Has anyone seen the ring $\Lambda[x\_0, x\_1, x\_2, \ldots]/(x\_i x\_j - (i+1) x\_0 x\_{i+j})$ in some natural context?


Here $\Lambda[x\_0, x\_1, x\_2, \ldots]$ is the (graded-)commutative algebra (either over the integers or the integers localized at 2) freely generated by elements $x\_0,x\_1,x\_2,\ldots$ of odd homological degrees, so that $x\_i x\_j = - x\_j x\_i$. In particular, we only get $2 x\_i^2 = 0$, not $x\_i^2 = 0$. I probably shouldn't write $\Lambda$ here: in characteristic $2$ or integrally, $\Lambda$ usually adds the relation $x\_i^2 = 0$.


ADDED NOTE: In the meantime, I have found a better way of solving the problem in which this arose, so it is merely a curiosity now. I am happy to delete it if people wish.


Forgive me for posting this to 'Commutative algebra', but as a topologist, commutative means $x y = (-1)^{(\deg x)(\deg y)} yx$, and my $x\_i$ are in odd degrees.


SECOND NOTE: This algebra has now shown up in another context, so Vladimir's answer below has been quite useful. Thanks to Vladimir and MO.



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 $d\ge 2$ in this algebra is certainly generated by $x\_0^{d-1}x\_k$, $k\ge 0$, 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 $i+j\le n$. In particular, if $n\ge 1$, 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 $d\ge 3$ .


