Porous numbers are numbers $k$ which are not multiples of 10 such that every m with sum of digits = $k$ and $k$ a divisor of both $m$ and rev($m$) has a zero in its digits. rev($m$) is the digit reversal of $m$ (e.g. rev(123) = 321).
Below 1000 there are only 11, 37, 74, 101 and 121 which fulfill these requirements (see OEIS sequence A337832).
Since $11$ and $11^2$ are porous, I wonder if $11^3$ and eventually all $11^n$ might be porous as well.
from Hot Weekly Questions - Mathematics Stack Exchange
Rüdi Jehn
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