  The palindromic length $\text{PL}(v)$ of a finite word $v$ is the minimal
number of palindromes whose concatenation is equal to $v$. In 2013, Frid,
Puzynina, and Zamboni conjectured that: If $w$ is an infinite word and $k$ is
an integer such that $\text{PL}(u)\leq k$ for every factor $u$ of $w$ then $w$
is ultimately periodic.
  Suppose that $w$ is an infinite word and $k$ is an integer such
$\text{PL}(u)\leq k$ for every factor $u$ of $w$. Let $\Omega(w,k)$ be the set
of all factors $u$ of $w$ that have more than $\sqrt[k]{k^{-1}\vert u\vert}$
palindromic prefixes. We show that $\Omega(w,k)$ is an infinite set and we show
that for each positive integer $j$ there are palindromes $a,b$ and a word $u\in
\Omega(w,k)$ such that $(ab)^j$ is a factor of $u$ and $b$ is nonempty. Note
that $(ab)^j$ is a periodic word and $(ab)^ia$ is a palindrome for each $i\leq
j$. These results justify the following question: What is the palindromic
length of a concatenation of a suffix of $b$ and a periodic word $(ab)^j$ with
"many" periodic palindromes?
  It is known that $\lvert\text{PL}(uv)-\text{PL}(u)\rvert\leq \text{PL}(v)$,
where $u$ and $v$ are nonempty words. The main result of our article shows that
if $a,b$ are palindromes, $b$ is nonempty, $u$ is a nonempty suffix of $b$,
$\vert ab\vert$ is the minimal period of $aba$, and $j$ is a positive integer
with $j\geq3\text{PL}(u)$ then $\text{PL}(u(ab)^j)-\text{PL}(u)\geq 0$.
