Recursive sequence modulo 4

Question

For all odd positive integers $k$, I define a recursive sequence by $$ d_k=2+ {k\choose 1}d_{k-2} + {k\choose 2}d_{k-4} + \dots +{k\choose \frac{k-1}{2}}d_1\\ d_1=2 $$

I want to study this sequence modulo $4$. By induction, it is easy to see that $d_k$ is either $0$ or $2$. Computing this sequence I get $$ 2,0,2,2,0,2,2,0,2,2,0,2,2,0,2\dots (\mod 4) $$ which made me think that $$ d_k\equiv 0 (\mod 4)\text{ if and only if } k\equiv 0 (\mod 3) $$

Do you have an idea how to prove that? I tried to prove but I don't find any nice behavior on the binomial coefficients that helps me.

Answer 1

Disclaimer: The following is more of a long comment than an answer. I will update it should I find out more.

Since saying that $d_k$ is $0$ or $2$ modulo $4$ is the same as saying that $d_k$ is even, allow me to rephrase your problem. Define $$ \begin{align} c_n &= 1 + \binom{2n+1}{1} c_{n-1} + \dotsc + \binom{2n+1}{n}c_0\\ c_0 &= 1 \end{align} $$ so that $d_{2n+1} = 2c_n$ for every $n \geq 0$. Then your question is equivalent to asking whether $$ c_n \text{ is even } \quad \text{iff} \quad n \equiv 1 \pmod 3. $$ This could be useful because, as a consequence of Lucas's theorem, $\binom{k}{i}$ is even if and only if at least one of the binary digits of $i$ is greater than the corresponding digit of $k$. In other words, $\binom{k}{i}$ is even if and only if the binary expansions of $k-i$ and $i$ have a $1$ in the same place (see also this answer on MSE). I used this to write some code to compute $c_n$ reasonably fast and checked that your conjecture holds at least for the first $10000$ terms of the sequence, but I don't have a proof, yet.

It might also be useful to note that the odd binomial coefficients, when arranged in Pascal's triangle, form an approximation of Sierpinski's triangle (you can find a proof here).

Answer 2

As in @A.P.'s answer, define $$ \begin{eqnarray} c_0 & = & 1 \\ c_n & = & 1 + \sum_{k = 1}^n {2n + 1 \choose k}c_{n - k} \end{eqnarray} $$ so that $ d_{2n + 1} = 2c_n $. Let's prove by induction that $ c_n $ is even iff $ n \equiv 1 \mod 3 $.

$ \bullet $ Initialisation: $ c_0 = 1 $ is odd.

$ \bullet $ Induction: We have $$ \begin{eqnarray} c_n & = & 1 + \sum_{k = 1}^n {2n + 1 \choose k}c_{n - k} \\ & \equiv & 1 + \sum_{k \in [\![1, n]\!], k \not\equiv n - 1\mod 3} {2n + 1 \choose k} \\ & = & 1 + \frac{\sum_{k = 1}^{2n} {2n + 1 \choose k}}2 - \sum_{k \in [\![1, n]\!], k \equiv n - 1\mod 3} {2n + 1 \choose k} \\ & = & 2^{2n} - \sum_{k \in [\![1, n]\!], k \equiv n - 1\mod 3} {2n + 1 \choose k} \\ & \equiv & \sum_{k \in [\![1, n]\!], k \equiv n - 1\mod 3} {2n + 1 \choose k} \mod 2 \\ & \equiv & \sum_{k \in [\![0, n]\!], k \equiv n - 1\mod 3} {2n + 1 \choose k} + \begin{cases}1 \text{ if $ 3 \mid n - 1 $} \\0 \text{ otherwise}\end{cases} \mod 2 \end{eqnarray} $$ Notice furthermore that $ k \equiv n - 1 \mod 3 \implies 2n + 1 - k \equiv n - 1 \mod 3 $. Hence $$ \begin{eqnarray} c_n \equiv \begin{cases}0 \text{ if $ 3 \mid n - 1 $} \\1 \text{ otherwise}\end{cases} \mod 2 & \iff & \sum_{k \in [\![0, n]\!], k \equiv n - 1\mod 3} {2n + 1 \choose k} \equiv 1 \mod 2 \\ & \iff & \sum_{k \in [\![0, 2n + 1]\!], k \equiv n - 1\mod 3} {2n + 1 \choose k} \equiv 2 \mod 4 \end{eqnarray} $$ Now denote $$ r_{a, m} = \sum_{k \in [\![0, m]\!], k \equiv a\mod 3} {m \choose k} $$ $ r $ behaves like a Pascal triangle rolled around. That is, $ r_{a + 1, m + 1} = r_{a, m} + r_{a + 1, m} $ and $ r_{a + 3, m} = r_{a, m} $. We can thus easily calculate $ r_{0, m}, r_{1, m}, r_{2, m} $ modulo $ 4 $ $$ \begin{eqnarray} m\quad & r_0 & r_1 & r_2 \\ 0\quad & 1 & 0 & 0 \\ 1\quad & 1 & 1 & 0 \\ 2\quad & 1 & 2 & 1 \\ 3\quad & 2 & 3 & 3 \\ 4\quad & 1 & 1 & 2 \end{eqnarray} $$ We see that it enters a cycle of period $ 6 $ starting from $ m = 2 $ and that $ r_{n - 1, 2n + 1} $ is always $ 2 $, as wanted.

Note: The pattern highly suggests that proving $ c_n \equiv c_{n - 1} + c_{n - 2} \mod 2 $ is an option, but I couldn't find a way to make it work.