# WARNING: The text below provides a guidance for a Project Euler problem.

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### A lagged Fibonacci sequence

The problem tasks us with solving a Fibonacci-like recurrence

for a large value of $k$. The first approach that comes to mind is to use matrix exponentiation in the same way as one does for calculating large values of the Fibonacci sequence; this however does not work by far fast enough due to the size of the matrix needed. There are, however, employable simplifications.

Let $M$ be the $2000 \times 2000$ matrix corresponding to the recurrence, that is:

The matrix $M$ has by definition the property that:

and we are thus looking for a way to compute $M^{10^{18}} \mathbf 1$, whereupon the bottommost entry of this will be the sought $g(10^{18})$. The power is clearly too large; and this is where the »Cayley-Hamilton theorem« comes to the rescue! The theorem states that every square matrix satisfies its characteristic equation. What is the characteristic equation of $M$? Well, this can be read off easily from the recurrence itself: $X^{2000} = 1 + X$.

From this it follows that $M^{10^{18}}$ may be rewritten as a polynomial of degree $% $, this being the corresponding representative in the quotient ring $\frac{\mathbb Z [X]} {(X^{2000} - 1 - X)}$. Let this polynomial be $\sum_{i = 0}^{1999} a_i M^i$ (This can without fancy terms be understood as the unique polynomial of degree $% $ such that $M^{10^{18}} = g(M) (X^{2000} - 1 - X) + \sum_{i = 0}^{1999} a_i M^i$ for some polynomial $g$; i.e. a polynomial version of the concept of remainder.). But then:

Reading off the bottommost entries in the above equation, we obtain:

Hence the only thing that we need to do is to find the polynomial representing $X^{10^{18}}$ in $\frac{\mathbb Z [X]} {(X^{2000} - 1 - X)}$ and sum up its coefficients! The following piece of Julia code returns the answer in about 0.1 seconds.