General formula for the denominator of Taylor expansion of $\sqrt{1-x}$

Question

The Taylor expansion of the square root $\sqrt{1-x}$ is

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

the first terms being

$$ 1-\frac x 2 -\frac{x^2}{8}-\frac{x^3}{16}-\frac{5x^4}{128}-\ldots $$

Is there a general formula for the $n^{\rm th}$ denominator? It is obvious that it will be of the form $2^m$, but it is not obvious what $m$ should be.

Answer 1

A simple OEIS search for $\,1, 2, 8, 16, 128, 256\,$ turns up the OEIS sequence A046161 from which the comment

Denominator of coefficient of $x^n$ in $(1+x)^{k/2}$ or $(1-x)^{k/2}$ for any odd integer $k$.

is applicable to your question with $k=1.$ A formula in the entry implies that $\, a(n) = 2^{b(n)} \,$ where $\, b(n)=A005187(n)\,$ and OEIS sequence A005187 has the formula $\, b(n) = \sum_{k\ge 0} \lfloor n/2^k \rfloor. \,$

Answer 2

I don't think that writing $$\binom{1/2}{n}$$ is the correct notation. If the power is a fraction then we just use the same idea but we write: \begin{align} \binom{1/2}{k} &=\frac{\frac12(\frac12-1)(\frac12-2)\cdots(\frac12-k+1)}{k!}\\ &=\frac{(-1)^{k-1}}{2^kk!}1\cdot3\cdot5\cdots(2k-3)\\ &=\frac{(-1)^{k-1}}{2^kk!}\frac{(2k-2)!}{2^{k-1}(k-1)!}\\ &=\frac{(-1)^{k-1}}{k2^{2k-1}}\binom{2k-2}{k-1} \end{align} This could give you a general expression.