Is there a nonzero solution to this infinite system of congruences? Is there a triple of nonzero even integers $(a,b,c)$ that satisfies the following infinite system of congruences?
$$
a+b+c\equiv 0 \pmod{4} \\
a+3b+3c\equiv 0 \pmod{8} \\
3a+5b+9c\equiv 0 \pmod{16} \\
9a+15b+19c\equiv 0 \pmod{32} \\
\vdots \\
s_na + t_nb + s_{n+1}c \equiv 0 \pmod{2^{n+1}} \\
\vdots
$$
where $(s_n)$ and $(t_n)$ are weighted tribonacci sequences defined by
$$
s_1=s_2=1, \\
s_3=3, \\
s_n = s_{n-1} +2s_{n-2} + 4s_{n-3} \text{ for } n>3,
$$
and
$$
t_1=1, \\
t_2=3, \\
t_3=5, \\
t_n = t_{n-1} +2t_{n-2} + 4t_{n-3} \text{ for } n>3.
$$
I think there are no nonzero solutions, but I haven't been able to prove this. Computationally, I found there are no nonzero solutions for integers $a$, $b$, and $c$ up to $1000$.
Note the $s_n$ and $t_n$ are always odd, and that the ratios $\frac{s_n}{s_{n-1}}$ and $\frac{t_n}{t_{n-1}}$ approach $2.4675...$.
 A: Let $u_n = a s_n + b t_n + c s_{n+1}$. The stronger claim is true: for large enough values of $n$,
the number $u_n$ will be exactly divisible
by a fixed power of $2$ that doesn't depend on $n$.
Let $u_n = a s_n + b t_n + c s_{n+1}$ then (by induction)
$$u_{n} = u_{n-1} + 2 u_{n-2} + 4 u_{n-3}.$$
The polynomial $x^3 - x^2 - 2 x - 4$  is irreducible and has three roots $\alpha_1$, $\alpha_2$, and $\alpha_3$ in $\overline{\mathbf{Q}}$.
By the general theory of recurrence relations,
$$u_n = A_1 \alpha^n_1 + A_2 \alpha^n_2 + A_3 \alpha^n_3$$
for constants $A_1$, $A_2$, $A_3$. Since $u_n \in \mathbf{Q}$, we may additionally deduce that $A_i$ lie in $\mathbf{Q}(\alpha_1,\alpha_2,\alpha_3)$.
That is because we can solve for $A_i$ using the equation
$$\left( \begin{matrix} \alpha_1 & \alpha_2 & \alpha_3 \\ \alpha^2_1 & \alpha^2_2 & \alpha^2_3 \\ \alpha^3_1 & \alpha^3_2 & \alpha^3_3 \end{matrix} \right)
\left( \begin{matrix} A_1 \\ A_2 \\ A_3 \end{matrix} \right) = \left( \begin{matrix} u_1 \\ u_2 \\ u_3 \end{matrix} \right)$$
and the matrix on the left is invertible (Vandermonde). In fact we deduce the stronger claim that
any Galois automorphism sending $\alpha_i$ to $\alpha_j$ sends $A_i$ to $A_j$.
(Simply consider the action of the Galois group on both sides of this equation,
noting that the $A_i$ are determined uniquely from this equation.) In particular, if one of the $A_i = 0$,
then all of the $A_i = 0$.
But now fix an embedding of $\overline{\mathbf{Q}}$ into $\overline{\mathbf{Q}}_2$. From the Newton Polygon,
we see that there is one root (call it $\alpha_1$) of valuation $0$, and the other two roots have valuation $1$. Hence
$$\|A_1 \alpha^n_1 \|_2 = \|A_1\|_2,  \quad \|A_2 \alpha^n_2 \|_2 = \|A_2\|_2 \cdot 2^{-n},  \quad  \|A_3 \alpha^n_3 \|_2 = \|A_3\|_2  \cdot 2^{-n}.$$
In particular,  if $A_1 \ne 0$, then (by the ultrametric inequality) $\| u_n \|_2 = \|A_1\|$ for $n$ large enough. Hence we deduce that either the $2$-adic valuation
of $u_n$ is eventually  constant (as claimed) or that $A_1 = 0$ and so $A_i = 0$ for all $i$, which implies that $u_n = 0$ for all $n$. But  if $u_1 = u_2 = u_3 = 0$,
then
$$\left( \begin{matrix} 1 &1 & 1 \\ 1 &3 & 3 \\ 3 & 5 & 9 \end{matrix} \right)
\left( \begin{matrix} a \\ b \\ c \end{matrix} \right) = \left( \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right)$$
The matrix on the left is invertible which implies that $a=b=c=0$.
A: $u_n=s_na + t_nb + s_{n+1}c$ satisfies the same recurrence relation as $s_n$ and $t_n$: $u_n = u_{n-1} +2u_{n-2} + 4u_{n-3}$. The question is whether $2^{n+1}\mid u_n$.
Since $v_n=u_n/2^{n+1}$ satisfies
$v_n = \displaystyle\frac{v_{n-1} +v_{n-2} + v_{n-3}}{2}$
the answer is affirmative only if there are $v_0, v_1, v_2$ (not all 0) such that $v_n$ is always integral.
EDIT.
As sharply noticed by the OP, the attempt below was wrong, since a matrix I claimed to be invertible (mod $2$) is in fact singular. A similar, more computational argument does work (mod $5$).

It's clear that (mod $2$) such a sequence $v_n$ must follow either one of the 3-periodic patterns $000$ and $110$ (up to shifts). In the $000$ case keep dividing entire sequence $(v_n)$ by $2$ until one term is odd, and then shift the sequence to start with that term, so it's back to the $110$ case.  Therefore it must be that 
$$\require{cancel}\cancel{\det\left (\begin{smallmatrix}v_1 & v_2 & v_3\\ v_2 & v_3 & v_4\\ v_3 & v_4 & v_5 \end{smallmatrix}\right )\equiv \det\left (\begin{smallmatrix}1 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 1 \end{smallmatrix}\right )\!\!\!\!\pmod{2} \ne0}$$
CORRECTED ARGUMENT.
Notice that
$$D=\det\left (\begin{matrix}v_1 & v_2 & v_3\\ v_2 & v_3 & v_4\\ v_3 & v_4 & v_5 \end{matrix}\right )=\det\left (\begin{matrix}v_1 & v_2 & v_3\\ v_2 & v_3 & \frac{v_1+v_2+v_3}{2}\\ v_3 & \frac{v_1+v_2+v_3}{2} & \frac{v_1+3v_2+3v_3}{4} \end{matrix}\right )\\=
\frac{-4v_3^3+4v_2 v_3^2+2v_1 v_3^2+v_2^2 v_3+5v_1 v_2 v_3-v_1^2 v_3-3v_2^3-2v_1 v_2^2-2v_1^2 v_2-v_1^3}{4}$$
is $\equiv 0 \pmod{5}$ if and only if $v_1\equiv v_2\equiv v_3\equiv 0 \pmod{5}$. This is proved by the following snippet of code:
awk -vp=5 'BEGIN {
    for(a=0; a<p; a++)
        for(b=0; b<p; b++)
            for(c=0; c<p; c++) {
                d=4*c^3-4*b*c^2-2*a*c^2-b^2*c-5*a*b*c+a^2*c+3*b^3+2*a*b^2+2*a^2*b+a^3;
                if(d%p==0) print a, b, c;
            }
}'

Now divide the entire sequence $(v_n)$ by a power of $5$ so that at least one term is is not $\equiv 0$, and shift it to start with that term, thus $v_1\not\equiv 0$ and therefore $D\ne0$.
Next this implies that there is an integral linear combination $(z_n)$ of $(v_n)$ and its shifts $(v_{n+1})$, $(v_{n+2})$ such that $z_1=z_2=0$, $z_3\ne 0$, and still $z_n=(z_{n-1} +z_{n-2} + z_{n-3})/2$ holds.
Finally write $z_3=2^m d$, with $d$ odd, and start the recursion from $0, 0, 2^m d$ to easily notice that it runs into a half-integer in $m+1$ steps.
