Abstract
For the multi-peg Tower of Hanoi problem with $k \geqslant 4$ pegs, so far the best solution is obtained by the Stewart's algorithm based on the the following recurrence relation: $\mathrm{S}_k(n)=\min_{1 \leqslant t \leqslant n} \left\{2 \cdot \mathrm{S}_k(n-t) + \mathrm{S}_{k-1}(t)\right\}$, $\mathrm{S}_3(n) = 2^n – 1$. In this paper, we generalize this recurrence relation to $\mathrm{G}_k(n) = \min_{1\leqslant t\leqslant n}\left\{ p_k\cdot \mathrm{G}_k(n-t) + q_k\cdot \mathrm{G}_{k-1}(t) \right\}$, $\mathrm{G}_3(n) = p_3\cdot \mathrm{G}_3(n-1) + q_3$, for two sequences of arbitrary positive integers $\left(p_i\right)_{i \geqslant 3}$ and $\left(q_i\right)_{i \geqslant 3}$ and we show that the sequence of differences $\left(\mathrm{G}_k(n)- \mathrm{G}_k(n-1)\right)_{n \geqslant 1}$ consists of numbers of the form $\left(\prod_{i=3}^{k}q_i\right) \cdot \left(\prod_{i=3}^{k}{p_i}^{\alpha_i}\right)$, with $\alpha_i\geqslant 0$ for all $i$, arranged in nondecreasing order. We also apply this result to analyze recurrence relations for the Tower of Hanoi problems on several graphs.