The infinite series representation
\begin{equation}
\frac1\pi\sum_{j=0}^\infty \biggl(j+\frac{1}{2}\biggr)^{n-1}\frac{2^{j+1/2}}{\binom{2j+1}{j+1/2}}, \quad n\ge0
\end{equation}
for the positive integer sequence
\begin{gather}
1, 1, 2, 7, 35, 226, 1787, 16717, 180560, 2211181, 30273047,\\
458186752, 7596317885, 136907048461, 2665084902482,\\
55726440112987, 1245661569161135, 29642264728189066,\\
748158516941653967, 19962900431638852297,\\
561472467839585937560, 16602088291822017588121, \dotsc
\end{gather}
appears at https://oeis.org/A014307.
But there is no reference given for this series representation.
Could you please help find a source (such as a paper, a monograph, a book, a handbook, and the like, but not a website) including this series representation or its proof? Thank you very much.
By the way, I established a closed-form expression
\begin{equation}
\sum_{k=0}^{n}(-1)^kS(n,k) (2k-1)!!, \quad n\ge0
\end{equation}
and others for the above sequence, where $S(n,k)$ denotes the Stirling numbers of the second kind.
Best Answer
I don't have a reference, but it is not hard to prove the result. You want to prove that if $$\left(\frac{e^x}{2-e^x}\right)^{1/2}=\sum_{n=0}^\infty a_n \frac{x^n}{n!}$$ then $$a_n=\frac 1\pi\sum_{j=0}^\infty \left(j+\frac 12\right)^{n-1}\frac{2^{j+1/2}}{\binom{2j+1}{j+1/2}}. $$ I am not a fan of non-integer binomial coefficients, so I write $$\frac 1{\binom{2j+1}{j+1/2}}=\frac{\pi(j+1/2)(1/2)_j}{2^{2j+1}j!}.$$ Here, $(a)_j=a(a+1)\dotsm(a+j-1)$ is the Pochhammer symbol. So we want to prove $$a_n=\sum_{j=0}^\infty\left(j+\frac 12\right)^{n}\frac{(1/2)_j}{j!2^{j+1/2}}.$$ Now simply plug one series into the other and change order of integration. I get $$\sum_{n=0}^\infty a_n \frac{x^n}{n!}=\sum_{j=0}^\infty\frac{(1/2)_j}{j!2^{j+1/2}}\sum_{n=0}^\infty \left(j+\frac 12\right)^{n} \frac{x^n}{n!} =\sum_{j=0}^\infty\frac{(1/2)_j}{j!2^{j+1/2}}e^{(j+1/2)x}. $$ Now we are reduced to the binomial series $$\frac 1{(1-t)^{1/2}}=\sum_{j=0}^\infty\frac{(1/2)_j}{j!}\,t^j$$ (with $t=e^{x}/2$).
(It would have been more pedagogical to write this backwards: First use the binomial theorem to expand your function in a series of $e^x$ and then use the Taylor series of the exponential function.)