I am curious about the validity of my claim concerning the expressions
$$\eqalign{E_1&:=(2k-1)t+1, \cr E_2&:=(2k^2-2k)t+(2k-1),\cr}$$
where $k=2,3,4,...$
My claim is that for almost all $k$ (or for infinitely many $k$) there exists a positive integer $t$ such that for this $t$ the two expressions simultaneously evaluate to a palindrome when written in decimal.
For instance when $k=3$ and $t=13$ then $E_1=66$ and $E_2=161$. If $k=4$ and $t=46$ then $E_1=323$ and $E_2=1111$. The same is also true when $k=7,8$.
Is my claim TRUE? Also any suggestion for the proof of my claim if it is true will be highly appreciated.
Let's estimate a very rough probability, just using the fact that there are about $10^{n/2}$ palindromes of length $n$, so the palindrome density around a large number $N$ scales as $1/\sqrt{N}$. If you fix $k$, then you're looking for a palindrome around $2kt$ and another around $2k^2 t$ for some $t$. For a given $t$, this may happen with probability about $(2kt)^{-1/2}(2k^2 t)^{-1/2}=(1/2) k^{-3/2} t^{-1}$. Because the sum of this probability over all $t$ diverges (logarithmically), it is likely that some value of $t$ will produce a double-palindrome just by chance. However, as $k$ becomes larger, you will expect to need to look at greater and greater values for $t$: since $\sum_{t=1}^{T}t^{-1} \sim \log T$, you'll need to look at values growing as $\exp(2k^{3/2})$.