A Combinatorial Problem Solved by a Meta-Fibonacci Recurrence Relation

TL;DR

This paper introduces a simple combinatorial problem whose solution is described by a specific meta-Fibonacci recurrence relation involving a prime number, expanding understanding of such relations in combinatorics.

Contribution

It presents a new, simpler combinatorial problem linked to a meta-Fibonacci recurrence, providing clearer insight compared to more general previous problems.

Findings

Solution characterized by a prime-based meta-Fibonacci recurrence

Simpler combinatorial problem with clear relation to the recurrence

Advances understanding of recurrence relations in combinatorics

Abstract

We present a natural, combinatorial problem whose solution is given by the meta-Fibonacci recurrence relation $a(n) = \sum_{i=1}^p a(n-i+1 - a(n-i))$, where $p$ is prime. This combinatorial problem is less general than those given in [3] (B. Jackson, F. Ruskey, 2006) and [4] (F. Ruskey, C. Deugau, 2009), but it has the advantage of having a simpler statement.

Equations

$$\lambda (n)=\lambda (n-\lambda (n-1))+\lambda (n-1-\lambda (n-2)).$$

$$c(M)=\underset{\vec{v}\in \mathbf{row}(M)}{\max }\Vert \vec{v}\Vert .$$

$${\cal M}_{n}^{*}=\{M\in{\mathbb{F}}_{p}^{m\times n}:1\leq m\leq p^{n}\mbox{ and no column of $M$ equals $\vec{0}$}\},$$

$${\lambda }_p(n)=\underset{M\in {\mathcal{M}}_n^{\ast }}{\min }c(M).$$

$${\lambda }_p(n)=\sum _{i=1}^p{\lambda }_p(n-i+1-{\lambda }_p(n-i)).$$

$$p\lambda (n-\Vert \vec{v}\Vert )>\Vert \vec{v}\Vert ,$$

$$\sum _{a=1}^{p-1}|S_a|\le |\mathrm{supp}({\vec{w}}_L)|=\Vert {\vec{w}}_L\Vert <(p-1)\lambda (n-k).$$

$$n=\sum _{j=\ell }^k{b}_j{\sigma }_j,$$

$$\lambda (n)\ge \sum _{j=\ell }^k{b}_j{p}^j.$$

$$n=\sum _{j=\ell }^m{b}_j{\sigma }_j,$$

$$\lambda (n)\ge \sum _{j=\ell }^m{b}_j{p}^j.$$

$$\Vert \vec{v}\Vert <\sum _{j=\ell }^k{b}_j{p}^j.$$

$$n-\Vert \vec{v}\Vert >n-\sum _{j=\ell }^k{b}_j{p}^j$$

$$=\sum _{j=\ell }^k{b}_j({\sigma }_j-p^j)$$

$$=\sum _{j=\ell }^k{b}_j{\sigma }_{j-1},$$

$$n-\Vert \vec{v}\Vert \ge \sum _{\ell -1\le j\le k-1}{b}_{j+1}{\sigma }_j+1.$$

$$\sum _{\ell -1\le j\le k-1}{b}_{j+1}{\sigma }_j+1$$

$$=\sum _{\ell \le j\le k-1}{b}_{j+1}{\sigma }_j+(p{\sigma }_{\ell -1}+1)$$

$$=\sum _{\ell \le j\le k-1}{b}_{j+1}{\sigma }_j+{\sigma }_{\ell }$$

$$=\sum _{\ell +1\le j\le k-1}{b}_{j+1}{\sigma }_j+b_{\ell +1}{\sigma }_{\ell }+{\sigma }_{\ell }$$

$$=\sum _{\ell +1\le j\le k-1}{b}_{j+1}{\sigma }_j+(b_{\ell +1}+1){\sigma }_{\ell }.$$

$$p\cdot \lambda \left(\sum _{\ell +1\le j\le k-1}{b}_{j+1}{\sigma }_j+(b_{\ell +1}+1){\sigma }_{\ell }\right)$$

$$=\sum _{\ell +1\le j\le k-1}{b}_{j+1}{p}^{j+1}+(b_{\ell +1}+1)p^{\ell +1}$$

$$=\sum _{\ell \le j\le k-1}{b}_{j+1}{p}^{j+1}+p\cdot p^{\ell }$$

$$=\sum _{\ell +1\le j\le k}{b}_j{p}^j+p\cdot p^{\ell }$$

$$=\sum _{\ell \le j\le k}{b}_j{p}^j,$$

$$p\lambda (n-\Vert \vec{v}\Vert )$$

$$\ge \sum _{\ell \le j\le k}{b}_j{p}^j>\Vert \vec{v}\Vert .$$

$$\sum _{\ell -1\le j\le k-1}{b}_{j+1}{\sigma }_j+1=\sum _{\ell -1\le j\le k-1}{b}_{j+1}{\sigma }_j+1\cdot {\sigma }_0.$$

$$p\cdot \lambda \left(\sum _{\ell -1\le j\le k-1}{b}_{j+1}{\sigma }_j+1\right)$$

$$=\sum _{\ell -1\le j\le k-1}{b}_{j+1}{p}^{j+1}+p$$

$$=\sum _{\ell \le j\le k}{b}_j{p}^j+p.$$

$$p\lambda (n-\Vert \vec{v}\Vert )$$

$$\ge \sum _{\ell \le j\le k}{b}_j{p}^j+p>\Vert \vec{v}\Vert .$$

$$p\cdot \lambda \left(\sum _{0\le j\le k-1}{b}_{j+1}{\sigma }_j+1\right)$$

$$\ge p\left(\sum _{1\le j\le k-1}{b}_{j+1}{p}^j+b_1+1\right)$$

$$=\sum _{1\le j\le k-1}{b}_{j+1}{p}^{j+1}+b_{1}p+p$$

$$=\sum _{1\le j\le k}{b}_j{p}^j+p.$$

$$p\lambda (n-\Vert \vec{v}\Vert )$$

Full text

1 Introduction

Let $M$ be a matrix with entries in ${\mathbb{Z}}_{2}$, such that every column contains at least one 1. We want to pick a subset of the rows such that when they are added together modulo 2, their sum $\vec{s}$ has as many 1’s as possible. If $M$ has $n$ columns, what is the largest number of 1’s we can guarantee $\vec{s}$ to have? For example, if $n=5$, we can always find a set of rows whose sum $\vec{s}$ contains at least four 1’s. Let $\lambda(n)$ denote the largest number of 1’s $\vec{s}$ can be guaranteed to have for any $M$ with $n$ nonzero columns. We will show that $\lambda(n)$ satisfies the recurrence relation

[TABLE]

More generally, for $p$ prime, let $\vec{v}=(v_{1},\ldots,v_{n})$ satisfy $v_{i}\in{\mathbb{F}}_{p}$ for $1\leq i\leq n$. Let $\operatorname{supp}(\vec{v})=\{i\in[n]:v_{i}\neq 0\}$ and let $\|\vec{v}\|=|\operatorname{supp}(\vec{v})|$, i.e., $\|\vec{v}\|$ is the number of nonzero terms in $\vec{v}$. Let $M$ be an $m\times n$ matrix whose entries are in ${\mathbb{F}}_{p}$. Let ${\bf row}(M)$ be the rowspace of $M$, i.e., the set of all linear combinations of the row vectors of $M$ over the field ${\mathbb{F}}_{p}$. Let $c(M)$ denote the capacity of $M$, which we define as follows,

[TABLE]

For each integer $n\geq 1$, let $\lambda_{p}(n)$ be the minimum possible capacity of an $\mathbb{F}_{p}$-matrix consisting of $n$ nonzero columns (i.e., no column equals $\vec{0}$). Restated, let

[TABLE]

then

[TABLE]

We will see that $\lambda_{p}$ satisfies the recurrence relation

[TABLE]

This type of recurrence relation is called a meta-Fibonacci relation.

Meta-Fibonacci sequences have been studied by various authors, dating at least as far back as 1985, when Hofstadter [2] apparently coined the term “meta-Fibonacci.” These are integer sequences defined by “nested, Fibonacci-like” recurrence relations, such as relation (1), which was studied by Conolly [1], and (2). Generalizations of (2) were shown in [3] and [4] to be solutions to certain combinatorial problems involving $k$-ary infinite trees, and compositions of integers. The “matrix capacity” problem described above is a different combinatorial problem whose solution is also given by relation (2). This combinatorial problem is “natural” in the sense that it arose while the first named author was working on a problem in spatial graph theory. It was only later that we learned (through the OEIS A046699) that it can be characterized as a meta-Fibonacci sequence.

2 Main Result

We begin with a lemma which allows us to produce a lower bound on $\lambda_{p}(n)$. For the remainder of this paper, instead of writing $\lambda_{p}$, we will simply write $\lambda$. For a matrix $M$, let ${\bf row}^{*}(M)={\bf row}(M)-\{\vec{0}\}.$

Lemma 1**.**

Let $M$ be an $\mathbb{F}_{p}$-matrix with $n$ nonzero columns, i.e., $M\in{\cal M}_{n}^{*}$. Let $\vec{v}\in{\bf row}^{*}(M).$ If

[TABLE]

then there is a vector $\vec{z}\in{\bf row}^{*}(M)$ such that $\|\vec{z}\|>\|\vec{v}\|.$

Proof.

Let $M$ be an $\mathbb{F}_{p}$-matrix with $n$ nonzero columns. Let $\vec{v}\in{\bf row}^{*}(M)$, and let $k=\|\vec{v}\|.$ Let $\vec{v}=(v_{1},\ldots,v_{n}).$ W.l.o.g., suppose $v_{i}\neq 0$ for $1\leq i\leq k$ and $v_{i}=0$ for $k+1\leq i\leq n$. Let $\vec{w}\in{\bf row}^{*}(M)$ be such that $w_{i}\neq 0$ for at least $\lambda(n-k)$ coordinates $i$, where $k+1\leq i\leq n$. In other words, if we let $\vec{w}_{L}=(w_{1},\ldots,w_{k})$ and $\vec{w}_{R}=(w_{k+1},\ldots,w_{n})$, then $\|\vec{w}_{R}\|\geq\lambda(n-k)$. Since $\|\vec{w}\|=\|\vec{w}_{L}\|+\|\vec{w}_{R}\|$, if $\|\vec{w}_{L}\|\geq(p-1)\lambda(n-k)$, then $\|\vec{w}\|\geq p\lambda(n-k)>\|\vec{v}\|$, and we are done. So we may assume that $\|\vec{w}_{L}\|<(p-1)\lambda(n-k).$

Our goal will be to prove that there exists a nonzero constant $c$ such that $\|c\vec{w}_{L}+\vec{v}_{L}\|>k-\lambda(n-k)$, where $\vec{v}_{L}=(v_{1},\ldots,v_{k}).$ Once we establish that such a constant exists, then we will be done, because we will have $\|c\vec{w}+\vec{v}\|=\|c\vec{w}_{L}+\vec{v}_{L}\|+\|\vec{w}_{R}\|>(k-\lambda(n-k))+\lambda(n-k)=k.$

For $1\leq a\leq p-1$, let $S_{a}=\{i\in[k]:aw_{i}+v_{i}=0\}$. Since $v_{i}\neq 0$ for $1\leq i\leq k$, then $S_{a}\subseteq\operatorname{supp}(\vec{w}_{L}).$ Thus, if $aw_{i}+v_{i}=0=bw_{i}+v_{i}$, then $w_{i}\neq 0$, which allows us to conclude that $a=b$. Therefore, if $a\neq b$, then $S_{a}\cap S_{b}=\emptyset$. Since $\bigcup_{a=1}^{p-1}S_{a}\subseteq\operatorname{supp}(\vec{w}_{L})$ and the $S_{a}$ are pairwise disjoint, we have

[TABLE]

Therefore, the average value of $|S_{a}|$ is strictly less than $\lambda(n-k)$, and if we let $c\in[p-1]$ be such that $|S_{c}|$ is minimum, then $|S_{c}|<\lambda(n-k)$. Thus, $\|c\vec{w}_{L}+\vec{v}_{L}\|=k-|S_{c}|>k-\lambda(n-k)$, and as noted above, we are done. Specifically, $\|c\vec{w}+\vec{v}\|>\|\vec{v}\|.$ ∎

It is easy to check that the following corollary holds.

Corollary 1**.**

If $1\leq n\leq p$, then $\lambda(n)=n.$

For an integer $k\geq 0$, let $\sigma_{k}=\sum_{j=0}^{k}p^{j}.$

Proposition 1**.**

Suppose

[TABLE]

where $b_{k}\geq 1$, and $0\leq b_{j}\leq p-1$ for $j\neq\ell$, and $1\leq b_{\ell}\leq p$. Then

[TABLE]

Proof of Proposition 1.

We proceed by induction on $k$. When $k=0$, then $n=b_{0}\sigma_{0}=b_{0}.$ Since $1\leq b_{0}\leq p$, then $\lambda(n)=b_{0}$ by Corollary 1, thus, $\lambda(n)=b_{0}p^{0}$ and the result holds. Now suppose $k\geq 1$. Our inductive hypothesis will be if

[TABLE]

where $b_{m}\geq 1$, and $0\leq b_{j}\leq p-1$ for $j\neq\ell$, and $1\leq b_{\ell}\leq p$, and $m<k$, then

[TABLE]

Let $M$ be an $\mathbb{F}_{p}$-matrix with $n$ nonzero columns. Suppose $\vec{v}\in{\bf row}^{*}(M)$ with

[TABLE]

Then

[TABLE]

where we define $\sigma_{-1}=0$ to handle the case $j=0$, since $\sigma_{0}-p^{0}=0.$ Thus,

[TABLE]

We want to determine a lower bound on $p\lambda\left(\sum_{j=\ell-1}^{k-1}b_{j+1}\sigma_{j}+1\right)$ that allows us to conclude that $p\lambda(n-\|\vec{v}\|)>\|\vec{v}\|$ so that we may use Lemma 1. We consider the case where $b_{\ell}=p$ and the case where $1\leq b_{\ell}\leq p-1$ separately.

Suppose $b_{\ell}=p$. Then

[TABLE]

Notice that our sum satisfies all of the criteria for the inductive hypothesis. Specifically, the coefficient of its lowest sigma-term $\sigma_{\ell}$ is $b_{\ell+1}+1$, which satisfies $1\leq b_{\ell+1}+1\leq p$; the coefficient of $\sigma_{j}$ is $b_{j+1}$ and $0\leq b_{j+1}\leq p-1$ for $j\neq\ell$; the coefficient of the largest sigma-term $\sigma_{k-1}$ is $b_{k}$, which satisfies $b_{k}\geq 1$; and finally, the index of its largest sigma term is $k-1$ which is strictly less than $k$. Therefore, by the inductive hypothesis,

[TABLE]

where the last equality holds because $b_{\ell}=p$. Since $\lambda$ is a nondecreasing function, our previous work implies

[TABLE]

Thus, by Lemma 1, there is a vector $\vec{z}\in{\bf row}^{*}(M)$ such that $\|\vec{z}\|>\|\vec{v}\|.$

Now suppose $1\leq b_{\ell}\leq p-1$. Recall that our sum is

[TABLE]

In this case, the smallest sigma-term is $\sigma_{0}$, and its coefficient is $b_{1}+1$, where $b_{1}=0$ if $\ell\geq 2$. We note that our sum satisfies all of the criteria for the inductive hypothesis. Since each $b_{j}$ satisfies $0\leq b_{j}\leq p-1$, then $1\leq b_{1}+1\leq p$; when $j\geq 1$, the coefficient of each $\sigma_{j}$ is $b_{j+1}$ and $0\leq b_{j+1}\leq p-1$; the coefficient of the largest sigma-term $\sigma_{k-1}$ is $b_{k}$, which satisfies $b_{k}\geq 1$; and finally, the index of its largest sigma term is $k-1$ which is strictly less than $k$.

When $\ell\geq 2$, the coefficient of $\sigma_{0}$ is 1, and we apply the inductive hypothesis to obtain

[TABLE]

Thus,

[TABLE]

When $\ell\in\{0,1\}$, our sum is $\sum_{j=0}^{k-1}b_{j+1}\sigma_{j}+1$, and we apply the inductive hypothesis to obtain

[TABLE]

Thus,

[TABLE]

Thus, by Lemma 1, there is a vector $\vec{z}\in{\bf row}^{*}(M)$ such that $\|\vec{z}\|>\|\vec{v}\|.$ Therefore $\lambda(n)\geq\sum_{j=\ell}^{k}b_{j}p^{j}.$ ∎

Now we show that every $n\geq 1$ can be written in the form described in Proposition 1.

Claim 1**.**

Let $n\in{\mathbb{Z}}^{+}$. Suppose $n<\sigma_{k+1}$. Let $n_{k+1}=n$, and for $0\leq j\leq k$, assuming $n_{j+1}$ is defined, let $b_{j}$ be the largest integer such that $b_{j}\sigma_{j}\leq n_{j+1}$, and let $n_{j}=n_{j+1}-b_{j}\sigma_{j}$. Then for $0\leq j\leq k$, we have $0\leq n_{j+1}\leq p\sigma_{j}$ and $0\leq b_{j}\leq p$. Moreover,

[TABLE]

and if $b_{j}=p$, then $b_{i}=0$ for $i<j$.

Proof of Claim 1.

Suppose $n\in{\mathbb{Z}}^{+}$ and $n<\sigma_{k+1}$. Then $n\leq\sigma_{k+1}-1=p\sigma_{k}$. Let $n_{k+1}=n$, and for $0\leq j\leq k$, assuming $n_{j+1}$ is defined, let $b_{j}$ be the largest integer such that $b_{j}\sigma_{j}\leq n_{j+1}$, and let $n_{j}=n_{j+1}-b_{j}\sigma_{j}$. We proceed by induction on $k-j$. Assume $0\leq n_{j+1}\leq p\sigma_{j}$ and let $b_{j}$ and $n_{j}$ be defined as above. Since $0\leq n_{j+1}$, then $b_{j}\geq 0$. Since $n_{j+1}\leq p\sigma_{j}$ and $b_{j}\sigma_{j}\leq n_{j+1}$, then $b_{j}\sigma_{j}\leq p\sigma_{j}$. Thus, since $\sigma_{j}\geq 1$, we have $b_{j}\leq p.$ Since $b_{j}\sigma_{j}\leq n_{j+1}$ and $n_{j}=n_{j+1}-b_{k}\sigma_{k}$, then $n_{j}\geq 0.$ Since $n_{j+1}<(b_{j}+1)\sigma_{j}$, then $n_{j+1}-b_{j}\sigma_{j}<\sigma_{j}$, i.e., $n_{j}\leq\sigma_{j}-1=p\sigma_{j-1}$. Therefore, by induction, $0\leq n_{j+1}\leq p\sigma_{j}$ and $0\leq b_{j}\leq p$ for $0\leq j\leq k$.

Now suppose $b_{j}=p$. Since $b_{j}\sigma_{j}\leq n_{j+1}\leq p\sigma_{j}$, then $n_{j+1}=p\sigma_{j}$ and $n_{j}=n_{j+1}-b_{j}\sigma_{j}=0$. Moreover, $b_{i}=0$ and $n_{i}=0$ for all $i<j$.

To see that $n=\sum_{j=0}^{k}b_{j}\sigma_{j},$ observe that $b_{j}\sigma_{j}=n_{j+1}-n_{j}$ for $0\leq j\leq k$, because of the definition of $n_{j}$. Thus,

[TABLE]

Since $0\leq n_{1}\leq p\sigma_{0}=p$, then, by definition, $b_{0}=n_{1}$ and $n_{0}=n_{1}-b_{0}\sigma_{0}=n_{1}-n_{1}(1)=0.$ Thus, $\sum_{j=0}^{k}b_{j}\sigma_{j}=n$ ∎

With Proposition 1 and Claim 1, we have established a lower bound on $\lambda(n)$ for all $n\geq 1$. We need to prove the corresponding upper bound. We will do so by constructing a matrix with $n$ columns whose capacity equals the lower bound given in Proposition 1. We begin by constructing such a matrix for certain values of $n$, namely, when $n=\sigma_{k}$ for some $k\geq 0$.

For each integer $k\geq 0$, we define a $(k+1)\times\sigma_{k}$ matrix $B_{k}$, recursively, as follows. The matrix $B_{0}$ is the $1\times 1$ matrix whose sole entry is 1. For $k\geq 1$, $B_{k}$ can be defined as a block matrix with a “row” consisting of $p$ copies of $B_{k-1}$ followed by a $k\times 1$ column of 0’s, then one more row of dimensions $1\times\sigma_{k}$ with its first $\sigma_{k-1}$ entries equal to 0 (below the first $B_{k-1}$), then $\sigma_{k-1}$ entries equal to 1 (below the next $B_{k-1}$), …, then $\sigma_{k-1}$ entries equal to $p-1$ (below the last $B_{k-1}$), and one last entry equal to 1, i.e.,

[TABLE]

For $k\geq 1$, let $B^{\prime}_{k}$ be the $k\times\sigma_{k}$ matrix obtained from $B_{k}$ by removing its last row, i.e.,

[TABLE]

Lemma 2**.**

For each $\vec{v}\in{\bf row}^{*}(B_{k})$, $\|\vec{v}\|=p^{k}$.

Proof.

We proceed by induction on $k$. When $k=0$, the result is trivial. Let $k\geq 1$. Assume the result for $j<k$. Let $\vec{v}\in{\bf row}^{*}(B_{k})$. We first consider the case where $\vec{v}\in{\bf row}^{*}(B^{\prime}_{k})$. Then we can write

[TABLE]

To shorten notation, we will write

[TABLE]

where $\vec{v}_{i}=(v^{(i)}_{1},\ldots,v^{(i)}_{\sigma_{k-1}})$ for $0\leq i\leq p-1$. Technically, in equation (3), $\vec{v}_{i}$ simply represents the coordinates $v^{(i)}_{1},\ldots,v^{(i)}_{\sigma_{k-1}}$. We observe that $\vec{v}_{0}=\vec{v}_{1}=\cdots=\vec{v}_{p-1}$ based on how $B^{\prime}_{k}$ and $\vec{v}$ are defined. We also observe that $\vec{v}_{i}\in{\bf row}^{*}(B_{k-1})$. By the inductive hypothesis, $\|\vec{v}_{i}\|=p^{k-1}$, therefore, $\|\vec{v}\|=p^{k}$.

We now show the result holds for $\vec{w}\in{\bf row}^{*}(B_{k})-{\bf row}^{*}(B^{\prime}_{k})$. Let $\vec{u}$ be the last row in $B_{k}$, i.e., $\vec{u}=(0,\ldots,0,1,\ldots,1,\ldots,p-1,\ldots,p-1,1)$. We observe that $\|\vec{u}\|=\sigma_{k}-\sigma_{k-1}=p^{k}$, thus, the result holds when $\vec{w}=\vec{u}.$ To illustrate our argument, we next consider the special case where $\vec{w}=\vec{v}+\vec{u}$ for some $\vec{v}\in{\bf row}^{*}(B^{\prime}_{k})$. Again, we slightly abuse notation and write $\vec{u}=(\vec{0},\vec{1},\ldots,(p-1)\vec{1},1)$, where $\vec{c}$ (or $c\vec{1}$) represents the $\sigma_{k-1}$-dimensional vector $(c,\ldots,c)$. Then we can write $\vec{v}+\vec{u}=(\vec{v}_{0}+\vec{0},\vec{v}_{1}+\vec{1},\ldots,\vec{v}_{p-1}+(p-1)\vec{1},1)$. Since we are working modulo $p$, a coordinate of $\vec{v}_{j}+j\vec{1}$ is congruent to 0 if and only if the corresponding coordinate of $\vec{v}_{j}$ is congruent to $p-j$. Thus, we can count the total number of coordinates that are congruent to 0 in $\vec{v}+\vec{u}$ as follows

[TABLE]

Since $\vec{v}_{0}=\vec{v}_{1}=\cdots=\vec{v}_{p-1}$, equation (4) reduces to

[TABLE]

Thus, $\|\vec{v}+\vec{u}\|=\sigma_{k}-\sigma_{k-1}=p^{k}.$ In general, $\vec{w}\in{\bf row}^{*}(B_{k})-{\bf row}^{*}(B^{\prime}_{k})$ satisfies $\vec{w}=\vec{v}+c\vec{u}$ for some $\vec{v}\in{\bf row}^{*}(B^{\prime}_{k})$ and $c\not\equiv 0\pmod{p}$. In this case, $\vec{w}=(\vec{v}_{0}+c\vec{0},\vec{v}_{1}+c\vec{1},\ldots,\vec{v}_{p-1}+c(p-1)\vec{1},1)$, and equation (4) becomes

[TABLE]

where arithmetic is modulo $p$. Since $\vec{v}_{0}=\vec{v}_{1}=\cdots=\vec{v}_{p-1}$, we obtain

[TABLE]

Since $p$ is prime and $c\not\equiv 0\pmod{p}$, then $\{p,p-c,p-2c,\ldots,p-(p-1)c\}$ is a equivalent to $\{0,1,\ldots,p-1\}$ modulo $p$, thus,

[TABLE]

Therefore, $\|\vec{w}\|=\sigma_{k}-\sigma_{k-1}=p^{k}$, and we can conclude that for each $\vec{v}\in{\bf row}^{*}(B_{k})$, $\|\vec{v}\|=p^{k}$. ∎

Since $B_{k}$ has $\sigma_{k}$ columns, Lemma 2 implies that $\lambda(n)\leq p^{k}$ when $n=\sigma_{k}$ for some nonnegative integer $k$. We would like a similar upper bound on $\lambda(n)$ for all positive integers $n$. Thus, we provide the following proposition.

Proposition 2**.**

If $n=\sum_{j=0}^{k}b_{j}\sigma_{j}$, then

[TABLE]

Proof of Proposition 2.

We will construct a matrix $M$ with $n$ columns such that $c(M)=\sum_{j=0}^{k}b_{j}p^{j}$. The matrix $M$ will essentially be a block matrix with $b_{j}$ copies of $B_{j}$ for $0\leq j\leq k$. However, the number of rows of $B_{j}$ does not equal the number of rows of $B_{\ell}$ when $j\neq\ell$. Thus, for $0\leq j\leq k$, we define the $(k+1)\times\sigma_{j}$ matrix $B^{(k)}_{j}$ where the first $j$ rows of $B^{(k)}_{j}$ match the first $j$ rows of $B_{j}$ and the last $k+1-j$ rows of $B^{(k)}_{j}$ all equal the last row of $B_{j}$. Thus, $B^{(k)}_{0}$ is a $(k+1)\times 1$ column of 1’s, and for $1\leq j\leq k$,

[TABLE]

where the last row is repeated $(k+1)-j$ times. After comparing $B^{(k)}_{j}$ with $B_{j}$, it is easy to see that ${\bf row}^{*}(B^{(k)}_{j})={\bf row}^{*}(B_{j})$.

Let $n$ be a positive integer such that $n=\sum_{j=0}^{k}b_{j}\sigma_{j}.$ Let $M$ be the $(k+1)\times n$ matrix defined as a block matrix with $b_{j}$ copies of $B^{(k)}_{j}$ for $0\leq j\leq k$, where the blocks appear in a single row in nondecreasing order according to their lower index, i.e.,

[TABLE]

Let $\vec{v}\in{\bf row}^{*}(M)$. Then we can (essentially) write

[TABLE]

where $\vec{v}_{i}^{(j)}\in{\bf row}^{*}(B_{j})$ for $0\leq j\leq k$ and $1\leq i\leq b_{j}$. Moreover, for $1\leq i\leq b_{j}$, we have $\vec{v}_{i}^{(j)}=\vec{v}_{b_{j}}^{(j)}$. Thus,

[TABLE]

Because $\vec{v}_{b_{j}}^{(j)}\in{\bf row}^{*}(B_{j})$, Lemma 2 implies $\|\vec{v}_{b_{j}}^{(j)}\|=p^{j}$, therefore,

[TABLE]

Thus, $c(M)=\sum_{j=0}^{k}b_{j}p^{j},$ and $\lambda(n)\leq\sum_{j=0}^{k}b_{j}p^{j}.$ ∎

Thus, we can combine Propositions 1 and 2 with Claim 1 to obtain the following corollary.

Corollary 2**.**

Let $n\in{\mathbb{Z}}^{+}$. Suppose $n<\sigma_{k+1}$. Then

[TABLE]

where $0\leq b_{j}\leq p$ for $0\leq j\leq k$, and if $b_{j}=p$, then $b_{i}=0$ for $i<j$. Moreover,

[TABLE]

Corollary 3**.**

The sequence $\lambda(n)$ satisfies the meta-Fibonacci recurrence relation

[TABLE]

Proof of Corollary 3..

We refer to Corollary 32 in [4], which implies that a sequence which is defined by the meta-Fibonacci recurrence relation (2) is also defined by the recurrence relation

[TABLE]

for $\sigma_{k}\leq n<\sigma_{k+1}$. Based on Corollary 2, it is clear that $\lambda(n)$ satisfies recurrence (6). Therefore, $\lambda(n)$ satisfies the meta-Fibonacci recurrence (2). ∎