Curious anti-commutative ring

I noticed this now, and I want to remark that the underlying abelian group can in fact be described very precisely. To do that, note that:

(1) the defining relations easily imply that the abelian group of elements of degree $dge 2$ in this algebra is certainly generated by $x_0^{d-1}x_k$, $kge 0$, and

(2) as discussed in the comments, there are the relations $(i+j+2)x_0x_{i+j}=0$ that follow from the defining relations and anticommutativity; effectively, these give just one relation for each $n$, namely $(n+2)x_0x_n=0$.

Now let me (inspired by typical Gröbner bases calculations) consider the following two chains of equalities: $$ x_ix_jx_k=(i+1)x_0x_{i+j}x_k=(i+1)(i+j+1)x_0^2x_{i+j+k} $$ and $$ x_ix_jx_k=(j+1)x_ix_0x_{j+k}=-(j+1)x_0x_ix_{j+k}=-(j+1)(i+1)x_0^2x_{i+j+k}. $$ They imply that $$ (i+1)(i+2j+2)x_0^2x_{i+j+k}=0 $$ for each choice of $i$ and $j$ with $i+jle n$. In particular, if $nge 1$, we may take $i=n-1$, $j=1$, obtaining $$ n(n+3)x_0^2x_n=(n-1+1)(n-1+2+2)x_0^2x_n=0. $$ But $(n+2)x_0x_n=0$ implies $(n+1)(n+2)x_0^2x_n=0$, so by subtraction we see that $2x_0^2x_n=0$. Moreover, no further relations can be obtained in a similar way, because once we have the 2-torsion property, we have $$ (i+1)(i+2j+2)x_0^2x_n=(i+1)ix_0^2x_n=0, $$ since $(i+1)i$ is always even.

In fact, using a version of Gröbner bases (or rewriting systems) for ideals in free anticommutative algebras, one can see that the system of all the defining relations thus obtained, namely $$ begin{cases} x_ix_j=(i+1)x_0x_{i+j},\ (n+2)x_0x_n=0,\ 2x_0^2x_n=0 end{cases} $$ is complete, and so your ring as an abelian group :

is freely generated by $1$ in degree $0$,

is freely generated by $x_0,x_1,ldots$ in degree $1$,

is the product of cyclic groups of orders $2,3,ldots$ generated by $x_0^2, x_0x_1, x_0x_2, ldots$ respectively in degree $2$,

is a product of countably many cyclic groups of order $2$ generated by $x_0^d, x_0^{d-1}x_1, x_0^{d-1}x_2, ldots$ in each degree $dge 3$ .