Chris Henson

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July 12, 2020
A Fibbonacci Power Series

This article follows question 137 and question 140 from Project Euler.

First, let us recall the definition of the Fibonacci numbers. We start with:

$$ F_0=0,\quad F_1= 1 $$

and define further numbers by the relation:

$$ F_n=F_{n-1} + F_{n-2} $$

This gives us the first few terms:

$$ 0,\;1,\;1,\;2,\;3,\;5,\;8,\;13,\;21,\;34,\;55,\;89,\;144,\; \ldots $$

(Please note that Project Euler omits the first index, but this will not change our calculations in any meaningful way, and I will try to be as explicit as possible about this.)

The natural generalization is to consider a different starting point. I will write

$$ G(a, b, n) = G(a, b, n-1) + G(a, b, n-2) $$

given the starting point:

$$ G(a, b, 0) = a,\quad G(a, b, 1) = b $$

and when we are working with a set $a$ and $b$ I will simply write $G_n$. You can see that $G(0, 1 , n)$ gives the Fibonacci numbers themselves.

These two Project Euler questions are concerned with the power series with coefficients given by the above sequences. That is, we are interested in the value of the summations:

$$ \sum_{k=1}^\infty G(a, b, k) x^k = G(a, b, 1)x + G(a, b, 2)x^2 + G(a, b, 3)x^3 + \dots $$

Again, if we take $a=0$ and $b=1$, we will have the Fibbonacci numbers:

\begin{align*} \sum_{k=1}^\infty G(0, 1, k) x^k &= F_1x + F_2x^2 + F_3x^3 + \dots \\ & = x + x^2 + 2x^3 + 3x^4 +5x^5 \dots \end{align*}

A Generating Function

Interestingly, it is not too difficult to derive a closed-form solution to these series. Let us fix $G_0 = a$ and $G_1 = b$. Then the power series is:

\begin{align*} \sum_{k=0}^\infty G_k x^k = G_0 + G_1 x + G_2 x^2 + \dots \end{align*}

Please note that I have included the initial term $G_0$ that Project Euler omits. We can correct for this at the end by simply subtracting this term.

First, I will pull out the first two given terms from the series:

\begin{align*} \sum_{k=0}^\infty G_k x^k = G_0 + G_1x + \sum_{k=2}^\infty G_k x^k \end{align*}

Now I can Replace $G_k$ with the relation that defines it, and split apart the summation further:

\begin{align*} \sum_{k=0}^\infty G_k x^k &= G_0 + G_1x + \sum_{k=2}^\infty (G_{k-1} + G_{k-2}) x^k \\ &= G_0 + G_1x + \sum_{k=2}^\infty G_{k-1} x^k + \sum_{k=2}^\infty G_{k-2} x^k \end{align*}

Next, I notice that multiplying my original series by $x$ and $x^2$ gives:

\begin{align*} x\sum_{k=0}^\infty G_k x^k &= G_0x + G_1 x^2 + G_2 x^3 + \dots\\ x^2\sum_{k=0}^\infty G_k x^k &= G_0x^2 + G_1 x^3 + G_2 x^4 + \dots \end{align*}

Substituting these into the above equation I have:

\begin{align*} \sum_{k=0}^\infty G_k x^k &= G_0 + G_1 x + x\sum_{k=0}^\infty G_k x^k - G_0 x + x^2\sum_{k=0}^\infty G_k x^k \end{align*}

Denoting the original summation as $s(x)$ this is:

$$ s(x) = G_0 + G_1 x + xs(x) - G_0 x + x^2 s(x) $$

and solving gives:

$$ s(x) = \dfrac{G_0 - G_0 x + G_1 x}{1 - x - x^2} $$

Remember that this includes our first term. So we have:

$$ \sum_{k=0}^\infty G_k x^k = G_0 + G_1 x + G_2 x^2 + \dots = \dfrac{G_0 - G_0 x + G_1 x}{1 - x - x^2} $$

For the Fibbonacci series we set $G_0 = 0$ and $G_1 = 1$ giving:

$$ \sum_{k=0}^\infty F_k x^k = F_0 + F_1 x + F_2 x^2 + \dots = \dfrac{x}{1 - x - x^2} $$

Rational Points that Give Integer Solutions

The next question under consideration is which rational values of x give an integer value for our power series. I will write:

$$ s(x) = \sum_{k=1}^\infty G_k x^k = G_0 + G_1 x + G_2 x^2 $$

Above we had the equation:

$$ s(x) = G_0 + G_1 x + xs(x) - G_0 x + x^2 s(x) $$

which solving for x using the quadratic formula gives us:

$$ x = \dfrac{-(G_1 - G_0 + s) + \sqrt{(G_1 - G_0 + s)^2 - 4s(G_0-s)}}{2s} $$

Note that we have two choices for roots, but one is outside the radius of convergence for the power series.

Remember that we included the term $G_0$. If we would like to correct for this, we can make the transformation $s \rightarrow S$ where:

\begin{align*} S(x)&= \sum_{k=1}^\infty G_k x^k \\ &= \sum_{k=0}^\infty G_k x^k - G_0\\ &= s(x) - G_0 \end{align*}

which gives instead:

$$ x = \dfrac{-(G_1 + S) + \sqrt{(G_1 + S)^2 - 4(S + G_0)(-S)}}{2(S + G_0)} $$

Using the second equation and taking $G_0 = 0$ and $G_1 = 1$ to give the Fibbonacci power series, we have have the following values:

\begin{align*} S\left(\sqrt{2} -1\right) &= 1\\ S\left(\dfrac{1}{2}\right) &= 2\\ S\left(\dfrac{\sqrt{13} - 2}{3}\right) &= 3\\ S\left(\dfrac{\sqrt{89} - 5}{8}\right) &= 4\\ \vdots \end{align*}

We can see for instance that $S = 2$ has the desired property that $x$ is rational. What conditions must be met for this? First, look at our equation for x where we substitute the particular values for $G_0$ and $G_1$:

$$ x = \dfrac{-(1 + S) + \sqrt{(1 + S)^2 - (2S)^2}}{2S} $$

This will be rational exactly when the discriminant $(1 + S)^2 - (2S)^2$ is a perfect square (The following arguement is directly from this article)

Since the discriminant is itself a difference of two squares, we can consider them as Pythagorean triples so that for integers m, n:

$$ S + 1 = m^2 - n^2,\quad 2S = 2mn $$

This gives the equation:

$$ m^2 - n^2 = mn + 1 $$

which can be written as

$$ (2m-n)^2 = 4 + 5n^2 $$

We need $4 + 5n^2$ to be a perfect square to ensure a rational x. Unexpectedly, using $n = F_{2j}$ (where j is a positive integer) ensures this is satisfied:

\begin{align*} 4 + 5n^2 &= 4 + 5F_{2j} = 4 + 5 \left( \frac{(\varphi^{2j}-\psi^{2j})^2}{5} \right)\\ &= 4 + \varphi^{4j} - 2(\varphi\psi)^{2j} + \psi^{4j}\\ &= \varphi^{4j} + 2 + \psi^{4j}\\ &= \varphi^{4j} + 2 (\varphi\psi)^{2j}+ \psi^{4j}\\ &= (\varphi^{2j} + \psi^{2j})^2 \end{align*}

where I have used the Binet formula in the first line.

Furthermore, substituing into the equation $(2m-n)^2 = 4 + 5n^2$ above gives:

\begin{align*} m &= \dfrac{1}{2} (\varphi^{2j} + \psi^{2j} + F_{2j})\\ &= \dfrac{1}{2} \left( \varphi^{2j} + \psi^{2j} + \frac{\varphi^{2j}-\psi^{2j}}{\sqrt{5}} \right)\\ &= \dfrac{\varphi\varphi^{2j} - \psi\psi^{2j}}{\sqrt5} = \dfrac{\varphi^{2j+1} - \psi^{2j+1}}{\sqrt5}\\ &= F_{2j+1} \end{align*}

Finally, substituting this back into the original series gives:

$$ \sum_{k=1}^\infty F_k (F_{2j}/F_{2j+1})^k = F_{2j}F_{2j+1}, \quad j \geq 1 $$

What an unexpected result! That we could consider successive coefficients in identifying these rational points is certainly not intuitive. We see that these numbers also grow very quickly:

\begin{align*} F_{2j}F_{2j+1}, n \in \mathbb{Z}^+= \{2, &15, 104, 714, 4895, 33552, 229970, 1576239, 10803704, 74049690,\\ & 507544127, 3478759200, 23843770274, 163427632719, 1120149658760, \dots\} \end{align*}

The last number in this list is the solution to problem 137. Problem 140 considers the initialization $G_0 = 3$ and $G_1 = 1$, which give a more complicated sort of equation that determines the rational points. Still, they can be written entirely in terms of the Fibonacci series.

Haskell implementation

Interestingly, the scale of the solutions makes checking finding rational solutions non-trivial in terms of computational time. My strategy was to find the first solutions, identify a general pattern, then find the remaining solutions.

Haskell makes it quite easy to define functions that themselves return functions, allowing us to build general solutions.

I define a function that returns a discriminant function for a particular generalized Fibonacci power series:

import Math.NumberTheory.Powers.Squares
import Data.Ratio
import Data.Sort

discriminant_check :: Int -> Int -> (Int -> Int)
discriminant_check a b = f where
                         f s = s^2 + 2*s*b + b^2 + 4*s^2 + 4*a*s

On my desktop I can retrieve the first seven rational solutions within a reasonable amount of time using the following:

let discriminant_fib = discriminant_check 0 1
let partial_nuggets = take 7 $ filter (\x -> isSquare (discriminant_fib x)) [1..]

For more solutions in a timely manner I first need the generating function:

generator_func :: Ratio Integer -> Ratio Integer -> (Ratio Integer -> Ratio Integer)
generator_func a b = f where
               f x = (a - a*x + b*x)/(1-x-x^2) - a

I also need the sequence itself:

gen_fib :: Int -> Int -> [Int]
gen_fib a b = seq where
              seq = a : b : zipWith (+) seq (tail seq)
              
fib = gen_fib 0 1

and can then generate the solutions:

let fib_generator = generator_func 0 1
let fib_nugget_2 = take 15 $
                   filter (>0) $
                   map (fib_generator) $ 
                   zipWith (%) (map (toInteger) [j | (i, j) <- zip [2..] fib, even i]) 
                               (map (toInteger) [j | (i, j) <- zip [2..] fib, odd i])

I can equivalently just use the relation that we showed above:

let fib_nugget = take 15 $ 
                 filter (>0) $ 
                 zipWith (*) [j | (i, j) <- zip [1..] fib, even i] 
                             [j | (i, j) <- zip [1..] fib, odd i]

The solution for problem 140 is very similar, but exploits a different pattern in the rational solutions which utilizes different sequences:

let gen = gen_fib 3 1

let discriminant_gen = discriminant_check 3 1
let partial_nuggets_gen = take 15 $ filter (\x -> isSquare (discriminant_gen x)) [1..]

let gen_func = generator_func 3 1

let g_nug = map (gen_func) $
            zipWith (%) (filter (>0) $ map (toInteger) [j | (i, j) <- zip [1..] (gen_fib 2 5), odd i]) 
                        (filter (>0) $ map (toInteger) [j | (i, j) <- zip [1..] (gen_fib 2 5), even i])

let f_nug = map (gen_func) $
            [j | (i, j) <- zip [1..] (zipWith (%) (map (toInteger) (gen_fib 1 2)) 
                                                  (map (toInteger) (gen_fib 2 3))), odd i]
            
let nug = take 30 $ merge g_nug f_nug

The full code with comments is availible here.