why no solutions here?

This can be done recursively. Call an array "valid" if it satisfies the condition in the problem. Note that when we have a valid array of size $n$, and delete the first column, we get a valid array of size $n-1$. Moreover, when we have a valid array of size $n-1$, there are only two valid arrays of size $n$ that's an extension of that array. (By extension I mean shift everything to the right by one and then populate the first column). That is, once you pick $a(1,1)$, it completely determines the rest of the first column. Definition: call a valid array "good" if it's a fixed point of $f$ A good array of size $n$ can be reduced to good array of size $n-1$ by deleting first column. But good array of size $n-1$ may or may not be extended to be a good array of size $n$. Definition: call a valid array "awesome" if it has an even number of ones in the first column. Lemma 1: A good array of size $n-1$ can only be extended to be a good array of size $n$ if and only if it is awesome. Furthermore, when we are extending an awesome array, there are two good arrays of size $n$ that are possible. Proof of lemma 1 is fairly easy. Just populate $a(1,1)$ and work your way down the first column making sure to keep the arrays valid. Each $1$ in the first column of former array (so now becomes column 2) would cause your entry to flip. So in order for you to have $a(n,1) = a(1,1)$ then you must encounter an even number of $1$s. If you do have an even number of ones then each choice of $a(1,1)$ would result in a good array. If you have an even number of ones, then you'll end up with $a(n,1)$ that is different from $a(1,1)$ making the resulting array valid but not good. Now, for each choice of $a(1,1)=0,1$ the resulting extension will be valid and good, so we have two good arrays that are possible. Lemma 2: Part 1: Every awesome array is also good. Part 2: A good array must have the first column palindromic (that is, when it's reversed it is the same). Proof of lemma 2 is also easy by induction. I put it 2 parts because we need both parts in the induction step (so that's why it's not lemma 2 and lemma 3). This is easily verified for small $n$. For induction step: a good array must be an extension of an awesome array of size $n-1$, and it is also good and therefore palindromic. Just observe what happens when you populate $a(1,1)$ first or $a(n,1)$ first, and the process is the same. So from the lemma 1 we know each awesome array of size $n-1$ can be extended to 2 good arrays of size $n$. If $n$ is odd, then out of these two good arrays, exactly one is awesome, one is good but not awesome. (Proof: each choice of $a(1,1)=0$ or $a(1,1)=1$ produces completely opposite result, so one of them must have an even number of one, the other one an odd number). But if $n$ is even, then both of these two good arrays are awesome. This can be shown by lemma 2, because the first column is palindromic. If a binary sequence of even length is palindromic then it must have an even number of zeros and ones. So putting it together, let $x(n)$ be the number of good arrays of size $n$, and $y(n)$ be the number of awesome arrays. In general, if $n$ is even: $x(n) = 2y(n-1), y(n) = 2y(n-1)$ because both extensions of awesome arrays will be awesome again. And if $n$ is odd: $x(n) = 2y(n-1), y(n) = y(n-1)$ because exactly one extension will be awesome. The first few values: $x(1) = 2, y(1) = 1$ $x(2) = 2, y(2) = 2$ $x(3) = 4, y(3) = 2$ $x(4) = 4, y(4) = 4$ $x(5) = 8, y(5) = 4$ Easily proved by induction: $x(n) = 2^{\lceil n/2 \rceil}, y(n) = 2^{\lfloor n/2 \rfloor}$ The final answer that you want is $x(n) = 2^{\lceil n/2 \rceil}$

ComplexPhi wrote: Given a positive integer $n$, consider a triangular array with entries $a_{ij}$ where $i$ ranges from $1$ to $n$ and $j$ ranges from $1$ to $n-i+1$. The entries of the array are all either $0$ or $1$, and, for all $i > 1$ and any associated $j$ , $a_{ij}$ is $0$ if $a_{i-1,j} = a_{i-1,j+1}$, and $a_{ij}$ is $1$ otherwise. Let $S$ denote the set of binary sequences of length $n$, and define a map $f \colon S \to S$ via $f \colon (a_{11}, a_{12},\cdots ,a_{1n}) \to (a_{n1}, a_{n-1,2}, \cdots , a_{1n})$. Determine the number of fixed points of $f$. Here's outline of a proof which involves linear algebra (not a lot really). For the sake of ease, call $a_{1i}$ as $x_i$. Throughout the steps, we will work in $\mathbb{F}_2$

The answer is $2^{\lceil \frac{n}{2} \rceil}$ which I will prove by induction. Note that the given condition is just addition mod $2$ of the previous two elements on the previous row. First, observe that every diagonal of the triangular array must be symmetric. This is easy to see by first looking at the first diagonal (with one element) and then inductively looking at the next diagonal and since the start and end elements are same and since its just adding a previously symmetric thing, the order of this diagonal shouldn't matter which way you start from. Also, see that if we want to add a new diagonal, then the bottom element of the new diagonal is going to be the top element added to the sum of numbers of the previous diagonal. Therefore, we must have that there are an even number of $1$'s on all but the last diagonal. Suppose we focus on only looking for arrays with all diagonals having an even number of $1$'s on its diagonals. If $n$ is odd, and we're looking for $n+1$, then since its symmetric, it has an even number of $1$'s anyway so it doesn't matter if we start with a $0$ or $1$ so we get two times the number of ways. If $n$ was even instead, then we want the "middle" element of the new diagonal to be a $0$, since changing the first element toggles this one, exactly one of $0$ and $1$ will work as the new element, so we have the same number of ways to do it. So inducting, we have that there are $2^{\lceil \frac{n+1}{2} \rceil} - 1$ arrays with all diagonals an even number of $1$'s so for the problem, we have twice these, so its $2^{\lceil \frac{n}{2} \rceil}$, as claimed. $\blacksquare$