On Generalized Zeckendorf Decompositions and Generalized Golden Strings
aa r X i v : . [ m a t h . N T ] M a y ON GENERALIZED ZECKENDORF DECOMPOSITIONS ANDGENERALIZED GOLDEN STRINGS
HÙNG VIÊ. T CHUA
BSTRACT . Zeckendorf proved that every positive integer has a unique representationas a sum of non-consecutive Fibonacci numbers. A natural generalization of this theo-rem is to look at the sequence defined as follows: for n ≥ , let F n, = F n, = · · · = F n,n = 1 and F n,m +1 = F n,m + F n,m +1 − n for all m ≥ n . It is known that everypositive integer has a unique representation as a sum of F n,m ’s where the indexes ofsummands are at least n apart. We call this the n -decomposition. Griffiths showed aninteresting relationship between the Zeckendorf decomposition and the golden string.In this paper, we continue the work to show a relationship between the n -decompositionand the generalized golden string.
1. I
NTRODUCTION
We define the Fibonacci sequence to be F = 1 , F = 1 , and F m = F m − + F m − for m ≥ . The Fibonacci numbers have fascinated mathematicians for centuries withmany interesting properties. A beautiful theorem of Zeckendorf [14] states that ev-ery positive integer m can be uniquely written as a sum of non-consecutive Fibonaccinumbers. This gives the so-called Zeckendorf decomposition of m . A more formalstatement of Zeckendorf’s theorem is as follows: Theorem 1.1.
For any m ∈ N , there exists a unique increasing sequence of positiveintegers ( c , c , . . . , c k ) such that c ≥ , c i ≥ c i − + 2 for i = 2 , , . . . , k , and m = P ki =1 F c i . Much work has been done to understand the structure of Zeckendorf decompositionsand their applications (see [1, 2, 3, 4, 7, 8, 9, 12]) and to generalize them (see [5,6, 10, 11, 13]). Before stating our main results, we mention several related resultsfrom the literature. Given n ∈ N ≥ , we define the sequence F n, = · · · = F n,n = 1 and F n,m +1 = F n,m + F n,m +1 − n for all m ≥ n . The following theorem, which isa generalization of Theorem 1.1, follows immediately from the proof of [5, Theorem1.3]. Theorem 1.2.
For any m ∈ N , there exists a unique increasing sequence of positiveintegers ( c , c , . . . , c k ) such that c ≥ n , c i ≥ c i − + n for i = 2 , , . . . , k , and m = P ki =1 F n,c i . For conciseness, we call the decomposition in Theorem 1.2 the n -decomposition.In [8], Griffiths made a connection between the golden string and the Zeckendorf de-composition. In particular, the golden string S ∞ = a a a a a a a a a a a a a . . . Date : June 5, 2020.2010
Mathematics Subject Classification.
Key words and phrases.
Zeckendorf decomposition, fixed term, Fibonacci. is defined to be the infinite string of a and a constructed recursively as follows. Let S = a and S = a , and for m ≥ , S m is defined to be the concatenation of thestrings S m − and S m − , which we denote by S m − ◦ S m − . Thus, S = S ◦ S = a ◦ a = a a ,S = S ◦ S = a a ◦ a = a a a , and so on. We generalize the golden string in an obvious way. Given n ∈ N ≥ , we let S = a , . . . , S n = a n ,S m = S m − ◦ S m − n for m ≥ n + 1 , We call S ∞ obtained from the recursive process the n -string. For example, when n = 2 ,we have the golden string; when n = 3 , we have the -string: a a a a a a a a a a a a a a . . . . Lemmas 3.1 and 3.2 in [8] show that the Zeckendorf decomposition ( -decomposition)is linked to the golden string ( -string). This fact might lead us to suspect that in general,the n -decomposition is linked to the n -string. Our first three theorems show that thesuspicion is indeed well-founded. Theorem 1.3.
Let n ≥ and m ≥ . The following items hold. (1) S m contains F n,m letters, of which F n,m − ( n − are a n ’s ,F n,m − n are a ’s ,F n,m − ( n +1) are a ’s , ... F n,m − (2 n − are a n − ’s . (2) For any m ∈ N ≥ , the concatenation S n +( n − m ◦ · · · ◦ S n +( n − ◦ S n gives the first F n,n + F n, n − + · · · + F n, ( m +1) n − m letters of S ∞ . Theorem 1.4.
Let n ≥ and m ≥ . Let F n,c + F n,c + · · · + F n,c k be the n -decomposition of m ∈ N . Then S c k ◦ S c k − ◦ · · · ◦ S c gives the first m letters of S ∞ . Remark 1.5.
In order that Theorem 1.3 item (1) makes sense, we need to extend thesequence F n,m to the left while following the recursive relation. It is straightforward thatfor n ≥ , we have F n, = · · · = F n, − n = 0 , F n, − n = 1 , and F − n = · · · = F − n = 0 .For each m ≥ , let N a i ( m ) denote the number of a i in the string S ∞ up to m . Theorem 1.6.
Let n ≥ and m ≥ . If m = F n,c + F n,c + · · · + F n,c k is an n -decomposition of m , then N a i ( m ) = ( F n,c − ( n + i − + F n,c − ( n + i − + · · · + F n,c k − ( n + i − , if ≤ i ≤ n − F n,c − ( n − + F n,c − ( n − + · · · + F n,c k − ( n − , if i = n. N GENERALIZED ZECKENDORF DECOMPOSITIONS AND GENERALIZED GOLDEN STRINGS 3
Our final result extends [7, Theorem 3.4], which describes the set of all positive inte-gers having the summand F k in their Zeckendorf decomposition. The following theo-rem sheds another light on the relationship between the n -string and the n -decomposition. Theorem 1.7.
For k ≥ n ≥ , the set of all positive integers having the summand F n,k in their n -decomposition is given by Z n ( k ) = ( j + F n,k + n X i =1 F n,k + i · N a i ( m ) : 0 ≤ j ≤ F n,k − ( n − − and m ≥ ) . Remark 1.8.
In [7, Theorem 3.4], Z ( k ) has a closed form thanks to [8, Theorem3.3], which provides a neat formula for N a i ( m ) in the case of the golden string. Theformula was deduced using the Binet’s formula for the Fibonacci numbers. However,the author of the present paper is unable to find such a closed form for Z n ( k ) when n ≥ . Thus, Theorem 1.7 only gives another (not quicker) way to find Z n ( k ) andshows a relationship between the n -string and the n -decomposition.As we proceed to the proof of the main theorems, a number of interesting immediateresults are encountered.2. R ELATIONSHIP BETWEEN THE n - DECOMPOSITION AND THE n - STRING
The following lemma will be used in due course.
Lemma 2.1.
Let n ≥ and S ∞ be the n -string. The following items hold. (1) Fix i and j such that n ≤ i ≤ j − ( n − . Then S j ◦ S i gives the first F n,i + F n,j letters of S ∞ . (2) Fix j ≥ i ≥ n . There exists S ∗ (possibly empty), a concatenation of some S i ’s,such that S i ◦ S ∗ = S j . (3) For m ≥ , there exists S ∗ (possibly empty), a concatenation of some S i ’s, suchthat ( S n +( n − m ◦ · · · ◦ S n +( n − ◦ S n ) ◦ S ∗ = S n +( n − m +1 ◦ S i for some n ≤ i ≤ n + ( n − m −
1) + 1 . (4) For m ≥ , there exists S ∗ (possibly empty), a concatenation of some S i ’s, suchthat ( S n +( n − m ◦ · · · ◦ S n +( n − ◦ S n ) ◦ S ∗ = S n +( n − m +2 . Proof.
We first prove item (1). By construction, we have S j +1 = S j ◦ S j − ( n − and S j +1 gives the first F n,j +1 letters of S ∞ . Hence, S j ◦ S j − ( n − gives the first F n,j +1 letters of S ∞ . Because n ≤ i ≤ j − ( n − , S i gives the first F n,i letters of S j − ( n − . Therefore, S j ◦ S i gives the first F n,i + F n,j letters of S ∞ , as desired.To prove item (2), it suffices to show that there exists S ∗ such that S i ◦ S ∗ = S i +1 .By construction and the fact that i ≥ n , we can let S ∗ = S i − ( n − .Next, we prove item (3). We proceed by induction on m . Base cases: for m = 2 , wehave S n +2( n − ◦ S n +( n − ◦ S n = S n +2( n − ◦ S n . Thus, letting S ∗ be the empty string, we have our claim hold. For m = 3 , we have S n +3( n − ◦ S n +2( n − ◦ S n +( n − ◦ S n = S n +3( n − ◦ S n +( n − . HÙNG VIÊ. T CHU
Again, letting S ∗ be the empty string, we have our claim hold. Inductive hypothesis: suppose that our claim holds for all ≤ m ≤ ℓ for some ℓ ≥ . We have S n +( n − ℓ +1) ◦ S n +( n − ℓ ◦ S n +( n − ℓ − ◦ · · · ◦ S n +( n − ◦ S n = S n +( n − ℓ +1)+1 ◦ ( S n +( n − ℓ − ◦ · · · ◦ S n +( n − ◦ S n )= S n +( n − ℓ +1)+1 ◦ S n +( n − ℓ − ◦ S i for some n ≤ i ≤ n + ( n − ℓ −
2) + 1 . The last equality is due to the inductive hypothesis. By item (2), there exists some S ′ , aconcatenation of some S i ’s, such that S i ◦ S ′ = S n +( n − ℓ − . Hence, ( S n +( n − ℓ +1) ◦ S n +( n − ℓ ◦ S n +( n − ℓ − ◦ · · · ◦ S n +( n − ◦ S n ) ◦ S ′ = S n +( n − ℓ +1)+1 ◦ S n +( n − ℓ − ◦ S i ◦ S ′ = S n +( n − ℓ +1)+1 ◦ S n +( n − ℓ − ◦ S n +( n − ℓ − = S n +( n − ℓ +1)+1 ◦ S n +( n − ℓ − . Therefore, our claim holds for m = ℓ + 1 . This completes our proof of item (3).Finally, item (4) follows immediately from items (2) and (3). (cid:3) We are ready to prove Theorems 1.3 and 1.4.
Proof of Theorem 1.3.
First, we prove item (1). The first part is clear by construction.We prove the second part by induction on m . Base cases : if ≤ m ≤ n − , then S m = a m by construction. All numbers in the set { m − ( n − , m − n, . . . , m − (2 n − } lie between − n and inclusive. By Remark 1.5, for all − n ≤ i ≤ except i = 1 − n , we have F n,i = 0 , while F n, − n = 1 . Write F n, − n = F n,m − ( m +( n − to seethat our claim holds. For m = n , it is easy to check that our claim also holds. Inductive hypothesis: suppose that our claim holds for all ≤ m ≤ ℓ for some ℓ ≥ n .Because S ℓ +1 = S ℓ ◦ S ℓ +1 − n , the number of a n ’s in S ℓ +1 is F n,ℓ − ( n − + F n,ℓ +1 − n − ( n − by the inductive hypothesis , = F n,ℓ − n +2 = F n, ( ℓ +1) − ( n − Similarly, the formulas for the number of a i ’s in S ℓ +1 are all correct. This completesour proof of item (1). (cid:3) Next, we prove item (2). For m = 1 , we have S n +( n − ◦ S n = S n , which gives thefirst F n, n = F n,n + F n, n − letters of S ∞ . For m ≥ , item (2) follows from the firstpart of item (1) and Lemma 2.1 item (4). Proof of Theorem 1.4.
We prove by induction on k , the number of terms in the n -decomposition. Base case: If k = 1 , the statement of the lemma is certainly true. Inductive hypothesis: assume that the statement is true for some k = ℓ ≥ . Considerthe n -decomposition m = F n,c + F n,c + · · · + F n,c ℓ + F n,c ℓ +1 . By the inductive hypothesis, we know that S c ℓ ◦ S c ℓ − ◦ · · · ◦ S c gives the first F n,c + F n,c + · · · + F n,c ℓ letters of S ∞ , and therefore, by Theorem 1.3item (2), of S c ℓ +1 − ◦ S c ℓ +1 − ◦ · · · ◦ S c ℓ +1 − ℓ . N GENERALIZED ZECKENDORF DECOMPOSITIONS AND GENERALIZED GOLDEN STRINGS 5
Hence, we know that S c ℓ +1 ◦ S c ℓ ◦ S c ℓ − ◦ · · · ◦ S c gives the first m letters of S ℓ +1 ◦ S c ℓ +1 − ◦ S c ℓ +1 − ◦ · · · ◦ S c ℓ +1 − ℓ , and hence, by Theorem 1.3 item (2), of S ∞ . (cid:3)
3. F
IXED TERM IN THE n - DECOMPOSITION
In this section, we fix k ≥ n ≥ . Our goal is to characterize the set Z n ( k ) , the set ofall positive integers having the summand F n,k in their n -decomposition. The followinglemma generalizes [7, Lemma 2.5]. Lemma 3.1.
Let m ≥ . For v ≥ and ≤ u ≤ n , we have F n,m + vn + u − ( F n,m + vn +( u − + · · · + F n,m + n +( u − ) = F n,m + u . (3.1) Proof. If v = 1 , the identity holds due to the linear recurrence relation of { F n,i } . As-sume that v ≥ . We have ( F n,m + vn +( u − + · · · + F n,m + n +( u − ) + F n,m + u = ( F n,m + vn +( u − + · · · + F n,m +2 n +( u − ) + ( F n,m + n +( u − + F n,m + u )= ( F n,m + vn +( u − + · · · + F n,m +2 n +( u − ) + F n,m + n + u ... = F n,m + vn +( u − + F n,m +( v − n + u = F n,m + vn + u . This completes our proof. (cid:3)
Lemma 3.2.
Let i and j ∈ N . If F n,u and F n,v are the largest summands in the n -decompositions of i and j , respectively, then u > v implies that i > j . The statement of the lemma is obvious since the n -decomposition of a number canbe found by the greedy algorithm. Lemma 3.3.
For j ≥ , the ( F n,nj ) th character of S ∞ is a n ; the ( F n,nj +1 ) th characterof S ∞ is a ; . . . ; ( F n,nj +( n − ) th character of S ∞ is a n − .Proof. Base case: By construction, we know that the statement is true for j = 1 . Inductive hypothesis: suppose the statement is true for j = m for some m ≥ . The ( F n,n ( m +1) ) th of S ∞ is the last letter of S n ( m +1) = S n ( m +1) − ◦ S nm . By the inductivehypothesis, the last letter of S nm is a n and so, the ( F n,n ( m +1) ) th letter of S ∞ is a n . Theproof is completed by similar arguments for the ( F n,n ( m +1)+ u ) th letter, where ≤ u ≤ n − . (cid:3) Let X n,k denote the set of all positive integers whose n -decompositions have F n,k asthe smallest summand. Next, let Q n,k = { q ( j ) } j ≥ be the strictly increasing infinite se-quence that results from arranging the elements of X n,k into ascending numerical order.Table 1 gives the ordered list of summands for each q ( j ) , where the r th row correspondsto the r th smallest element from X n,k . (Lemma 3.2 helps explain the ordering of rows.) HÙNG VIÊ. T CHU
Row1 F n,k F n,k F n,k + n F n,k F n,k + n +1 ... ( n + 1) F n,k F n,k +2 n − ( n + 2) F n,k F n,k +2 n ( n + 3) F n,k F n,k + n F n,k +2 n ( n + 4) F n,k F n,k +2 n +1 ( n + 5) F n,k F n,k + n F n,k +2 n +1 ( n + 6) F n,k F n,k + n +1 F n,k +2 n +1 ...Table 1. The n -decompositions, in numerical order, of the positive integers having F n,k as their smallest summand. Lemma 3.4.
For j ≥ n , the rows of Table 1 for which F n,k + j is the largest summandare those numbered from F n,j + 1 to F n,j +1 inclusive.Proof. We prove by induction on j . Base cases: for n ≤ j ≤ n − , there is exactlyone row with F n,k + j being the largest summand. In particular, the row with F n,k + j beingthe largest summand is the ( j − ( n − th . Hence, the base cases are done if we provethe two following claims:(1) F n,j +1 − F n,j = 1 ;(2) F n,j + 1 = j − ( n − .For the first claim, we have F n,j +1 − F n,j = F n,j − ( n − = 1 for all n ≤ j ≤ n − .This is due to the construction of { F n,j } . To prove the second claim, observe that as j goes from n to n − , we have F n,j increase from to n . Inductive hypothesis: suppose that the statement of the lemma holds for n ≤ j ≤ m for some m ≥ n − . We want to show that the rows with F n,k + m +1 being the largestsummand is from F n,m +1 + 1 to F n,m +2 . By the inductive hypothesis, the number ofrows with the largest summand not larger than F n,k + m +1 − n is m +1 − n X j = n ( F n,j +1 − F n,j ) = 1 + F n,m +2 − n − F n,n = F n,m +2 − n , which is also the number of rows with F n,k + m +1 being the largest summand.By the inductive hypothesis, the rows with F k + m being the largest is from F n,m + 1 to F n,m +1 . Therefore, the rows with F n,k + m +1 being the largest is from F n,m +1 + 1 to F n,m +1 + F n,m +2 − n = F n,m +2 . (cid:3) Lemma 3.5.
For j ≥ , we have q ( j + 1) − q ( j ) = F n,k +1 , if the j th character of S ∞ is a ; F n,k +2 , if the j th character of S ∞ is a ; ... F n,k + n , if the j th character of S ∞ is a n . (3.2) N GENERALIZED ZECKENDORF DECOMPOSITIONS AND GENERALIZED GOLDEN STRINGS 7
Proof.
We prove by induction on j . Base cases: consider ≤ j ≤ F n, n − n . FromTable 1, we know that q ( j + 1) − q ( j ) = ( F n,k + n , if j = 1; F n,k +( j − , if ≤ j ≤ n. (3.3)Using the fact that S ∞ = a n a a . . . a n − . . . and (3.3), we have proved the base casesfor ≤ j ≤ F n, n − . Inductive hypothesis:
Assume that the statement is true for ≤ j ≤ F n,m − forsome m ≥ n . By Lemma 3.4, the first F n,m − n +1 rows of Table 1 are those for whichthe largest summand is no greater than F n,k + m − n , and the rows for which F n,k + m is thelargest summand are from F n,m + 1 to F n,m +1 .Due to the ordering of rows in Table 1, we have q ( i ) + F n,k + m = q ( i + F n,m ) for ≤ i ≤ F n,m − n +1 . Hence, for ≤ i ≤ F n,m − n +1 − , we have q ( i + 1 + F n,m ) − q ( i + F n,m ) = ( q ( i + 1) + F n,k + m ) − ( q ( i ) + F n,k + m )= q ( i + 1) − q ( i ) . By the construction of S ∞ , the string of the first F n,m − n +1 characters is identical to thestring of characters from the ( F n,m +1) th to the F n,m +1 th inclusive. Therefore, we knowthat (3.2) is true for F n,m ≤ j ≤ F n,m +1 − . It remains to show that (3.2) is truefor j = F n,m . We have q ( F n,m + 1) − q ( F n,m ) = F n,k + m − ( F n,k + m − + F n,k + m − − n + · · · + F n,k + m − − ℓn ) , where ℓ satisfies n ≤ m − − ℓn < n . Write m = vn + u for some ≤ u ≤ n . Itfollows that − ( u − /n ≤ v − ℓ < − ( u − /n , so v − ℓ = 1 . Hence, q ( F n,m + 1) − q ( F n,m )= F n,k + vn + u − ( F n,k + vn + u − + F n,k +( v − n + u − + · · · + F n,k + n + u − ) = F n,k + u due to Lemma 3.1. Using Lemma 3.3, we conclude that (3.2) is true for j = F n,m . Thiscompletes our proof. (cid:3) Finally, we prove Theorem 1.7.
Proof of Theorem 1.7.
By Lemma 3.5, we can write X n,k = ( F n,k + n X i =1 F n,k + i · N a i ( m ) : m ≥ ) . The numbers in { F n,n , F n,n +1 , . . . , F n,k − n } are used to obtain the n -decompositions ofall integers for which the largest summand is no greater than F n,k − n . In particular, such n -decompositions generate all integers from to F n,k − n +1 − inclusive. Furthermore,such decompositions can be appended to any n -decomposition having F n,k as its small-est summand to produce another n -decomposition. Therefore, we know that Z n ( k ) = ( j + F n,k + n X i =1 F n,k + i · N a i ( m ) : 0 ≤ j ≤ F n,k − ( n − − and m ≥ ) , as desired. (cid:3) HÙNG VIÊ. T CHU R EFERENCES [1] A. Best, P. Dynes, X. Edelsbrunner, B. McDonald, S. Miller, C. Turnage-Butterbaugh, and M.Weinstein, Benford behavior of Zeckendorf’s decompositions,
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NIVERSITY OF I LLINOIS AT U RBANA -C HAMPAIGN , U