Recall that the Riemann tensor for any connection is alternating in its last two lower indices. This is just the very general fact that the curvature of a connection $\nabla$ in a vector bundle $(E,\pi,M)$ is an $\text{End}(E)$-valued $2$-form; the $2$-form nature is responsible for this alternating property. However, the Bianchi identity in $E=TM$ is only true for connections whose torsions are ‘nice enough’. The general equation is that if $\nabla$ is a connection on $TM$ and $d_{\nabla}$ denotes the exterior covariant derivative, then for any vectors $x,y,z\in T_pM$
\begin{align}
R(x,y)z+R(y,z)x+R(z,x)y&=d_{\nabla}(\text{Tor}_{\nabla})(x,y,z),
\end{align}
where $\text{Tor}_{\nabla}:=d_{\nabla}(\text{id}_{TM})$ is the torsion of the connection (which is a $TM$-valued $2$-form on $M$). So, the first Bianchi identity holds if and only if $d_{\nabla}(\text{Tor}_{\nabla})=0$, i.e if and only if the torsion is $d_{\nabla}$-closed. This is of course the case if the torsion itself vanishes identically.
None of these things beyond my first sentence is really relevant to your question. I only mention them to highlight that the algebraic/first Bianchi identity is really a statement about the vanishing of the (exterior covariant) derivative of the torsion. In particular, this property is more special than the curvature simply being alternating in its last two lower slots (i.e it’s more special than simply being (an endomorphism-valued) $2$-form).
Now, let’s consider an arbitrary $(0,3)$ tensor $T$ which is alternating in its last two slots. Then,
\begin{align}
T_{[abc]}&:=\frac{1}{6}\left(T_{\mu\nu\rho}+T_{\nu\rho\mu}+T_{\rho\mu\nu}-T_{\nu\mu\rho}-T_{\mu\rho\nu} - T_{\rho\nu\mu}\right)\\
&=\frac{1}{6}\left(2T_{\mu\nu\rho}+2T_{\nu\rho\mu}+2T_{\rho\mu\nu}\right)\\
&=\frac{1}{3}\left(T_{\mu\nu\rho}+T_{\nu\rho\mu}+T_{\rho\mu\nu}\right).
\end{align}
On the other hand, $T_{(abc)}=0$ because there are three pairs of terms which cancel (e.g the $T_{\mu\nu\rho}+T_{\mu\rho\nu}$ adds to $0$). Now, if we define the ‘first Bianchi-ization’ of a tensor $T$ that is alternating in the last two slots to be
\begin{align}
B[T]_{abc}:= T_{\mu\nu\rho}+T_{\nu\rho\mu}+T_{\rho\mu\nu},
\end{align}
then we have that $B[T]_{abc}=3 T_{[abc]}$ and $T_{(abc)}=0$. Note that the second equality is a triviality due to $T$ being alternating in its last two slots, however $T_{[abc]}$ need not always vanish. Hence, the vanishing of $B[T]_{abc}$ is an extra non-trivial condition on $T$.
So, the first Bianchi identity $R_{[\mu\nu\rho]}^{a}=3 B[R]^a_{\,\mu\nu\rho}=0$ is a non-trivial property of the curvature which goes beyond simply $R^a_{\,\mu\nu\rho}=-R^a_{\,\mu\rho\nu}$, whereas the assertion $R^a_{\,(\mu\nu\rho)}=0$ is a simple corollary of the alternating nature. So, long story short, (A) is the Bianchi identity (for tensors alternating in their last two slots), while (S) is always true (for tensors alternating in their last two slots).
Edit: Remarks about indexing conventions.
If $\nabla$ is a connection in a vector bundle $(E,\pi,M)$, then the formula, where $X,Y$ are vector fields on $M$ and $\psi$ a smooth section of $E$,
\begin{align}
R(X,Y)\psi:=\nabla_X\nabla_Y\psi-\nabla_Y\nabla_X\psi-\nabla_{[X,Y]}\psi\tag{$*$}
\end{align}
defines $R$ as an $\text{End}(E)$-valued $2$-form on $M$. There are some books which define $R$ with an overall minus sign. So, at the level of endomorphism-valued $2$-forms, there are only two possible conventions.
In order to extract components, one should choose:
- a local frame $\{\xi_1,\dots, \xi_n\}$ of vector fields on $M$ (often people use coordinate vector fields $\frac{\partial}{\partial x^{\mu}}$)
- a local frame $\{e_1,\dots, e_k\}$ for the vector bundle $E$ (in the special case where $E=TM$, one has $k=n$, and by default one takes $\{e_1,\dots, e_n\}=\{\xi_1,\dots,\xi_n\}$). Also, let $\{\epsilon^1,\dots,\epsilon^k\}$ denote the dual coframe (i.e a local frame for $E^*$), defined such that $\epsilon^{i}(e_j)=\delta^i_{\,j}$.
With these local frames at hand one extracts the curvature components
\begin{align}
\epsilon^a\bigg(R(\xi_{\mu},\xi_{\nu})\cdot e_b\bigg).\tag{$**$}
\end{align}
There are now a bunch of different ways of typographically dressing up $R$ with indices in order to denote this particular expression… and this is of course super annoying when going from source to source (but atleast all the ones below use $(*)$):
- $R^a_{\,\,b\mu\nu}$ - Hawking and Ellis, MTW, Sachs and Wu use this convention (and this is the convention I’ve adopted in writing my answer).
- $R^{\,\,\,\,\,a}_{\mu\nu\,\,\,b}$ - this looks horrible to write (and type) which is why I’d never use this convention, but this is the convention adopted by Freedman, Van Proeyen.
- $R_{\mu\nu b}^{\,\,\,\,\,\,\, a}$ - this is the indexing convention used in John Lee’s Introduction to Riemannian Manifolds
- $R^a_{\,\,\mu\nu b}$ - this would have been my natural inclination to index things just based on how $(**)$ is written, and reading left-to-right, but I haven’t seen (or atleast don’t recall) any book using this convention. But convention (1) is the closest to this, hence why I prefer (1) over (2) and (3).
So, as @naturallyInconsistent mentions in the comment, with Freedman, Van Proeyen’s conventions, you should replace all instances of ‘last two lower indices’ in my answer above with ‘first two lower indices’. But, anyway, don’t let this distract you from the conceptual distinction of requiring $T_{(abc)}=0$ vs $B[T]_{abc}=0$; the former is trivially true, while the latter is not.
