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A new combinatorial interpretation of partial sums of $m$-step Fibonacci numbers

Erik Bates, Blan Morrison, Mason Rogers, Arianna Serafini, Anav Sood

TL;DR

The paper provides a new combinatorial interpretation for the partial sums $\sum_{k=0}^n F^{(m)}_k$ of the $m$-step Fibonacci numbers by introducing a keyword-driven equivalence $\sim_a$ on binary words and showing that the number of equivalence classes in $\{0,1\}^n$ depends only on the length of the keyword and equals the partial sum. The authors develop a framework of simple maps that flip occurrences of $a$ and $\neg a$, establish a concatenation lemma, and derive a recurrence that matches the $m$-step Fibonacci partial sums, thus proving the main theorem. They also analyze the sizes and internal structure of equivalence classes, provide a graph-theoretic perspective via $\mathcal{G}_n^{(a)}$, prove bipartiteness, and study isomorphisms among graphs under keyword transformations, while highlighting open problems about the full classification of keyword-induced structures. The results connect to Zeckendorf representations for the Fibonacci case and open avenues for understanding the growth and diversity of class sizes across different keywords and $m$, with potential links to pattern substitution and numeration systems.

Abstract

The sequence of partial sums of Fibonacci numbers, beginning with $2$, $4$, $7$, $12$, $20$, $33,\dots$, has several combinatorial interpretations (OEIS A000071). For instance, the $n$-th term in this sequence is the number of length-$n$ binary words that avoid $110$. This paper proves a related but new interpretation: given a length-$3$ binary word -- called the keyword -- we say two length-$n$ binary words are equivalent if one can be obtained from the other by some sequence of substitutions: each substitution replaces an instance of the keyword with its negation, or vice versa. We prove that the number of induced equivalence classes is again the $n$-th term in the aforementioned sequence. When the keyword has length $m+1$ (instead of $3$), the same result holds with $m$-step Fibonacci numbers. What makes this result surprising -- and distinct from the previous interpretation -- is that it does not depend on the keyword, despite the fact that the sizes of the equivalence classes do. On this final point, we prove several results on the structure of equivalence classes, and also pose a variety of open problems.

A new combinatorial interpretation of partial sums of $m$-step Fibonacci numbers

TL;DR

The paper provides a new combinatorial interpretation for the partial sums of the -step Fibonacci numbers by introducing a keyword-driven equivalence on binary words and showing that the number of equivalence classes in depends only on the length of the keyword and equals the partial sum. The authors develop a framework of simple maps that flip occurrences of and , establish a concatenation lemma, and derive a recurrence that matches the -step Fibonacci partial sums, thus proving the main theorem. They also analyze the sizes and internal structure of equivalence classes, provide a graph-theoretic perspective via , prove bipartiteness, and study isomorphisms among graphs under keyword transformations, while highlighting open problems about the full classification of keyword-induced structures. The results connect to Zeckendorf representations for the Fibonacci case and open avenues for understanding the growth and diversity of class sizes across different keywords and , with potential links to pattern substitution and numeration systems.

Abstract

The sequence of partial sums of Fibonacci numbers, beginning with , , , , , , has several combinatorial interpretations (OEIS A000071). For instance, the -th term in this sequence is the number of length- binary words that avoid . This paper proves a related but new interpretation: given a length- binary word -- called the keyword -- we say two length- binary words are equivalent if one can be obtained from the other by some sequence of substitutions: each substitution replaces an instance of the keyword with its negation, or vice versa. We prove that the number of induced equivalence classes is again the -th term in the aforementioned sequence. When the keyword has length (instead of ), the same result holds with -step Fibonacci numbers. What makes this result surprising -- and distinct from the previous interpretation -- is that it does not depend on the keyword, despite the fact that the sizes of the equivalence classes do. On this final point, we prove several results on the structure of equivalence classes, and also pose a variety of open problems.

Paper Structure

This paper contains 13 sections, 9 theorems, 65 equations, 1 figure, 2 tables.

Table of Contents

  1. Introduction
  2. Main result
  3. Discussion of main result
  4. Related literature
  5. Proof of main result
  6. Notation and definitions
  7. Preliminary observation
  8. Counting equivalence classes
  9. Sizes and structures of equivalence classes
  10. Equivalence classes of size $1$
  11. Graph structure within equivalence classes
  12. Open problems
  13. Acknowledgments

Key Result

Theorem 1.1

For any $n,m\ge 1$ and any keyword $a\in\{0,1\}^{m+1}$, the number of equivalence classes on $\{0,1\}^n$ induced by the equivalence relation $\sim_a$ is equal to $F_0^{(m)}+F_1^{(m)}+\cdots+F_{n}^{(m)}$.

Figures (1)

  • Figure 1: The unique size-8 component of $\mathcal{G}_6^{(101)}$ from Table \ref{['differentsizes']} (right).

Theorems & Definitions (26)

  • Theorem 1.1
  • Remark 1.2: Other relevant sequences for Theorem \ref{['wordversion']}
  • Remark 1.3: Technical point concerning the Fibonacci numeration system
  • Definition 2.1: Substitutions involving the keyword
  • Definition 2.2: Equivalence relation induced by the keyword
  • Lemma 2.3
  • Definition 2.4: Action by lists of simple maps
  • Lemma 2.5
  • proof
  • proof : Proof of Lemma \ref{['tack_on']}
  • ...and 16 more