The pebbling number of Fibonacci cubes

TL;DR

This paper determines the pebbling number of Fibonacci cubes, showing it equals that of hypercubes for small n and conjecturing it holds for all n, using combinatorial and computational methods.

Contribution

It establishes the pebbling number of Fibonacci cubes as equal to that of hypercubes for small n and proposes a conjecture for all n, supported by computational verification.

Findings

Pebbling number of Fibonacci cubes equals 2^n for n ≤ 6.

Standard potential argument provides a lower bound.

Exhaustive MILP verification confirms the upper bound.

Abstract

The $n$-th Fibonacci cube $\Gamma_n$ is the subgraph of the hypercube $Q_n$ induced by binary strings with no two consecutive ones. We determine $\pi(\Gamma_n) = 2^n$ for $n \le 6$, so the pebbling number of $\Gamma_n$ equals that of the ambient hypercube $Q_n$ despite $\Gamma_n$ having far fewer vertices. The lower bound is a standard potential argument. For the upper bound, the Weight Function Lemma yields $2^n+1$ -- one too many -- so we close the gap by exhaustive MILP verification. We conjecture $\pi(\Gamma_n) = 2^n$ for all $n$.