Counting unitaries of T-depth one

We show that the number of T-depth one unitaries on $n$ qubits is $\sum_{m=1}^{n}\tfrac{1}{m!}\prod_{k=0}^{m-1}(4^n/2^k - 2^k) \times \# \mathcal{C}_n,$ where $\#\mathcal{C}_n$ is the size of the $n$-qubit Clifford group, that is the number of unitaries of T-depth zero. The number of T-depth one unitaries on $n$ qubits grows as $2^{Ω(n^2)} \cdot \# \mathcal{C}_n$.

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