It is certainly mentioned matter-of-factly as completely well known in "Coequalizers in categories of algebras" of Linton (link.springer.com/chapter/10.1007/BFb0083082). One may argue that this is already in "Functorial semantics of algebraic theories" of Lawvere (pnas.org/doi/abs/10.1073/pnas.50.5.869), though of course there it is mentioned (as obvious) for equational algebraic theories only.

Can you solve the original recursion in the direction of $\beta$ and $\gamma$ also? Perhaps knowing all of them would help somehow. Or is $\alpha$ somehow special?

If one substitutes this in the original recurrence and denotes $g(b,c)=f(0,b,c)$, we get, I believe, $g(b,c)=-(4b+6c-2)(4b+6c-3)g(b-1,c)-8(b+1)g(b+1,c-1)-12(c+1)g(b-2,c+1)$ - perhaps Mathematica can do miracles for this recurrence once again?

@ZachTeitler what do you mean? Clicking on either of the links brings me to the correct PDF file.

Dear Neil, I have been telling a lot of people about your argument, and I am informed that one of those conversations was fruitful, and there is an arXiv preprint that proves a much more intricate thing this way: arxiv.org/abs/2409.05605 !

On the other hand, if one is concerned with the whole matrix algebra $M_n(\mathbb{R}$, the answer to the question is rather trivially "yes". Concretely, one can take the basis that consists of all off-diagonal matrix units $E_{ij}$ [which all square to $0$], the identity matrix $I_n$, and the matrices $A_i=\mathrm{diag}(1,1,\ldots,-1,\ldots,-1)$, that is, $i$ copies of $1$ followed by $n-i$ copies of $-1$, for $i=1,\ldots,n-1$ [which all square to $I_n$], since these $n$ diagonal matrices clearly form a basis in the space of all diagonal matrices.

Well, if you realize this should be an MSE question, it is especially inappropriate to post it on MO.

This is not research level.

I'd say "reference-request" would be a more appropriate tag than "analytic-number-theory", but perhaps I am missing something.

@valle a useful keyword would be en.wikipedia.org/wiki/Young_symmetrizer

@valle what is your background? it is not easy to give this kind of advice without any context available

@valle this is one of many motivations for representation theory of symmetric groups $S_n$. The subject is vast, but almost every treatment of that theory will have the answer shine through.

I am a bit surprised that you take $A/[A,A]$ as coefficients - at least if you want to imitate the divergence statement, since the conceptually meaningful divergence of a derivation takes values in the commutator quotient of the universal enveloping algebra, not in the commutator quotient of the algebra itself.

That vague statement does not really promise a pairing, it promises some analogue of a pairing. I do not think there is anything reasonable to expect. Already in the classical case of symmetric/exterior algebras, what would such a pairing be, in your opinion? Symmetric powers do not really pair with exterior powers.

Perhaps you can clarify what you mean by "formula". Expanding a determinant using $n!$ terms is also a formula. Already for $n=3$ the corresponding polynomial is irreducible, as I just checked, so no cute factorization to be expected.