The Sierpiński triangle and the number of $(0,1)$-polynomials $p(x)$ where $p(x)^2$ has largest coefficient $k$. My Twitter bot @oeisTriangles randomly selects an OEIS "table"-style sequence and draws an image, where even terms are light-colored and odd terms are dark-colored. Today it tweeted an image for OEIS sequence A169950:

[T]riangle read by rows, in which $T(n,k)$ [...] is the number of [monic degree $n$ (0,1)-]polynomials of thickness $k$ ($1 \leq k \leq n+1$).

where

The thickness of a polynomial $f(x)$ is the magnitude of the largest coefficient in the expansion of $f(x)^2$.

The table begins:
$$
1 \\
1~~~~1 \\
1~~~~2~~~~1 \\
1~~~~5~~~~1~~~~1 \\
1~~~~8~~~~4~~~~2~~~~1 \\
1~~~~13~~~~8~~~~8~~~~1~~~~1 \\
1~~~20~~~15~~~18~~~~7~~~~2~~~~1
$$

Conjecture
This "parity triangle" looks like a Sierpiński triangle, which makes me conjecture that for all $n, k \in \mathbb N_{\geq 0},$ $$T(n,k+1) \equiv \binom{n}{k} \bmod 2.$$

Is this conjecture true? If so, is there a recurrence that explains this?
 A: Let $U(n,k)$ be the number of monic degree $n$ $(0,1)$-polynomials with constant term $1$ and thickness $k$. Every monic degree $n$ $(0,1)$-polynomial $f(x)$ factors as $x^j g(x)$ where $g$ is monic and has constant term $1$, and $f$ and $g$ have the same thickness, so we see that
$$T(n,k) = \sum_{m=0}^n U(n,k).$$
It is equivalent (by the hockey stick identity) to prove that
$$U(n,k) \equiv \binom{n-1}{k-2} \bmod 2.$$
Now, the reason I like $U$ better than $T$ is that, if $f_n x^n + \cdots + f_1 x + f_0$ is a monic degree $n$ $(0,1)$-polynomial with constant term $1$ then so is the reversal $f_0 x^n + \cdots + f_{n-1} x + f_n$. And a polynomial and its reversal have the same thickness.
Thus, all polynomials that are not their own reversals contribute twice and don't matter modulo $2$, and we can restrict our attention to palindromic polynomials. Let $V(n,k)$ be the number of monic degree $n$ palidromic $(0,1)$-polynomials with thickness $k$. So it is equivalent to prove
$$V(n,k) \equiv \binom{n-1}{k-2} \bmod 2.$$
I computed $V(n,k)$ up to $n=20$, and I observe the following formula:
$$V(n,k) = \begin{cases} 0 & n \equiv k \equiv 1 \bmod 2 \\ \binom{\lfloor (n-1)/2 \rfloor}{\lfloor (k-2)/2 \rfloor} & \text{otherwise} \\ \end{cases}. \qquad (\ast).$$
Define the right hand side of $(\ast)$ to be $W(n,k)$.
Lemma We have $W(n,k) \equiv \binom{n-1}{k-2} \bmod 2$.
Proof We use Lucas' formula. Let the base two expressions of $n-1$ and $k-2$ be $n-1 = (n_r n_{r-1} \cdots n_0)_2$ and $k-2 = (k_r k_{r-1} \cdots k_0)_2$. If $n \equiv k \equiv 1 \bmod 2$ then $n_0=0$ and $n_1=1$, so Lucas' formula has a factor of $\binom{0}{1}=0$ and gives $0$. Otherwise, the $\binom{n_0}{k_0}$ factor is $1$, so we deduce that
$$\binom{n-1}{k-2} \equiv \prod_{j=0}^r \binom{n_j}{k_j} \equiv \prod_{j=1}^r \binom{n_j}{k_j} \equiv \binom{\lfloor (n-1)/2 \rfloor}{\lfloor (k-2)/2 \rfloor} \bmod 2.$$
Either way, we get $W(n,k) \equiv \binom{n-1}{k-2} \bmod 2$. $\square$
Lemma Let $f(x)$ be a monic degree $n$ palidromic $(0,1)$-polynomial. Then the thickness of $f^2$ is just equal to the number of monomials with coefficient $1$ in $f$.
Proof For each monomial $x^j$ in $f$, the monomial $x^{n-j}$ also appears in $f$, so their product contributes to the coefficient of $x^n$ in $f^2$. We deduce that the coefficient of $x^n$ in $f^2$ is equal to the number monomials with coefficient $1$ in $f$. On the other hand, no other monomial in $f^2$ can have a higher coefficient than this, since any terms of $f^2$ is given by a sum over monomials of $f$. $\square$.
Lemma The number of monic degree $n$ palidromic $(0,1)$-polynomials with $k$ terms of coefficient $1$ is $W(n,k)$.
Proof If $n$ is odd, then the monomials of $f$ occur in distinct pairs, $x^j$ and $x^{n-j}$, so the number of them is even. This proves that we have the right formula when $n$ and $k$ are odd. When $n$ is odd and $k$ is even, we are required to include the pair $(1, x^n)$ and can choose $(k-2)/2$ other pairs to include; this gives the formula $\binom{(n-1)/2}{(k-2)/2}$ in this case.
The analysis when $n$ is even is similar; the parity of $k$ tells us whether or not to include $x^{n/2}$ and then the rest is as before. $\square$
This completes the proof.
