The line - codes we are looking at consist of black and red lines. These lines can have width 1 or 2. Black and red lines are taking turns, black line, red line, black line, ... The code ends and begins with a black line (width 1 or 2). The width of the full code is 14.


How many codes are possible ? Give a proof.

Example: enter image description here

  • 3
    $\begingroup$ This is just a normal combinatorics problem. It belongs on math.stackexchange.com, not here. $\endgroup$ Jun 22 '19 at 20:39

Let $R(n)$ be the number of codes of total width $n$, so the question asks for $R(14)$.

Consider any code that contains at least $3$ stripes. You can strip off the last two stripes, and still have a valid code that begins and ends in a black stripe, and conversely any valid code can be extended by appending a red and a black stripe to it. The two stripes can be of width $\color{red}{1}+1$, $\color{red}{1}+2$, $\color{red}{2}+1$, or $\color{red}{2}+2$. This means that the $R(n)$ satisfies the recurrence relation: $$R(n)=R(n-2)+2R(n-3)+R(n-4)$$ It is fairly obvious that $$R(0)=0\\R(1)=1\\R(2)=1\\R(3)=1\\R(4)=3$$ We can simply extend the sequence by applying the recurrence relation: $$0, 1, 1, 1, 3, 4, 6, 11, 17, 27, 45, 72, 116, 189, 305, 493, 799, ...$$ So $R(14)=305$.

Some additional maths: If there were no colours, so that the codes could have any number of stripes, then the sequence would be the Fibonacci numbers. In this case you almost have a Fibonacci sequence too, as each term is at most one away from being the sum of the previous two terms. In fact, $$R(n)=R(n-1)+R(n-2)+c_n$$ where $c_n$ is $-1$, $1$, or $0$, when $n$ is $0$, $1$, $2$ modulo $3$ respectively.
The ratio of successive terms is the golden ratio, just like Fibonacci. If you look at the associated polynomial, $x^4=x^2+2x+1$, it has the roots $\phi$, $1/\phi$, (like Fibonacci) but also the complex roots $\omega$ and $\omega^2$ which causes those offsets of $-1$, $0$, and $1$.

  • $\begingroup$ Note that the polynomial reduces to (x^2)^2 = (x+1)^2. If the code has n colors in sequence, the characteristic polynomial becomes (x^2)^n = (x+1)^n, providing additional roots. $\endgroup$ Jun 30 '19 at 14:11

There are:

$305$ ways to do this.


Each code belongs to one of the following generating functions.
$$\color{black}{(x+x^2)}\color{red}{(x+x^2)}\color{black}{(x+x^2)}\color{red}{(x+x^2)}\color{black}{(x+x^2)}\color{red}{(x+x^2)}\color{black}{(x+x^2)}$$ $$\color{black}{(x+x^2)}\color{red}{(x+x^2)}\color{black}{(x+x^2)}\color{red}{(x+x^2)}\color{black}{(x+x^2)}\color{red}{(x+x^2)}\color{black}{(x+x^2)}\color{red}{(x+x^2)}\color{black}{(x+x^2)}$$ $$\color{black}{(x+x^2)}\color{red}{(x+x^2)}\color{black}{(x+x^2)}\color{red}{(x+x^2)}\color{black}{(x+x^2)}\color{red}{(x+x^2)}\color{black}{(x+x^2)}\color{red}{(x+x^2)}\color{black}{(x+x^2)}\color{red}{(x+x^2)}\color{black}{(x+x^2)}$$ $$\color{black}{(x+x^2)}\color{red}{(x+x^2)}\color{black}{(x+x^2)}\color{red}{(x+x^2)}\color{black}{(x+x^2)}\color{red}{(x+x^2)}\color{black}{(x+x^2)}\color{red}{(x+x^2)}\color{black}{(x+x^2)}\color{red}{(x+x^2)}\color{black}{(x+x^2)}\color{red}{(x+x^2)}\color{black}{(x+x^2)}$$


Each generating function must contain a certain number of $x^2$ (regardless of color) to hit a code length of $14$. Hence the answer is:

  • $\begingroup$ I have an other solution with a little bit more codes. $\endgroup$
    – Matti
    Jun 22 '19 at 10:58
  • 2
    $\begingroup$ 9; 5 = 126 not 105, if you sum that up correctly then you get the same result as Jaap Scherphuis. $\endgroup$ Jun 22 '19 at 14:08

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