# Calculating the "nth term" in terms of taylor series coefficients

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.

• OEIS A011779 suggests this is related to Project Euler problem 577. There will be a 9th order recurrence Nov 21, 2021 at 11:24
• 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. Nov 21, 2021 at 11:33

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 $$\lfloor k/3\rfloor$$.
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 $$a_{10^6}=4629712963476853138890$$