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How to prove that symmetric traceless "transverse" tensor rank $s$ in 4 dimensions has $ 2s + 1$ independent components?

Let's have tensor $$ F^{\mu_{1}\dots \mu_{s}}, \quad {F^{\quad \mu}_{\mu}}^{\mu_{1}\dots \mu_{s - 2}}, \quad \partial_{\mu}F^{\mu ... \mu_{s - 1}} = 0 . $$ Symmetric tensor rank $s$ has $C^{s}_{s + 3}$ independent components. The trace fixed two indices, so it decrease the number of components to $(s + 1)^{2}$. But I don't understand, how the transverse $\partial_{\mu}F^{\mu ... \mu_{s - 1}} = 0$ decrease the number of components to $2s + 1$.

Addition. The correct answer is $C^{s}_{s + 3} - C_{s + 1}^{s - 2} - (C_{s + 2}^{s - 1} - C_{s}^{s - 3})$, where the second summand corresponds to the number of independence conditions, which refer to transverse symmetrical tensor.

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