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.

  • $\begingroup$ Why would you want to know this?:) $\endgroup$ – Valentin May 26 '15 at 1:50
  • $\begingroup$ @Valentine, I just want to know that my claim is true. I am interested with palindromes that's why I am able to create the problem. $\endgroup$ – Jr Antalan May 26 '15 at 23:41
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    $\begingroup$ I decided to test this theorem by running a computer simulation of it. So far I've gotten the following combinations: $(k,t) = (2,1), (3,13), (4,14), (7,17), (8,118), (14,31), (15,2197715), (20, 1312826), (21,12924364)$. The program is supposed to output the minimum $t$ required for each $k$ as well... lol $\endgroup$ – 2012ssohn May 29 '15 at 3:17
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    $\begingroup$ Number theory conjectures such as this are notorious for being misleading at relatively small values. In particular, there was one conjecture I recall which involved sorting primes into residue classes modulo a particular constant. It seemed that one class always contained the most elements, that was until you reached a billion or so. $\endgroup$ – eloiprime May 31 '15 at 16:23
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    $\begingroup$ @eloiPrime Sounds like you're referring to Chebyshev's bias. The closest fit I know of for "a billion or so" is the mod 3 race which crosses over at $23.3$ billion. But that's small compared to the first crossover for Littlewood's phenomenon, whose first example is not known, but could well be as large as $e^{727}$. $\endgroup$ – Erick Wong Jun 2 '15 at 0:58

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 $\exp(2k^{3/2})$.

  • $\begingroup$ Thanks Sir @mjqxxxx. Will have a follow up question in a few days. $\endgroup$ – Jr Antalan Jun 10 '15 at 11:19

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