 Solution

For the defined shuffle, count 52 = 8. The printed assertion about seven shuffles is inconsistent with the actual permutation. Number positions from zero. On an even length $n>2$, a position $i<n-1$ moves to $2i\mathrm{mod}(n-1)$, while the last position stays fixed. Indeed, positions in the first half move to $2i$, and positions in the second half move to $2i-(n-1)$. Consequently the number sought is the multiplicative order of $2$ modulo $n-1$:Here the identity's period means the least positive number of applications, so the empty and two-element lists have period one. Since $n-1$ is odd, $2$ is invertible modulo $n-1$, ensuring termination. Using Integer for residues avoids overflow in doubling; the mathematical correctness assumes the returned period fits Int. The algorithm uses one modular update per shuffle in the period, rather than constructing each shuffled deck.

For $n=52$, $2^8\equiv 1(\mathrm{mod}51)$, whereas $2$, $2^2$ and $2^4$ are not congruent to one, ruling out every proper divisor of eight. After seven shuffles, the original position $1$ is at position $26$, not back at $1$.

This single position already disproves a seven-step return; the modular calculation proves that the whole permutation returns after eight.