Step recursion is a form of bounded recursion in which each recursive call moves from an input y to a prescribed predecessor $ρ_g(y)$. Its depth $D_g(y)$ is the exact number of such moves needed to reach zero. A natural question is whether knowing this depth for every input determines the expressive power of the resulting function algebra. We prove that it does not. We first construct two simple generators with exactly the same depth map but different step-recursion algebras. One gives ordinary binary halving, $b(x)=2x+1$; the other is $p(0)=1$, $p(x)=x+2^{λ(x)}$ for $x>0$, where $λ$ is binary length. Although $D_p=D_b$ pointwise, the predecessor $ρ_p$ cannot be defined from any fixed-stride binary-halving descent at basis zero. Thus two recursion schemes may take exactly the same number of steps on every input and still have different expressive power. The phenomenon is much larger than this example. Whenever infinitely many depth levels allow more than one predecessor arrangement, a single exact depth profile supports $2^{\aleph_0}$ distinct step-recursion algebras over every countable basis containing zero and the projections. In the computable setting the corresponding effective family has exactly $\aleph_0$ distinct algebras. Hence exact recursion depth is an informative resource measure, but it is not a complete invariant: the geometry of predecessor choices inside each depth level carries additional algebraic information.

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