[Math] Number of Fibonacci numbers in a range

fibonacci-numbers

The definition of the Fibonacci numbers is given by:

$$\begin{align}f_1 &= 1;\\ f_2 &= 2;\\
f_n &= f_{n-1} + f_{n-2},\qquad (n >= 3);
\end{align}$$

now we are given two numbers $a$ and $b$, and we have to calculate how many Fibonacci numbers are in the range $[a,b]$. How can we calculate it?

Best Answer

We know that $F_n\approx \frac {\phi^n}{\sqrt 5}$, so given $a$, the next larger Fibonacci number is $F_k$, where $k= \left \lceil\frac {\log (a\sqrt 5)}{\log \phi }\right \rceil$. Similarly the $F_m$ below $b$ is $m= \left \lfloor\frac {\log (b\sqrt 5)}{\log \phi }\right \rfloor$, then there are $m-k+1$ between $a$ and $b$. You have to think about what you want if $a$ or $b$ are themselves Fibonacci numbers.

Related Solutions

Fibonacci, Tribonacci, and Similar Sequences – Recurrence Relations

In addition to André's notes, another means of calculating solutions to these recurrence relations is to rephrase them using linear algebra as a single matrix multiply and then apply the standard algorithms for computing large powers of numbers (i.e., via binary representation of the exponent) to computing powers of the matrix; this allows for the $n$th member of the sequence to be computed with $O(\log(n))$ multiplies (of potentially exponentially-large numbers, but the multiplication can also be sped up through more complicated means).

In the Fibonacci case, this comes by forming the vector $\mathfrak{F}_n = {F_n\choose F_{n-1}}$ and recognizing that the recurrence relation can be expressed by multiplying this vector with a suitably-chosen matrix: $$\mathfrak{F}_{n+1} = \begin{pmatrix}F_{n+1} \\\\ F_n \end{pmatrix} = \begin{pmatrix}F_n + F_{n-1} \\\\ F_n \end{pmatrix} = \begin{pmatrix} 1&1 \\\\ 1&0 \end{pmatrix} \begin{pmatrix} F_n \\\\ F_{n-1} \end{pmatrix} = M_F\mathfrak{F}_n $$

where $M_F$ is the $2\times2$ matrix $\begin{pmatrix} 1&1 \\\\ 1&0 \end{pmatrix}$. This lets us find $F_n$ by finding $M_F^n\mathfrak{F}_0$, and as I noted above the matrix power is easily computed by finding $M_F^2, M_F^4=(M_F^2)^2, \ldots$ (note that this also gives an easy way of proving the formulas for $F_{2n}$ in terms of $F_n$ and $F_{n-1}$, which are just the matrix multiplication written out explicitly; similarly, the Binet formula itself can be derived by finding the eigenvalues of the matrix $M_F$ and diagonalizing it).

Similarly, for the Tribonacci numbers the same concept applies, except that the matrix is 3x3: $$\mathfrak{T}_{n+1} = \begin{pmatrix} T_{n+1} \\\\ T_n \\\\ T_{n-1} \end{pmatrix} = \begin{pmatrix} T_n+T_{n-1}+T_{n-2} \\\\ T_n \\\\ T_{n-1} \end{pmatrix} = \begin{pmatrix} 1&1&1 \\\\ 1&0&0 \\\\ 0&1&0 \end{pmatrix} \begin{pmatrix} T_n \\\\ T_{n-1} \\\\ T_{n-2} \end{pmatrix} = M_T\mathfrak{T}_n$$ with $M_T$ the $3\times3$ matrix that appears there; this is (probably) the most efficient all-integer means of finding $T_n$ for large values of $n$, and again it provides a convenient way of proving various properties of these numbers.

[Math] Sum of product of Fibonacci numbers

Let's use $$ F_n = \frac{1}{\sqrt{5}} \left( \phi^n - (-1)^n \phi^{-n} \right) $$ where $\phi$ is a Golden ratio constant $\phi = \frac{\sqrt{5}+1}{2}$. Now the sum reduces to the combination of geometric sums: $$ \begin{eqnarray} \sum_{k=1}^{n-1} F_k F_{n-k} &=& \frac{1}{5}\sum_{k=1}^{n-1} \left(\phi^k - (-1)^k \phi^{-k}\right)\left(\phi^{n-k} - (-1)^{n-k} \phi^{k-n}\right) \\ &=& \frac{1}{5}\sum_{k=1}^{n-1} \left(\phi^n - (-1)^k \phi^{n-2k} - (-1)^{n-k} \phi^{2k-n} + (-1)^n \phi^{-n} \right) \\ &=& \frac{n-1}{5} \left(\phi^{n} + (-1)^n \phi^{-n}\right) - \frac{\phi^{n}}{5} \sum_{k=1}^{n-1} \left( -\phi^{-2}\right)^{k} - \frac{(-1)^{n}}{5} \phi^{-n} \sum_{k=1}^{n-1} \left(-\phi^2\right)^k \end{eqnarray} $$ Using the geometric sum $\sum_{k=1}^{n-1} x^k = \frac{x-x^n}{1-x}$ we get: $$\begin{eqnarray} \sum_{k=1}^{n-1} F_k F_{n-k} &=& \frac{n-1}{5} \left(\phi^{n} + (-1)^n \phi^{-n}\right) + \frac{2}{5} \frac{\phi^{n-1} + (-1)^n \phi^{1-n}}{\phi + \phi^{-1}} \\ &=& \frac{n-1}{5} L_n + \frac{2}{5} F_{n-1} \end{eqnarray} $$ where $L_n$ denotes $n$-th Lucas number.

Quick sanity check in Mathematica:

In[84]:= Table[
 Sum[Fibonacci[k] Fibonacci[n - k], {k, 1, n - 1}] == (n - 1)/
    5 LucasL[n] + 2/5 Fibonacci[n - 1], {n, 1, 6}]

Out[84]= {True, True, True, True, True, True}

The sum stated in the edited question is obtained by replacing $n$ with $n+1$ in the sum evaluated above: $$ \sum_{k=1}^{n} F_k F_{n+1-k} = \frac{n}{5} L_{n+1} + \frac{2}{5} F_n $$

Best Answer

Related Solutions

Fibonacci, Tribonacci, and Similar Sequences – Recurrence Relations

[Math] Sum of product of Fibonacci numbers

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