The finitistic dimension of an algebra is defined as the supreme of all the projective dimensions of the modules that have a finite projective dimension.

For finite dimension algebras. $ A $ with the radical zero cube it is known that for the finitist dimension $ fd (A) $ one has $ fd (A) leq dim (A) ^ 2 + 1 $ (See https://www.sciencedirect.com/science/article/pii/S0021869383712056).

Question: we have $ fd (A) leq dim (A) ^ {r-1} + 1 $ for algebras $ A $ with $ J ^ r = 0 $ when $ J $ is Jacobson's radical of $ A $ Y $ r geq 3 $?

(Of course, one can probably only expect a counterexample in an answer here)