Arbitrarily long palindromes in two consecutive number bases

Solution 1:

I have not worked out a proof yet , but it seems that $$n:=\frac{b^k-1}{b+1}$$ with even $k\ge 2$ is palindrome in bases $b$ and $b+1$ for sufficient large $b$. For example , $b=10^{99}$ and $k=108$ does the job.

Solution 2:

Thanks to @Peter's answer for conjecturing a pattern that should give such a sequence.

Here, I managed to prove his proposed identity.

The linked answer proposed that the following gives $(b,b+1)$ 2-palindromes for even $k$ and large $b$:

$$ \frac{b^k-1}{b+1} $$

For large $k$, we have arbitrarily large amount of digits in those two number bases.

It is not hard to see that the given expression is palindromic in base $b$.

What is needed to prove, is it being palindromic in $b+1$ for sufficiently large $b$, for infinitely many $k$.

More specifically, what we needed to prove was the following:

For all $n,b\in\mathbb N$, if $b\ge \sum_{k=1}^n \binom{n}{k}^2$, then there exits $A_n(i)$ such that following identity is true:

$$ \frac{b^{2n}-1}{b+1}=\sum_{i=1}^{2n-1}A_n(i)(b+1)^{2n-1-i}\\ A_n(i)=A_n(2n-i),i=1,\dots,2n-1 $$

That is if $k=2n$, the expression is a $d=2n-1$ digit palindrome in base $b+1$ for all $b\ge \sum_{k=1}^n \binom{n}{k}^2$.

Initially, my conjectured pattern for $A_n(i)$ that holds so far was:

$$ A_n(i)=\begin{cases}b-a_n(i), && i\text{ is odd}\\a_n(i), && i\text{ is even}\end{cases} $$

Where $a_n(i)$ is given by: ($n$th row, $i$th element)

$$\newcommand\s[]{\space} 1\\ 3\s\s\s\s\s\s 5\s\s\s\s\s\s 3\\ 5\s\s\s\s\s\s 14\s\s\s\s\s 19\s\s\s\s\s 14\s\s\s\s\s 5\\ 7\s\s\s\s\s\s 27\s\s\s\s\s 55\s\s\s\s\s 69\s\s\s\s\s 55\s\s\s\s\s 27\s\s\s\s\s 7\\ 9\s\s\s\s\s\s 44\s\s\s\s\s 119\s\s\s\s 209\s\s\s\s 251\s\s\s\s 209\s\s\s\s 119\s\s\s\s 44\s\s\s\s\s 9\\ 11\s\s\s\s\s 65\s\s\s\s\s 219\s\s\s\s 494\s\s\s\s 791\s\s\s\s 923\s\s\s\s 791\s\s\s\s 494\s\s\s\s 219\s\s\s\s 65\s\s\s\s\s 11\\ 13\s\s\s\s\s 90\s\s\s\s\s 363\s\s\s\s 1000\s\s\s 2001\s\s\s 3002\s\s\s 3431\s\s\s 3002\s\s\s 2001\s\s\s 1000\s\s\s 363\s\s\s\s 90\s\s\s\s\s 13\\ 15\s\s\s\s\s 119\s\s\s\s 559\s\s\s\s 1819\s\s\s 4367\s\s\s 8007\s\s\s 11439\s\s 12869\s\s 11439\s\s 8007\s\s\s 4367\s\s\s 1819\s\s\s 559\s\s\s 119\s\s\s 15\\ \dots $$

Some patterns are clear, like the middle column being $\sum_{k=1}^n \binom{n}{k}^2$, for example.

After closer examination, we can notice that the diagonal elements are given by:

$$ D(r,q)=\binom{2(r+q-1)}{q}-1 $$

And when solving for $n,i$ we obtain:

$$ a_n(i)=\binom{2n}{2n-i}-1 $$

And this is indeed the correct pattern. Now we simply sum the initial sum and show the identity is true.

We can use Mathematica:

FullSimplify[Sum[(b ((-1)^(i + 1) + 1)/2 + (-1)^i (Binomial[2 n, -i + 2 n] - 1)) (b + 1)^(2 n - 1 - i), {i, 1, 2 n - 1}]  - (b^(2 n) - 1)/(b + 1), Element[n, Integers]]

To obtain RHS-LHS=0. We are done!