  A positive semi-definite (PSD) tensor which is not a sum-of-squares (SOS)
tensor is called a PSD non-SOS (PNS) tensor. Is there a fourth order four
dimensional PNS Hankel tensor? Until now, this question is still an open
problem. Its answer has both theoretical and practical meanings. We assume that
the generating vector $v$ of the Hankel tensor $A$ is symmetric. Under this
assumption, we may fix the fifth element $v_4$ of $v$ at $1$. We show that
there are two surfaces $M_0$ and $N_0$ with the elements $v_2, v_6, v_1, v_3,
v_5$ of $v$ as variables, such that $M_0 \ge N_0$, $A$ is SOS if and only if
$v_0 \ge M_0$, and $A$ is PSD if and only if $v_0 \ge N_0$, where $v_0$ is the
first element of $v$. If $M_0 = N_0$ for a point $P = (v_2, v_6, v_1, v_3,
v_5)^\top$, then there are no fourth order four dimensional PNS Hankel tensors
with symmetric generating vectors for such $v_2, v_6, v_1, v_3, v_5$. Then, we
call such a point $P$ PNS-free. We show that a $45$-degree planar closed convex
cone, a segment, a ray and an additional point are PNS-free. Numerical tests
check various grid points, and find that they are also PNS-free.
