Benedict Irwin edited section_Further_Series_Let_the__.tex  over 8 years ago

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\end{equation}  where all the top row terms are $1$. This means when $x=1$, we should have $h_3(1)=\phi$, the golden ratio.  We end up with a few more predictable/obvious terms giving \begin{equation}  h_3(x)=\frac{1}{x}-\frac{1}{x^3}+\frac{2}{x^5}-\frac{5}{x^7}+\frac{14}{x^9}-\frac{42?}{x^11}+\cdots h_3(x)=\frac{1}{x}-\frac{1}{x^3}+\frac{2}{x^5}-\frac{5}{x^7}+\frac{14}{x^9}-\frac{42?}{x^{11}}+\cdots  \end{equation}  this suggests the Catalan Numbers $C(n)=A000108$. In the OEIS, it is listed that the Hankel Transform of these numbers is the sequence $1,1,1,1,1...$ therefore, it may be that this is the inverse Hankel Transform.