Every positive integer is a sum of three palindromes

For integer $g\ge 5$, we prove that any positive integer can be written as a sum of three palindromes in base $g$

Palindromic Length of Words with Many Periodic Palindromes

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 $\text{PL}(u(ab)^j)-\text{PL}(u)\geq 0$

Enumerating Palindromes and Primitives in Rank Two Free Groups

Let $F=$ be a rank two free group. A word $W(a,b)$ in $F$ is {\sl primitive} if it, along with another group element, generates the group. It is a {\sl palindrome} (with respect to $a$ and $b$) if it reads the same forwards and backwards. It is known that in a rank two free group any primitive element is conjugate either to a palindrome or to the product of two palindromes, but known iteration schemes for all primitive words give only a representative for the conjugacy class. Here we derive a new iteration scheme that gives either the unique palindrome in the conjugacy class or expresses the word as a unique product of two unique palindromes. We denote these words by $E_{p/q}$ where $p/q$ is rational number expressed in lowest terms. We prove that $E_{p/q}$ is a palindrome if $pq$ is even and the unique product of two unique palindromes if $pq$ is odd. We prove that the pairs $(E_{p/q},E_{r/s})$ generate the group when $|ps-rq|=1$. This improves the previously known result that held only for $pq$ and $rs$ both even. The derivation of the enumeration scheme also gives a new proof of the known results about primitives.Comment: Final revisions, to appear J Algebr

Palindromic continued fractions

In the present work, we investigate real numbers whose sequence of partial quotients enjoys some combinatorial properties involving the notion of palindrome. We provide three new transendence criteria, that apply to a broad class of continued fraction expansions, including expansions with unbounded partial quotients. Their proofs heavily depend on the Schmidt Subspace Theorem