$$ P(x) = \frac{4x^6}{(1-x)^2(1-2x)} $$
How do I make a $$ P(n) $$ in closed form of expression? Any tips would be appreciated, thank you.
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1$P(n)$ is already in closed form... "not closed form" usually means "involving integrals", like here. – Parcly Taxel Oct 14 '20 at 16:01
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2not even sure what you are looking for – gt6989b Oct 14 '20 at 16:01
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1I guess $P(n)=\frac{4n^6}{(1-n)^2(1-2n)}$. – Oct 14 '20 at 16:02
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What about $\displaystyle\frac{4n^6}{(1-n)^2(1-2n)}?$. – José Carlos Santos Oct 14 '20 at 16:04
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3Do you want to find the sequence that is related to $P(x)$ as a generating function? – player3236 Oct 14 '20 at 16:06
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1You may find OEIS sequence A258547 of interest. The sequence begins: $16,44,104,228,480,988,2008,\dots$. – Somos Oct 14 '20 at 16:29
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Let $$\sum_{n\ge 0}p_n x^n= \frac{4 x^6}{(1 - 2 x) (1 - x)^2}.$$ Via partial fraction decomposition, \begin{align} \frac{4 x^6}{(1 - 2 x) (1 - x)^2} &= -\frac{49}{4} - \frac{17}{2}x - 5 x^2 - 2 x^3 + \frac{16}{1-x} - \frac{4}{(1 - x)^2} + \frac{1}{4 (1-2 x)} \\ &= -\frac{49}{4} - \frac{17}{2}x - 5 x^2 - 2 x^3 + 16\sum_{n\ge 0} x^n - 4\sum_{n\ge 0} \binom{n+1}{1} x^n + \frac{1}{4} \sum_{n\ge 0} (2x)^n \\ &= -\frac{49}{4} - \frac{17}{2}x - 5 x^2 - 2 x^3 + \sum_{n\ge 0} \left(12 - 4n + 2^{n-2}\right) x^n, \end{align} which immediately implies that \begin{align} p_n &= -\frac{49}{4}[n=0] - \frac{17}{2}[n=1] - 5[n=2] - 2[n=3] + 12 - 4n + 2^{n-2} \\ &= \begin{cases} 0 & \text{if $0 \le n \le 3$} \\ 12 - 4n + 2^{n-2} &\text{if $n > 3$}\end{cases} \end{align}
RobPratt
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