Jan 26, 2026 · That is, there seems to be fairly strong symbolic evidence that for $n=4$, if $ABBA-BAAB = A-B$ and $A$ is nilpotent, then $B^4 = \lambda I$ for some $\lambda$.

Nov 21, 2013 · Truly lost here, I know abba could look anything like 1221 or even 9999. However how do I prove 11 divides all of the possiblities?

Because abab is the same as aabb. I was how to solve these problems with the blank slot method, i.e. _ _ _ _. If I do this manually, it's clear to me the answer is 6, aabb abab abba baba bbaa baab Which is …

Oct 4, 2016 · The algorithm is normally created by taking AB, then inverting each 2-state 'digit' and sticking it on the end (ABBA). You then take this entire sequence and repeat the process (ABBABAAB).

Apr 19, 2022 · Although both belong to a much broad combination of N=2 and n=4 (AAAA, ABBA, BBBB...), where order matters and repetition is allowed, both can be rearranged in different ways: …

There must be something missing since taking $B$ to be the zero matrix will work for any $A$.

May 13, 2016 · This is nice work and an interesting enrichment. (+1). I realized when I had solved most of it that the OP seems to know how to compute the generating function but is looking for a way to …

I get the trick. Use the fact that matrices "commute under determinants". +1

Feb 28, 2018 · Hint: in digits the number is $abba$ with $2 (a+b)$ divisible by $3$.

I've found and proven the following extensions to palindromes of the usual divisibility rules for 3 and 9: A palindrome is divisible by 27 if and only if its digit sum is. A palindrome is divisible by 81 if and only if …