Williamson's problem: can $f\in U_{2p}$ have $2p+2$ consecutive zero Taylor coefficients? (Problem 2.74)

Among the real entire functions of finite order having only real zeros, define for each integer $p\ge0$ the class $V_{2p}$ of functions of the form $f(z)=g(z)\,e^{-a z^{2p+2}}$, where $a\ge0$ and $g$ is a constant multiple of a real entire function of genus at most $2p+1$ with only real zeros; set $U_0=V_0$ and $U_{2p}=V_{2p}\setminus V_{2p-2}$ for $p\ge1$. Consider $f(z)=1+a_1z+a_2z^2+\cdots\in U_{2p}$. For $p=0$ it is classical that a non-polynomial $f\in U_0$ cannot have two consecutive Taylor coefficients equal to zero. Question (Williamson): can a non-polynomial $f\in U_{2p}$ with $p\ge1$ have $2p+2$ consecutive Taylor coefficients equal to zero? Equivalently, what is the maximal length of a run of zero Taylor coefficients possible for a non-polynomial member of $U_{2p}$? The first open case is $p=1$: can a non-polynomial $f\in U_2$ have $4$ consecutive zero Taylor coefficients?

Acceptance. FULLY RESOLVES (for a given $p\ge1$): EITHER an explicit non-polynomial $f\in U_{2p}$ exhibiting $2p+2$ (or more) consecutive zero Taylor coefficients, with a certificate that $f\in U_{2p}$ (genus / real-zeros / exponential-factor data), OR a proof that no non-polynomial $f\in U_{2p}$ can have $2p+2$ consecutive zero coefficients (e.g. a Turan/Wronskian-type inequality among consecutive coefficients generalizing the $p=0$ argument). ADVANCES: settle the first case $p=1$ (can $U_2$ admit $4$ consecutive zeros?); determine the exact maximal run length for some $p$; or a verified real-algebraic / SDP feasibility result for genus-bounded subfamilies.