The pebbling number of Fibonacci cubes
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arXiv
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| Format: | Preprint |
| Published: |
2026
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| _version_ | 1866913150740201472 |
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| author | Niu, Tong |
| author_facet | Niu, Tong |
| contents | The $n$-th Fibonacci cube $Γ_n$ is the subgraph of the hypercube $Q_n$ induced by binary strings with no two consecutive ones. We determine $π(Γ_n) = 2^n$ for $n \le 6$, so the pebbling number of $Γ_n$ equals that of the ambient hypercube $Q_n$ despite $Γ_n$ 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 $π(Γ_n) = 2^n$ for all $n$. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2605_04328 |
| institution | arXiv |
| publishDate | 2026 |
| record_format | arxiv |
| spellingShingle | The pebbling number of Fibonacci cubes Niu, Tong Combinatorics 05C99, 05C57, 68R10 The $n$-th Fibonacci cube $Γ_n$ is the subgraph of the hypercube $Q_n$ induced by binary strings with no two consecutive ones. We determine $π(Γ_n) = 2^n$ for $n \le 6$, so the pebbling number of $Γ_n$ equals that of the ambient hypercube $Q_n$ despite $Γ_n$ 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 $π(Γ_n) = 2^n$ for all $n$. |
| title | The pebbling number of Fibonacci cubes |
| topic | Combinatorics 05C99, 05C57, 68R10 |
| url | https://arxiv.org/abs/2605.04328 |