constant $\Gamma(\frac{1}{3})$ that vanishes in the ratio: $$ \frac{ 1+(\frac{1}{6})^2+ (\frac{1.7}{6.12} )^2 + (\frac{1.7.13}{6.12.18} )^2 + \dots }{ 1+ ( \frac{1}{3} )^2 + ( \frac{1.4}{3.6} )^2 + ( \frac{1.4.7}{3.6.9} )^2 + \dots } = \frac{2^{\frac{1}{3}}}{3^{\frac{1}{2}}} $$ It is quite surprising that the ratio take such a nice value.

Many more identities can be obtained in this way of extrapolating simple counting problems. As it turns out, hypergeometric functions provide vast generalizations of these analytic continuations, but until now I haven't recognized their value.

Many well-known combinatorial identities has an analytic version. For example, the following identities $$ 2^n = \sum_{k=0}^n \binom{n}{k} $$ $$ \binom{2n}{n} = \sum_{k=1}^n \binom{n}{k}^2 $$ can be generalized to $$ 2^z = \sum_{n=0}^{\infty} \binom{z}{n} $$ and $$ \frac{\Gamma(1+2z)}{\Gamma(1+z)^2} = \sum_{n=0}^{\infty} \binom{z}{n}^2 $$ where $z$ is in a small neighborhood of zero in $\mathbb C$ and $\binom{z}{n}=\frac{z(z-1)\dots (z-n+1)}{n!}$. For the Dixon's identity, $$ \sum_{k=0}^n (-1)^k \binom{2n}{k}^3 = (-1)^n \frac{(3n)!}{(n!)^3}, $$ it is less obvious to guess the analytic version of it. After some trial and error, I found $$ \sum_{n=0}^{\infty} (-1)^n \binom{z}{n}^3 = \cos \left(\frac{\pi z}{2}\right) \frac{\Gamma(1+3z/2)}{\Gamma(1+z/2)^3} $$ To prove this, my idea is to find a functional equation satisfied by both sides of the formula. However, before delving into it, I want to ask if this formula exists somewhere in the literature and in which context.

Remarks. Although my question has been completely answered in the comment section, I want to add here some motivations and remarks. My goal is to derive formulas by extrapolating simple combinatorial identities. For example, with the analytic version of the formula involving $\binom{2n}{n}$, by taking $z=1/2$, we obtain $$ \sum_{k \geq 0} \frac{(C_n)^2}{16^n} = \frac{16}{\pi}-4 $$ where $C_n$ is the $n$-th Catalan number. Taking the derivative at $z=1/2$, we get $$ \sum_{n=1}^\infty \frac{(C_n)^2}{16^n} \left( 1+\frac{1}{3}+\dots+\frac{1}{2n-1}\right) = \frac{24-16 \ln 2}{\pi} - 4 $$ Taking the third derivative at $z=0$, we have $$ \sum_{n \geq 1} \frac{1}{n^2} \left(1+\frac{1}{2}+\dots+\frac{1}{n}\right)= 2\zeta(3) $$ Taking $z=-1/3$ and $z=-1/6$, we obtain infinite series with values depending on the constant $\Gamma(\frac{1}{3})$ that vanishes in the ratio: $$ \frac{ 1+(\frac{1}{6})^2+ (\frac{1.7}{6.12} )^2 + (\frac{1.7.13}{6.12.18} )^2 + \dots }{ 1+ ( \frac{1}{3} )^2 + ( \frac{1.4}{3.6} )^2 + ( \frac{1.4.7}{3.6.9} )^2 + \dots } = \frac{2^{\frac{1}{3}}}{3^{\frac{1}{2}}} $$ It is quite surprising that the ratio take such a nice value.

Many more identities can be obtained in this way of extrapolating simple counting problems. As it turns out, hypergeometric functions provide vast generalizations of these analytic continuations, but until now I haven't recognized their value.

Many well-known combinatorial identities has an analytic version. For example, the following identities $$ 2^n = \sum_{k=0}^n \binom{n}{k} $$ $$ \binom{2n}{n} = \sum_{k=1}^n \binom{n}{k}^2 $$ can be generalized to $$ 2^z = \sum_{n=0}^{\infty} \binom{z}{n} $$ and $$ \frac{\Gamma(1+2z)}{\Gamma(1+z)^2} = \sum_{n=0}^{\infty} \binom{z}{n}^2 $$ where $z$ is in a small neighborhood of zero in $\mathbb C$ and $\binom{z}{n}=\frac{z(z-1)\dots (z-n+1)}{n!}$. For the Dixon's identity, $$ \sum_{k=0}^n (-1)^k \binom{2n}{k}^3 = (-1)^n \frac{(3n)!}{(n!)^3}, $$ it is less obvious to guess the analytic version of it. After some trial and error, I found $$ \sum_{n=0}^{\infty} (-1)^n \binom{z}{n}^3 = \cos \left(\frac{\pi z}{2}\right) \frac{\Gamma(1+3z/2)}{\Gamma(1+z/2)^3} $$ To prove this, my idea is to find a functional equation satisfied by both sides of the formula. However, before delving into it, I want to ask if this formula exists somewhere in the literature and in which context.

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Pluviophile
Pluviophile
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Many well-known combinatorial identities has an analytic version. For example, the following identities $$ 2^n = \sum_{k=0}^n \binom{n}{k} $$ $$ \binom{2n}{n} = \sum_{k=1}^n \binom{n}{k}^2 $$ can be generalized to $$ 2^z = \sum_{n=0}^{\infty} \binom{z}{n} $$ and $$ \frac{\Gamma(1+2z)}{\Gamma(1+z)^2} = \sum_{n=0}^{\infty} \binom{z}{n}^2 $$ where $z$ is in a small neighborhood of zero in $\mathbb C$ and $\binom{z}{n}=\frac{z(z-1)\dots (z-n+1)}{n!}$. For the Dixon's identity, $$ \sum_{k=0}^n (-1)^k \binom{2n}{k}^3 = (-1)^n \frac{(3n)!}{(n!)^3}, $$ it is less obvious to guess the analytic version of it. After some trial and error, I found $$ \sum_{n=0}^{\infty} (-1)^n \binom{z}{n}^3 = \cos \left(\frac{\pi z}{2}\right) \frac{\Gamma(1+3z/2)}{\Gamma(1+z/2)^3} $$ To prove this, my idea is to find a functional equation satisfied by both sides of the formula. However, before delving into it, I want to ask if this formula exists somewhere in the literature and in which context.

Remarks. Although my question has been completely answered in the comment section, I want to add here some motivations and remarks. My goal is to derive formulas by extrapolating simple combinatorial identities. For example, with the analytic version of the formula involving $\binom{2n}{n}$, by taking $z=1/2$, we obtain $$ \sum_{k \geq 0} \frac{(C_n)^2}{16^n} = \frac{16}{\pi}-4 $$ where $C_n$ is the $n$-th Catalan number. Taking the derivative at $z=1/2$, we get $$ \sum_{n=1}^\infty \frac{(C_n)^2}{16^n} \left( 1+\frac{1}{3}+\dots+\frac{1}{2n-1}\right) = \frac{24-16 \ln 2}{\pi} - 4 $$ Taking the third derivative at $z=0$, we have $$ \sum_{n \geq 1} \frac{1}{n^2} \left(1+\frac{1}{2}+\dots+\frac{1}{n}\right)= 2\zeta(3) $$ Taking $z=-1/3$ and $z=-1/6$, we obtain infinite series with values depending on the constant $\Gamma(\frac{1}{3})$ that vanishes in the ratio: $$ \frac{ 1+(\frac{1}{6})^2+ (\frac{1.7}{6.12} )^2 + (\frac{1.7.13}{6.12.18} )^2 + \dots }{ 1+ ( \frac{1}{3} )^2 + ( \frac{1.4}{3.6} )^2 + ( \frac{1.4.7}{3.6.9} )^2 + \dots } = \frac{2^{\frac{1}{3}}}{3^{\frac{1}{2}}} $$ It is quite surprising that the ratio take such a nice value.

Many more identities can be obtained in this way of extrapolating simple counting problems. As it turns out, hypergeometric functions provide vast generalizations of these analytic continuations, but until now I haven't recognized their value.

Many well-known combinatorial identities has an analytic version. For example, the following identities $$ 2^n = \sum_{k=0}^n \binom{n}{k} $$ $$ \binom{2n}{n} = \sum_{k=1}^n \binom{n}{k}^2 $$ can be generalized to $$ 2^z = \sum_{n=0}^{\infty} \binom{z}{n} $$ and $$ \frac{\Gamma(1+2z)}{\Gamma(1+z)^2} = \sum_{n=0}^{\infty} \binom{z}{n}^2 $$ where $z$ is in a small neighborhood of zero in $\mathbb C$ and $\binom{z}{n}=\frac{z(z-1)\dots (z-n+1)}{n!}$. For the Dixon's identity, $$ \sum_{k=0}^n (-1)^k \binom{2n}{k}^3 = (-1)^n \frac{(3n)!}{(n!)^3}, $$ it is less obvious to guess the analytic version of it. After some trial and error, I found $$ \sum_{n=0}^{\infty} (-1)^n \binom{z}{n}^3 = \cos \left(\frac{\pi z}{2}\right) \frac{\Gamma(1+3z/2)}{\Gamma(1+z/2)^3} $$ To prove this, my idea is to find a functional equation satisfied by both sides of the formula. However, before delving into it, I want to ask if this formula exists somewhere in the literature and in which context.

Many well-known combinatorial identities has an analytic version. For example, the following identities $$ 2^n = \sum_{k=0}^n \binom{n}{k} $$ $$ \binom{2n}{n} = \sum_{k=1}^n \binom{n}{k}^2 $$ can be generalized to $$ 2^z = \sum_{n=0}^{\infty} \binom{z}{n} $$ and $$ \frac{\Gamma(1+2z)}{\Gamma(1+z)^2} = \sum_{n=0}^{\infty} \binom{z}{n}^2 $$ where $z$ is in a small neighborhood of zero in $\mathbb C$ and $\binom{z}{n}=\frac{z(z-1)\dots (z-n+1)}{n!}$. For the Dixon's identity, $$ \sum_{k=0}^n (-1)^k \binom{2n}{k}^3 = (-1)^n \frac{(3n)!}{(n!)^3}, $$ it is less obvious to guess the analytic version of it. After some trial and error, I found $$ \sum_{n=0}^{\infty} (-1)^n \binom{z}{n}^3 = \cos \left(\frac{\pi z}{2}\right) \frac{\Gamma(1+3z/2)}{\Gamma(1+z/2)^3} $$ To prove this, my idea is to find a functional equation satisfied by both sides of the formula. However, before delving into it, I want to ask if this formula exists somewhere in the literature and in which context.

Remarks. Although my question has been completely answered in the comment section, I want to add here some motivations and remarks. My goal is to derive formulas by extrapolating simple combinatorial identities. For example, with the analytic version of the formula involving $\binom{2n}{n}$, by taking $z=1/2$, we obtain $$ \sum_{k \geq 0} \frac{(C_n)^2}{16^n} = \frac{16}{\pi}-4 $$ where $C_n$ is the $n$-th Catalan number. Taking the derivative at $z=1/2$, we get $$ \sum_{n=1}^\infty \frac{(C_n)^2}{16^n} \left( 1+\frac{1}{3}+\dots+\frac{1}{2n-1}\right) = \frac{24-16 \ln 2}{\pi} - 4 $$ Taking the third derivative at $z=0$, we have $$ \sum_{n \geq 1} \frac{1}{n^2} \left(1+\frac{1}{2}+\dots+\frac{1}{n}\right)= 2\zeta(3) $$ Taking $z=-1/3$ and $z=-1/6$, we obtain infinite series with values depending on the constant $\Gamma(\frac{1}{3})$ that vanishes in the ratio: $$ \frac{ 1+(\frac{1}{6})^2+ (\frac{1.7}{6.12} )^2 + (\frac{1.7.13}{6.12.18} )^2 + \dots }{ 1+ ( \frac{1}{3} )^2 + ( \frac{1.4}{3.6} )^2 + ( \frac{1.4.7}{3.6.9} )^2 + \dots } = \frac{2^{\frac{1}{3}}}{3^{\frac{1}{2}}} $$ It is quite surprising that the ratio take such a nice value.

Many more identities can be obtained in this way of extrapolating simple counting problems. As it turns out, hypergeometric functions provide vast generalizations of these analytic continuations, but until now I haven't recognized their value.

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Pluviophile
Pluviophile
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Analytic continuation of Dixon's identity

Many well-known combinatorial identities has an analytic version. For example, the following identities $$ 2^n = \sum_{k=0}^n \binom{n}{k} $$ $$ \binom{2n}{n} = \sum_{k=1}^n \binom{n}{k}^2 $$ can be generalized to $$ 2^z = \sum_{n=0}^{\infty} \binom{z}{n} $$ and $$ \frac{\Gamma(1+2z)}{\Gamma(1+z)^2} = \sum_{n=0}^{\infty} \binom{z}{n}^2 $$ where $z$ is in a small neighborhood of zero in $\mathbb C$ and $\binom{z}{n}=\frac{z(z-1)\dots (z-n+1)}{n!}$. For the Dixon's identity, $$ \sum_{k=0}^n (-1)^k \binom{2n}{k}^3 = (-1)^n \frac{(3n)!}{(n!)^3}, $$ it is less obvious to guess the analytic version of it. After some trial and error, I found $$ \sum_{n=0}^{\infty} (-1)^n \binom{z}{n}^3 = \cos \left(\frac{\pi z}{2}\right) \frac{\Gamma(1+3z/2)}{\Gamma(1+z/2)^3} $$ To prove this, my idea is to find a functional equation satisfied by both sides of the formula. However, before delving into it, I want to ask if this formula exists somewhere in the literature and in which context.