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Suppose I want to calculate the coefficient of a certain term (modulo a prime $p$) in the series expansion of $$f(x) = \frac{1}{1 - x^u - x^v}$$ where $u,v$ are positive integers. We know that $$\frac{1}{1 - x} = \sum_{k = 0}^\infty x^k$$ Therefore $$f(x) = \sum_{k = 0}^\infty \left(x^u + x^v\right)^k$$ The expansion of $(x^u + x^v)^k$ contains a term of degree $m$, if and only if there exist integers $r,s$ satisfying $$\left\{\begin{aligned}r + s &= k\\ru + sv &= m\end{aligned}\right.$$ In this case the coefficient of that term is $\binom{k}{r}$.

In general all integral solutions to the equation $ru + sv = m$ can be represented as $$\left\{\begin{aligned}r &= a + ck\\s &= b + dk\end{aligned}\right.$$ where $a,b,c,d$ are integer constants and $k$ is an integer variable.

Therefore what I want to calculate is $$\sum_{k = k_1}^{k_2} \binom{r + s}{r} \quad (\mathrm{mod}\; p)$$ where $k_1,k_2$ are necessary bounds on $k$ to ensure both $r,s$ are positive. If $k_2 - k_1$ is small I could just use Lucas's theorem to calculate the residue for each term. However in my problem $k_2 - k_1$ is really large (around $10^{11}$). Is there a better algorithm to calculate this sum?

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There is: note that your coefficients $c_n$ satisfy a fibonnaci-style recurrence relation $c_n=c_{n-u}+c_{n-v}$ (this can be proven by classic generating function methods - note that $(1-x^u-x^v)f(x)=1$ and expand the series term-by-term on both sides; initial conditions should be easy to figure out). From there, there are plenty of methods that can be used, but the best for your purposes (large-$n$ coefficients modulo a prime) are probably the matrix multiplication methods - you can create a vector of coefficients $\mathfrak{C_n}=\langle c_n-1, c_{n-1}, \ldots, c_{n-v}\rangle$ and then the coefficients of $\mathfrak{C_{n+1}}$ are derivable by just taking $\mathfrak{C_{n+1}}=M\mathfrak{C_n}$ for some matrix $M$. This means that $\mathfrak{C}_{n+k}=M^k\mathfrak{C}_n$, and so by computing large powers of $M$ (for which the usual binary method can be used), large values of $c_n$ can be computed. If your $v$ is large then the matrices may be large, but since all your work can be done mod $p$, then none of the coefficients should get out of hand.

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  • $\begingroup$ Yes, I know this problem can be brute forced by the matrix method. I'm just curious if there is a faster method for large u,v. This problem is abstracted from a competitive programming problem. $\endgroup$ – user2249675 Mar 11 at 5:02
  • $\begingroup$ @user2249675 What are the specific parameters on your problem? You mentioned values of $k_2-k_1$ but it'd be good to have that directly translated back into (rough) $u$ / $v$ and $n$ ranges. $\endgroup$ – Steven Stadnicki Mar 13 at 20:50
  • $\begingroup$ Thank you. I have solved the original problem (ProjectEuler 258). In that problem $u = 1999,v = 2000,n = 10^{18}$. The trick is to reduce matrix multiplication via Cayley-Hamilton theorem. $\endgroup$ – user2249675 Mar 13 at 20:56
  • $\begingroup$ It seems at first blush that using Cayley-Hamilton is basically just equivalent to solving the original problem, though it's very possible I'm missing something there. In any case, though, an $n$ of $10^{18}$ means about 60 squaring operations of the matrix, and kerrywong.com/2009/03/07/matrix-multiplication-performance-in-c suggests that a $2000\times2000$ multiply in naive code takes a bit less than a minute, so even the relatively naive approach here is pretty feasible in the grand scheme of things. $\endgroup$ – Steven Stadnicki Mar 13 at 21:47

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