Summand minimality and asymptotic convergence of generalized Zeckendorf decompositions

Abstract

Given a recurrence sequence H, with $H_n = c_1 H_{n-1} + \dots + c_t H_{n-t}$ where $c_i \in \mathbb {N}_0$ for all i and $c_1, c_t \ge 1$, the generalized Zeckendorf decomposition (gzd) of $m \in \mathbb {N}_0$ is the unique representation of m using H composed of blocks lexicographically less than $\sigma = (c_1, \dots , c_t)$. We prove that the gzd of m uses the fewest number of summands among all representations of m using H, for all m, if and only if $\sigma$ is weakly decreasing. We develop an algorithm for moving from any representation of m to the gzd, the analysis of which proves that $\sigma$ weakly decreasing implies summand minimality. We prove that the gzds of numbers of the form $v_0 H_n + \dots + v_\ell H_{n-\ell }$ converge in a suitable sense as $n \rightarrow \infty$; furthermore we classify three distinct behaviors for this convergence. We use this result, together with the irreducibility of certain families of polynomials, to exhibit a representation with fewer summands than the gzd if $\sigma$ is not weakly decreasing.

Appendices

Appendix A: Algorithm from any representation to the gzd

Definition A.1

Let $\rho = (r_n, \dots , r_0, \infty _{t-1})$ be a representation using $H_\sigma$. We say $\rho$ is legal up to $\mathbf {s}$ if $(r_n, \dots , r_s)$ can be expressed as $\Lambda _1 \oplus \dots \oplus \Lambda _j$ with each $\Lambda _i$ an allowable block and $\Lambda _1 \ne (0)$ (where $\oplus$ represents concatenation).

Definition A.2

The minimum legal index (m.l.i.) of $\rho$ is the smallest index s such that $\rho$ is legal up to s.

Notice that if $\sigma = (c_1, \dots , c_t)$ and $\rho$ is a representation whose m.l.i. is s, then $r_{s-1} \ge c_1$. If $r_{s-1} = c_1$, then $r_{s-2} \ge c_2$. At some point, we must either have that $r_{s-j} > c_j$, or for all $1 \le j < t$, we have $r_{s-j} = c_j$ and $r_{s-t} \ge c_t$. This motivates the following definition.

Definition A.3

Suppose $\rho$ has m.l.i. equal to s. Let j be the smallest index such that $r_{s-j} >c_j$, or if $r_{s-i}=c_i$ for all $1 \le i \le t$, then let $j=t$. The violation index is $s-j$ and we call $r_{s-j}$ the violation.

Definition A.4

Suppose $\rho$ has m.l.i. equal to s and violation index equal to j. Then $(r_{s-1}, \dots ,$ $r_{s-(j-1)}) = (c_1,\dots ,c_{j-1})$ is called the violation prefix. The prefix and the violation together comprise the violation block.

Remark A.1

A representation $\rho$ can be decomposed as

$$\begin{aligned} \rho \ = \ \Lambda _1 \oplus \cdots \oplus \Lambda _j \oplus \Psi \oplus (r_d, \dots , r_0, \infty _{t-1}), \end{aligned}$$

(A.14)

with each $\Lambda _i$ a valid block, and $\Psi$ the violation block. For any $\Lambda _i$ (or $\Psi$), we call the block immediately to the left of $\Lambda _i$ (or $\Psi$) as in Eq. (A.14) the left neighbor block of $\Lambda _i$ (or $\Psi$). The left neighbor block of $\Lambda _1$ is defined to be (0).

Definition A.5

Suppose $\rho$ has m.l.i. s and violation index j. We say that $\rho$ is semi-legal up to q if $q = s - j + 1$. We call q the semi-legal index (s.l.i.).

Remark A.2

We note that the s.l.i. is the index to the left of the violation index, i.e., the s.l.i. is the violation index plus one. Furthermore, the difference between the m.l.i. and the s.l.i. is exactly the length of the violation prefix, which is 0 if the violation prefix is empty.

Remark A.3

The m.l.i. and the s.l.i. give us a way to quantify how “close” a representation of m is to $GZD (m)$. Definitions A.1, A.2 and A.5 imply that the m.l.i. and the s.l.i. of $\rho$ are both equal to 0 if and only if $\rho$ is allowable.

Example A.1

Let $\sigma =(3,2,4)$ and $\rho = (3, 2, 1, 1, 3, 0, 3, 3, 5, \infty _2)$. Then

$$\begin{aligned} \rho \ =\ \left( \, \boxed {3, 2, 1}, \boxed {1}, \boxed {3, 0}, \begin{array}{|l}\hline 3,3\\ \hline \end{array}, 5, \infty _2 \, \right) , \end{aligned}$$

where each closed box represents a valid block, and the right-opened box ($\begin{array}{|l}\hline 3,3\\ \hline \end{array}$) represents the violation block. The violation index is 0, the s.l.i. is 1, and the m.l.i is 3. We are able to carry to 3 (the m.l.i.) because $r_2=3=c_1$, $r_1=3>2=c_2$ and $r_0=5>4=c_3$. After carrying, we get

$$\begin{aligned} \Big ( \, \boxed {3, 2, 1}, \boxed {1}, \boxed {3, 1}, \boxed {0},\boxed {1}, \boxed {1}, \infty _2 \,\Big ), \end{aligned}$$

which is the gzd. We note that the m.l.i. and s.l.i. equal 0.

Now consider the same signature as Example A.1 but with $\rho =(3,2,1,1,3,0,3,3,1, \infty _2)$. The violation index, m.l.i., and s.l.i. are still the same but we are not able to carry because $r_0=1<4=c_3$. This motivates the following definitions.

Definition A.6

Suppose $\rho$ has m.l.i. equal to s. We call $s-\ell$ the carry obstruction index (c.o.i.) if for all $1 \le i < \ell$, we have $r_{s-i} \ge c_i$ and $r_{s-\ell } < c_\ell$.

Definition A.7

Suppose $\rho$ has m.l.i. equal to s. Then $s-e$ is called the rightmost excess index (r.e.i.) if e is the largest index such that for all $i < e$, we have $r_{s-i} \ge c_i$ and $r_{s-e} > c_e$.

Example A.2

Consider the aforementioned example with $\sigma =(3,2,4)$ and $\rho = (3, 2, 1, 1, 3,$ $0, 3, 3, 1, \infty _2)$. Here, m.l.i. $=3$, s.l.i. $=2$ and the violation index is 1. The c.o.i. is 0 and the r.e.i. is 1 because $r_1=3>c_2=2$. We can borrow from the r.e.i. to make our c.o.i. large enough that we are able to carry. We demonstrate this in Table 3.

Table 3 Sequence of borrows and carries to move to the gzd

Remark A.3 explains how a representation can be thought of as being “far” from the gzd if its m.l.i. and s.l.i. are large. As such, a potential way to turn any representation into the gzd is to try decreasing the m.l.i. and s.l.i. of the representation to zero. To do so, one may start by finding the first violation and attempting to “fix” it. Because any valid block is lexicographically less than $\sigma$ (see Definition 2.2), the entry at the violation index is always “too large.” This suggests that we may try to carry in order to fix it, as in Example A.1. In the case where we are not able to carry, which is to say that the c.o.i. exists, we would first borrow from the r.e.i. in order to carry, as in Example A.2. As one performs these borrows and carries to fix all possible violations, one would expect to decrease the m.l.i. and the s.l.i. to zero to reach the gzd. Motivated by this intuition, we introduce Algorithm A.1. Although there is some subtlety in the algorithm, this intuition is the key idea in the proof of validity.

Algorithm A.1

Example A.3

Let $\sigma = (3, 2, 5)$. We apply Algorithm A.1 to $\rho = (2, 3, 2, 4, 4, 3, 0, 0, \infty _2)$.

Remark	Index
	7	6	5	4	3	2	1	0	$-$ 1	$-$ 2
m.l.i. $=4$, s.l.i. $=4$, c.o.i. $=1$, r.e.i. $=2$	2	3	2	4	4	3	0	0	$\infty$	$\infty$
Borrow from 2						$-$ 1	3	2	5
m.l.i. $=4$, s.l.i. $=4$, c.o.i. $=1$, r.e.i. $=3$	2	3	2	4	4	2	3	2	$\infty$	$\infty$
Borrow from 3					$-$ 1	3	2	5
Able to carry	2	3	2	4	3	5	5	7	$\infty$	$\infty$
Carry to 4				1	$-$ 3	$-$ 2	$-$ 5
Left neighbor block $= \sigma$	2	3	2	5	0	3	0	7	$\infty$	$\infty$
Carry to 7	1	$-$ 3	$-$ 2	$-$ 5
m.l.i. $=1$, s.l.i. $=1$, able to carry	3	0	0	0	0	3	0	7	$\infty$	$\infty$
Carry to 1							1	$-$ 3	$-$ 2	$-$ 5
m.l.i. $=1$, s.l.i. $=1$, able to carry	3	0	0	0	0	3	1	4	$\infty$	$\infty$
Carry to 1							1	$-$ 3	$-$ 2	$-$ 5
m.l.i. $=0$. We’ve reached the gzd	3	0	0	0	0	3	2	1	$\infty$	$\infty$

To assist in the proof that Algorithm A.1 terminates in the gzd, we state and prove a few lemmas. We first show that the s.l.i. weakly decreases in Lemma A.1. Then, we show that the s.l.i. strictly decreases after finitely many steps in Lemma A.2. Finally, using those two lemmas, we prove that the m.l.i. decreases to 0 in finitely many steps.

Lemma A.1

The semi-legal index (s.l.i.) monotonically decreases during Algorithm A.1.

Proof

When performing Algorithm A.1, if we are not able to carry to the m.l.i., then the s.l.i. either stays the same or decreases. Thus, for the purposes of proving the lemma, we may suppose that we are able to carry.

Suppose the left neighbor block of the violation block is $(c_1, \dots , c_{\ell -1}, d_\ell )$ with $d_\ell < c_\ell$. If $d_\ell < c_\ell -1$, then we still have a valid block after carrying. The entries of $\rho$ with indices strictly between the old m.l.i. and the old s.l.i. become zero after carrying, and thus become valid blocks. Therefore the s.l.i. will have either stayed the same or decreased.

Now instead suppose that $d_\ell = c_\ell - 1$ and $\ell < t$. After carrying, our representation is of the form

$$\begin{aligned} \Lambda _1 \oplus \dots \oplus \Lambda _j \oplus (c_1, \dots , c_{\ell -1}, c_{\ell }, 0, \dots , 0, v-c_{k+1}, \dots ), \end{aligned}$$

where v is the original violation and the length of $0,\dots , 0$ in the middle is the length of the violation prefix before carrying.

We cannot have a violation before $v-c_{k+1}$ because a violation requires that an entry is greater than its corresponding entry in the signature; however, the first $\ell$ terms agree with the signature and the remaining terms are all zero, so they are either equal to or less than the corresponding terms in the signature. Therefore the earliest possible violation is at $v-c_{k+1}$, so in this case the s.l.i. either stays the same or decreases.

Finally, suppose $\ell = t$ and $d_\ell = c_t - 1$. After we carry, the left neighbor block of the violation block is now $(c_1, \dots , c_t)$, so we immediately carry again by Algorithm A.1. After the carry, our representation looks like

$$\begin{aligned} \Lambda _1 \oplus \dots \oplus \Lambda _{j-1} \oplus (c_1, c_2, \dots , c_m - r, \underbrace{0, \dots , 0}_{t}, \dots ), \end{aligned}$$

with $r \ge 0$. If $r \ge 1$, then clearly the s.l.i. has not increased. If $m < t$ and $r = 0$, then since we have t trailing zeros, the s.l.i. still has not increased ($(c_1, \dots , c_m, \underbrace{0, \dots , 0}_{t})$ is composed of allowable blocks). If $r = 0$ and $m = t$, then by Algorithm A.1, we must carry yet again to the next left neighbor block and repeat the above arguments. We know that at some point the left neighbor block is not $(c_1, \dots , c_t - 1)$ since there are finitely many nonzero blocks. As such, at some point this process terminates without increasing the s.l.i. $\square$

Lemma A.2

Suppose s.l.i. $\ge 1$. Then, after finitely many steps of Algorithm A.1, the s.l.i. decreases.

Proof

Let $\rho$ be

$$\begin{aligned} \rho \ =\ \Lambda _1 \oplus \dots \oplus \Lambda _j \oplus (c_1, \dots , c_m, v, \dots ), \end{aligned}$$

where $0\le m<t$, the violation is v, and each $\Lambda _i$ is a valid block. Suppose $\Lambda _j=(c_1,\dots ,c_{\ell -1},d_\ell )$, with $d_\ell <c_\ell$.

We now show that after finitely many steps, the value of v decreases. In performing Algorithm A.1, every time we borrow, the value in the rightmost excess index (r.e.i.) decreases by 1. If we keep borrowing and are never able to carry, then at some point we must have decreased the value at the original r.e.i. to the point where it is no longer “excess”. At this point, the r.e.i. increases, and so if we never carry, then after finite time the r.e.i. becomes equal to the violation index. In that case, the next time we borrow, v decreases.

We are left to handle the case when we are able to carry. When we carry, v decreases by $c_{m+1} \ge 0$. If this amount is $c_{m+1} \ge 1$, then clearly v decreases. Otherwise, $c_{m+1} = 0$ (implying $m \ge 1$). After carrying, we obtain

$$\begin{aligned} \Lambda _1 \oplus \dots \oplus \Lambda _{j-1} \oplus (c_1, \dots , c_{\ell -1}, d_{\ell }+1, \underbrace{0,\dots ,0}_{m},v,\dots ). \end{aligned}$$

There are now $m \ge 1$ zeros to the left of v. By Lemma A.1, after performing any possible carries in the left neighbor blocks, the s.l.i. will at worst stay the same, in which case either the s.l.i. decreases (as we ultimately want) or v remains the violation.

Assuming v remains the violation, then repeating the above arguments, we either have that at some point we borrow from v, or else we carry. In the former case, v obviously decreases; in the latter case, either v decreases or the violation block is of the form $(c_1, \dots , c_q, v)$ with $c_{q+1} = 0$. However at this point, since we’ve already carried once before (resulting in m zeros to the left of v), we have that $c_{q-m+1} = \dots = c_q = 0$. Since $c_1 \ne 0$, we must have that $q - m \ge 1$. Then, after the second carry, the number of zeros to the left of v increases by $q-m \ge 1$.

As Algorithm A.1 continues, we may repeat the above arguments implying that either the s.l.i. decreases, or v eventually decreases, or the number of zeros preceding v grows larger and larger. In the last case, eventually the number of zeros preceding v grows larger than t. Then at the next step, either the s.l.i. decreases, or the violation prefix has zero length, in which case v decreases by $c_1 \ge 1$ after the next carry.

Thus in all cases, either the s.l.i. decreases, or v decreases. However, if v continues to decrease, then after finitely many steps v must be small enough that the s.l.i. decreases. $\square$

Theorem A.1

Algorithm A.1 terminates in the gzd.

Proof

By Lemma A.2, the s.l.i. decreases to zero. Theorem A.1 holds if and only if the m.l.i. decreases to zero after finitely many steps. Therefore, we need only show that when the s.l.i. is zero, the m.l.i. goes to zero.

Suppose the s.l.i. is zero. If the m.l.i. is not equal to zero, then our representation is of the form $\rho =\Lambda _1 \oplus \dots \oplus \Lambda _j \oplus (c_1, \dots , c_m, \infty _{t-1})$. Suppose $\Lambda _j = (c_1, \dots , c_{\ell -1}, d_\ell )$ with $d_\ell <c_\ell$. We can immediately carry to the m.l.i. using the $\infty$ places, resulting in

$$\begin{aligned} \Lambda _1 \oplus \dots \oplus \Lambda _{j-1} \oplus (c_1, \dots , c_{\ell -1}, d_\ell +1, \underbrace{0,\dots ,0}_{m},\infty _{t-1}). \end{aligned}$$

If $d_\ell +1<c_\ell$, then the m.l.i. is zero.

Otherwise, we have $d_\ell +1=c_\ell$. If $\ell = t$, then we carry to the left neighbor block as needed which ultimately decreases the m.l.i. to zero. If $\ell < t$, then unless $c_{\ell +1} = \dots = c_{\ell +m} = 0$, we are done. Otherwise, we must carry again, and the violation prefix grows larger. Repeating the above arguments, we see that either the algorithm terminates with the m.l.i. equal to zero, or else the violation prefix grows arbitrarily large. However, the violation prefix cannot grow larger than t, so we must eventually end up in the former case. $\square$

Appendix B: Proof of Proposition 4.1

Definition B.1

Let H be a sequence. Define $H_{n, k}$ by

$$\begin{aligned} H_{n, k}\ =\ \begin{bmatrix} H_n&H_{n+1}&\cdots&H_{n+k} \\ H_{n+1}&H_{n+2}&\cdots&H_{n+k+1}\\ \vdots&\vdots&\ddots&\vdots \\ H_{n+k}&H_{n+k+1}&\cdots&H_{n+2k}\\ \end{bmatrix}. \end{aligned}$$

Matrices of the form $H_{n, k}$ are known as Hankel matrices.

We need the following result of [21].

Lemma B.1

(Lemma 3 of [21]) A sequence satisfies some order-k linear recurrence if and only if $\det (H_{n, k}) = 0$ for all n.

Let $\overline{H}_\ell = \{H_n\}_{n \ge \ell }$ denote the $\ell$th truncation of H, i.e., the sequence obtained by removing the first $\ell -1$ terms.

Proposition B.1

Suppose f is the minimal polynomial for H. Then f is the minimal polynomial for all truncations $\overline{H}_\ell$.

Proof

Let H be some linear recurrence sequence and let f be the minimal polynomial for H. Let $f = x^t - c_1 x^{t-1} - \dots - c_t$. Note that $c_t \ne 0$. We shall show that $\det (H_{n, t-1}) \ne 0$ for all n. Let $D = \det (H_{1, t-1})$. In particular we will show that $|\det (H_{n, t-1})| = |c_t^n D|$.

We proceed by induction. Suppose we have

$$\begin{aligned} H_{n, t-1} \ =\ \begin{bmatrix} H_n&H_{n+1}&\cdots&H_{n+t-1} \\ H_{n+1}&H_{n+2}&\cdots&H_{n+t}\\ \vdots&\vdots&\ddots&\vdots \\ H_{n+t-1}&H_{n+t}&\cdots&H_{n+2t-2}\\ \end{bmatrix}. \end{aligned}$$

We index our columns starting from zero. Notice that the 1st through $(t-1)$th columns appear as columns in $H_{n+1, t-1}$. Furthermore, notice that if we multiply the zeroth column by $c_t$ and add to it $c_{t-1}$ times the first column, plus $c_{t-2}$ times the second column, plus etc., plus $c_1$ times the $t-1$ column, then the columns of the resulting matrix agree with the columns of $H_{n+1, t-1}$. The determinant has gone up by a factor of $c_t$. In order to move the resulting matrix to the form of $H_{n+1, t-1}$, we need to permute some columns, which may change the sign of the determinant, but not the magnitude.

We know that $f \in \mathcal {I}(\overline{H}_n)$. However, if f did not generate $\mathcal {I}(\overline{H_n})$, then there must be some polynomial of lower degree in $\mathcal {I}(\overline{H}_n)$. In particular, there must be some polynomial of degree $t-1$ in $\mathcal {I}(\overline{H}_n)$. If this were so, then $\det (H_{m, t-1})$ would be zero for all $m \ge n$ by Lemma B.1, which is a contradiction. Therefore, we must have that f is the lowest degree polynomial in $\mathcal {I}(\overline{H}_n)$ for all n, and thus is the minimal polynomial for all $\overline{H}_n$. $\square$

Proof of Proposition 4.1

By Proposition B.1, it suffices to show that $\det {H_{-(t-1), t-1}} \ne 0$. It is immediate from writing out the matrix $H_{-(t-1),t-1}$ that by switching columns, we can make it lower triangular with diagonal entries all equal to 1, and hence $\det ({H_{-(t-1), t-1}})=1$. $\square$

Appendix C: Proof of Proposition 4.2

Proof of Proposition 4.2

Let A denote the companion matrix corresponding to f. All entries in A are non-negative. It is straightforward to check that the tth power of A has all entries positive. Hence A is a primitive matrix. Hence we may apply the Perron-Frobenius theorem to conclude that f has a unique real dominating root of multiplicity one, call it $\beta$.

Notice that $\beta \ge c_1$ since $f(c_1) = -c_2 c_1^{t-2} - \dots - c_t \le 0$. If $t \ge 2$, then $\beta > c_1 = 1$. If $c_1 \ge 2$, then $\beta \ge 2$. If $c_1 = 1$ and $t = 1$, then $f = x - 1$ which we do not consider to be of positive type.

Furthermore, notice that the coefficients of f have exactly one sign change. Therefore by Descartes’ rule of signs, f must have at most one positive root, and hence exactly one. $\square$

Acknowledgements

Steven J. Miller was partially supported by NSF Grants DMS1561945 and DMS1265673, while the Yen Nhi Truong Vu was supported by Professor Amanda Folsom and her NSF Grant DMS1449679. The authors thank the SMALL REU Program at Williams College, which is supported by NSF Grant DMS1347804 and the Williams College Science Center. The authors are also thankful for the detailed and helpful comments of the referee, which have greatly improved the exposition of this paper and allowed us to better connect our results with the existing literature. We would also like to thank Jeffrey Lagarias for recommending several relevant papers to us.