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I've got a sequence that's defined by the coefficients of a Taylor series (of a function) centred at x=0. The function in question is : $$f(x) = \frac{1}{(1-x)^3(1-x^3)^2}$$ So, taking the nth order derivative (at x=0) and dividing by n! leaves us with $$1 + 3 x + 6 x^2 + 12 x^3 + 21 x^4 + 33 x^5...$$ Cool, so I can work out the first few terms easily. How about if I wanted the 1000th term, for example? Is there an easier pattern or do I really have to compute (and sub zero into the derivative) : $$\frac{\frac{d^{1000}}{dx^{1000}}(f(x))}{1000!}$$ I failed to find a pattern with the coefficients that gets me around this calculation, so am I missing the pattern or is this problem "impossible" to solve this way? I noticed that because we need to sub in zero, any term that involves x on the numerator of any part of the expansion is also 0, but it doesn't get around calculating the derivative in the first place.

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    $\begingroup$ OEIS A011779 suggests this is related to Project Euler problem 577. There will be a 9th order recurrence $\endgroup$
    – Henry
    Nov 21, 2021 at 11:24
  • $\begingroup$ Your best bet would be to partial fraction the expression to known series, but not "all the way down" to irreducible polynomials, such as $$f(x) = \frac{A}{(1-x)^3} + \frac{B}{(1-x^3)^2} + \frac{C}{(1-x)^2} + \frac{D}{1-x^3} + \frac{E}{1-x} + F$$ and see if it works out nicely. $\endgroup$ Nov 21, 2021 at 11:33

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First, note that $$ \frac{1}{1 - x} = 1 + x + x^{2} + x^{3} + \cdots. $$ By differentiating both sides, we also get $$ \frac{1}{(1 - x)^{2}} = 1 + 2x + 3x^{2} + \cdots + (n+1)x^{n} + \cdots $$ and $$ \frac{1}{(1 - x)^{3}} = 2 + 6x + 12x^{2} + \cdots + (n+2)(n+1)x^{n} + \cdots. $$ From these, we can express $f(x)$ as $$ f(x) = \frac{1}{(1 - x)^{3}}\frac{1}{(1 - x^{3})^{2}} = \left(\sum_{n\geq 1}(n+2)(n+1)x^{n}\right) \left(\sum_{m\geq 0} (m+1)x^{3m}\right) = \sum_{k\geq 0}a_{k}x^{k} $$ where $$ a_{k} = \sum_{n, m \geq 0, n + 3m = k} (n+2)(n+1)(m+1) = \sum_{0\leq m \leq \lfloor k / 3\rfloor}(k-3m+2)(k-3m+1)(m+1) $$ which is also possible to find a closed form in terms of $\lfloor k/3\rfloor$.

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Working a little the coefficients (which make sequence $A011779$ in $OEIS$, we can write $$648\,a_n=3 n^4+54 n^3+333 n^2+n \left(810+16 \sqrt{3} \sin \left(\frac{2 \pi n}{3}\right)\right)+$$ $$624+24 \left(3 \sqrt{3} \sin \left(\frac{2 \pi n}{3}\right)+\cos \left(\frac{2 \pi n}{3}\right)\right)$$ which I did not find elsewhere.

So $a_{10^3}=4713478140$ and $a_{10^6}=4629712963476853138890$

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