The last comment appearing under Qiaochu's answer, by Qfwfq, asks Qiaochu if he could back up his claim that that's where the term "torsion" came from. It seems QY didn't respond. It is a nice story, and I have no reason to doubt it, but I think maybe it's not yet substantiated.

Are your new models published in some form?

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Generatingfunctionology is a lot of fun, more so to play than as a spectator sport. Also, you taught me a new word: pedice/pedices. I found that in an Italian-English dictionary. The top hit from Google that I got, though, directed me to Wiktionary, and while I'm always glad to learn new things, what I learned there is essentially unrepeatable here. (The singular is pēdīcō, which the curious can look up.)

Putting $j = n-k$, it looks like $\left(\sum_{k \geq 0} \frac1{k!} t^k X^k\right)\left(\sum_{j \geq 0} \frac{D_j X^j}{j!}\right)$, which is $e^{tX} \frac{e^{-X}}{1 - X}$.

Well, any generator and relations description of an operad will give you a generators and relations description of an algebraic theory, but we still can't get the equations you wrote down if the OP accepts any monoid as a particular type of $\Delta$-algebra. I can make sense of your remark if the $\Delta_n$ are supposed to be (opposite to) finitely generated free algebras of a Lawvere theory; for example, $\Delta_1$ has just one element, so of course we would get idempotency, but of course that would radically change the question -- even if OP is secretly also interested in that question.

Thanks, Tobias; interesting observations. It's certainly not true that all $\Delta$-algebras will manifest idempotency, mediality, etc. of one of these $\ast$ operations, as the example of monoids being $\Delta$-algebras makes clear enough. (By the way, you mentioned my (3), but it's (2) and (3) together that give the relations.)

@JamesHanson Golly, I'm aware of that, and it's why I didn't post my comment as an answer, but just as a possible point of interest.

The nLab article on Goedel incompleteness takes a hyperdoctrinal approach.

The Schanuel topos is a cotopos?

By the way, Sridhar: this question was instigated by a conversation still going on at the Category Theory Zulip, under the thread "distributive laws and monadic functors". A lot of interesting observations are being made. (I see that you're a member, but you might not have tuned in recently.)

Yes, of course. I wish I'd had the idea before; it's simple enough.

Yes, I guess this does work. Thanks!

I think this works. You get isomorphic monads for this case iff $\eta F'$ is an isomorphism, i.e. iff free groups are in fact torsionfree abelian groups, which they are not.