# Generators and relations for the Steenrod algebra in terms of $Sq^{2^i}$'s?

One can show inductively that the Steenrod algebra is generated as an algebra by the $Sq^{2^i}$'s, but can one write down explicitly the Adem relations in terms of this presentation?

Of course, the standard presentation of the Steenrod algebra $\mathcal{A}^\ast$ describes it as the free associative $\mathbb{F}_2$-algebra on the generating set $\{Sq^n \mid n \in \mathbb{N}\}$, modulo the Adem relations plus the identity $Sq^0 = 1$. But it's a standard fact (following from the Adem relations) that this generating set is redundant: $\mathcal{A}^\ast$ is already generated by the restricted generating set $\{Sq^{2^i} \mid i \in \mathbb{N}\}$. However, it's far from obvious how to express a complete set of relations in terms of the restricted generating set; for example, the Adem relations for the product $Sq^{2^i} Sq^{2^j}$ introduce terms of the form $Sq^{2^i+2^j-2^r}Sq^{2^r}$ which must be inductively written in terms of the restricted generating set.

Of course, there are other presentations of the Steenrod algebra, for example in terms of the Milnor basis. But I'm still curious: what are the relations among the $Sq^{2^i}$'s explicitly?

Of course, one could ask a similar question at odd primes.