I was browsing through the OEIS when I noticed the name of the sequence A052634 "Expansion of e.g.f. 1/((1-2x^2)(1-x))".
I would like to know what is the expansion of e.g.f and how can I calculate it for any other case. (The only information I got was e.g.f= exponential generating function, but in the definition a succession of functions is required.)
Thanks in advance.
As Joriki stated, a sequence is related to its exponential generating function by the fact that the elements of the sequence appear as coefficients in the Taylor series expansion of the generating function, i.e. in the present case : $$ A(x) := \sum_{n=0}^\infty a_n\frac{x^n}{n!} = \frac1{(1-2x^2)(1-x)} $$ You can recover the elements with the relation $a_n = A^{(n)}(0)$, where $(n)$ denotes the $n^\mathrm{th}$ derivative; however, given that the present generating function is a rational expression, it is easier to use partial fraction decomposition and the geometric series : $$ \begin{array}{rcl} \displaystyle\frac1{\left(1-2x^2\right)(1-x)} &=& \displaystyle -\frac{1}{1-x} + \frac{2(1+x)}{1-2x^2} \\ &=& \displaystyle -\sum_{k\ge0}x^k + 2(1+x)\sum_{k\ge0}2^kx^{2k} \\ &=& \displaystyle -\sum_{k\ge0}(x^{2k}+x^{2k+1}) + \sum_{k\ge0}2^{k+1}(x^{2k}+x^{2k+1}) \\ &=& \displaystyle \sum_{k\ge0}\left((2k)!(2^{k+1}-1)\frac{x^{2k}}{(2k)!} + (2k+1)!(2^{k+1}-1)\frac{x^{2k+1}}{(2k+1)!}\right) \end{array} $$ hence $$ \begin{cases} \,\,\,a_{2k}\; = (2^{k+1}-1)\cdot(2k)! \\ a_{2k+1} = (2^{k+1}-1)\cdot(2k+1)! \end{cases} \verb+ +\forall k\ge0 $$