Edit $(2020)$: Update is included at the end of the post.
$4$ consecutive bases?
Are there numbers that are a palindrome in $4$ consecutive number bases?
I'm not counting a one digit palindrome as a palindrome. (Discarding trivial solutions.)
After testing some of my plots of palindromic numbers & number systems, I noticed that I couldn't find any numbers which are a palindrome in more than $3$ consecutive bases. I was curious to find out why is this the case.
I ran a simple code to check numbers up to $10^{7}$ (and all relevant bases), and didn't find any numbers that are a palindrome in $4$ or more consecutive bases. For reference, here are the smallest numbers which are palindromic in $1,2,3$ consecutive bases:
$$3 = 11_2$$ $$10 = 101_3=22_4$$ $$178 = 454_6 =343_7 = 262_8$$
For example, $3=1\cdot 2^1 + 1\cdot 2^0=11_2$ is a binary palindrome.
I strongly suspect that a solution for four consecutive bases does not exits, but I do not know how to prove this observation. For comparison, there are infinitely many numbers that are palindromic in $3$ consecutive number bases.
Almost $4$ consecutive bases
Lets examine numbers which are "almost palindromic in four consecutive bases". That is, the numbers palindromic in bases $b$ and $b+3$, and in either $b+1$ or $b+2$ number base.
Checking separately some $d$ digit palindromes up to some number base $b$, I found:
($b\le6000$) For $2$ digits, there are no examples.
($b\le900$) For $3$ digits, there are $1484$ examples.
($b\le800$) For $4$ digits, there is only one example at $b=10$.
($b\le150$) For $5$ digits, only two examples at $b=16$ and at $b=17$
($b\le100$) For $6$ digits, there are no examples.
And etc.
Notice that other than the three exceptions, all other palindromes (examples) of this type have exactly $3$ digits in their palindromic bases.
If we can prove this observation, then our solution should have exactly $3$ digits in its palindromic bases. This in fact would solve the problem, because $3$ digit numbers cannot be palindromic in more than $3$ consecutive number bases.
That is,
Two smallest three digit numbers that are a palindrome in three consecutive are: $$178 = 454_6 =343_7 = 262_8$$ $$300 = 606_7 = 454_8 = 363_9$$
All other three digit palindromes which are palindromic in three consecutive number bases are given by (Also mentioned in the OEIS sequence) the following expression using $n\ge7$ and is odd:
$$\frac{1}{2}(n^3 + 6n^2 + 14n + 11)$$
Each term given by this is palindromic in bases $n+1, n+2, n+3$ and is $3$ digits long.
$373$ is the first number given by this equation, and is palindromic in bases $8,9,10$.
This three digit pattern will never extend to a fourth consecutive base as TMM said in the comments; which Ross Millikan posted later in his partial answer.
It remains to prove the observation that "almost 4 consecutive base palindromes" can't have more than $3$ digits if they are sufficiently large.
This was also cross-posted on Math Overflow, with patterns for $5$ and $7$ digits also presented there; but nothing new came up so far.
Update
Thanks to Max Alekseyev's method, we know that if a palindrome in $4$ consecutive number bases exists, then either:
It has an equal number of digits in all corresponding number bases and also has $15$ or more digits in those number bases (see this answer and corresponding OEIS sequence A323742).
It does not have an equal number of digits in corresponding bases and is larger than $10^{12}$ (i.e. is a term of the OEIS sequence A327810.)
Can we rule out at least one of these two cases?