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AN INTEGER SEQUENCE WITH A DIVISIBILITY

PROPERTY

MUHAREM AVDISPAHI´

C AND FARUK ZEJNULAHI

Abstract. Modifying the divisibility condition, we introduce a

new integer sequence and establish its representation in terms of

Fibonacci numbers and the golden mean.

1. Introduction

A permutation {ai}∞

i=1 of positive integers is said to be divisible if

n

X

i=1

ai≡0 (mod n) for all n∈N.

A concrete example of a permutation with such a property is the

O.E.I.S. sequence A019444. Our point of interest in the present note

is a case of a modiﬁed divisibility condition.

Let the sequence {zn}∞

n=1 be deﬁned as follows: z1= 1 and zn(n > 1)

is the least positive integer distinct from z1, z2, . . . , zn−1such that

Xn=

n

X

i=1

zi≡1 (mod n+ 1) for all n∈N.

We shall prove that it possesses the properties given in the next

theorem.

Theorem 1.1. The sequence {zn}∞

n=1 is a permutation of the set of

positive integers and the following formulas hold:

zn=

1for n= 1;

Fk+1 for n=Fk(k > 2);

Fk−1−1for n=Fk−1(k > 4);

bkτ cfor n=bkτ 2c,n6=Fk(k > 2) and n6=Fk−1(k > 4);

bkτ 2cfor n=bkτ c,n6=Fk(k > 2) and n6=Fk−1(k > 4).

Here {Fk}∞

k=1 denotes the Fibonacci sequence and τ=1+√5

2is the

golden mean.

2010 Mathematics Subject Classiﬁcation. 05A17, 11P83, 11B39.

Key words and phrases. Fibonacci numbers, partitions of integers, congruences

and congruential restrictions.

1

2 MUHAREM AVDISPAHI ´

C AND FARUK ZEJNULAHI

2. Preliminaries

In order to prove Theorem 1.1, we shall need several facts about the

Fibonacci sequence and the golden mean. Some of these results are new

or possibly new as indicated in Remark 2.1 at the end of this section.

We put τ1=√5−1

2and note that τ2=3+√5

2. The straightforward

identities τ= 1 + τ1,τ2= 1 + τ,τ τ1= 1 will be used in the sequel

without special mention.

Lemma 2.A. For any k∈N, the following identities are valid:

(i) bbkτ cτc=bkτ c+k−1;

(ii) b(bkτ c+ 1) τc=bkτc+k+ 1;

(iii) b(bkτ c+k)τc= 2 bkτc+k;

(iv) bkτ c=bjτ cfor j∈Nif and only if k=j.

Proof. Let us check (iii) and (iv). The ﬁrst two identities can be de-

rived analogously. Now, the proof of (iii) is accomplished through the

following chain of equivalent statements, the last one being obviously

true:

k+ 2 bkτc<(bkτc+k)τ < k + 2 bkτc+ 1

⇐⇒ bkτc<(bkτ c+k) (τ−1) <bkτ c+ 1

⇐⇒ bkτc<(bkτ c+k)τ1<bkτ c+ 1

⇐⇒ bkτcτ < bkτ c+k < (bkτc+ 1) τ

⇐⇒ bkτc(τ−1) < k < (bkτc+ 1) (τ−1) + 1

⇐⇒ bkτcτ1< k < (bkτc+ 1) τ1+ 1

⇐⇒ bkτc< kτ < bkτ c+1+τ.

For (iv) suppose n=bkτ c=bjτ c. Adding up relations n≤j τ < n + 1

and −n−1<−kτ ≤ −n, we get −1<(j−k)τ < 1. Multiplication

by τ1gives −τ1< j −k < τ1. This yields j−k= 0 since 0 < τ1<1.

For a further reference, we adopt the convention

x≺y⇐⇒ x+ 2 ≤y.

Theorem 2.B. (Zeckendorf) Every positive integer nhas the unique

representation in the form

(∆) n=Fi1+Fi2+· · · +Fir

where i1≺i2≺ ·· · ≺ ir, and i1≥2.

The relation (∆) is known as the Fibonacci representation of n.

AN INTEGER SEQUENCE WITH A DIVISIBILITY PROPERTY 3

Deﬁnition 2.1. With respect to (∆), we introduce sets A1,A2,A3

and A4as follows:

n∈A1⇔i1= 2 and i2is odd or n= 1;

n∈A2⇔i1= 2 and i2is even;

n∈A3⇔i1>2 and i1is even;

n∈A4⇔i1>2 and i1is odd.

We also make use of the function e:N→Ndeﬁned by

e(n) = Fi1−1+Fi2−1+· · · +Fir−1,

where nis represented through (∆).

For the sake of completeness of exposition, we recall the following

theorem.

Theorem 2.C. (Beatty-Skolem) If αand βare positive irrational

numbers such that 1

α+1

β= 1, then the sets Sα={bnαc:n∈N}

and Sβ={bnβc:n∈N}form a disjoint decomposition of the set of

positive integers, i.e., Sα∪Sβ=Nand Sα∩Sβ=∅.

The next lemma provides some useful information about the sets A1,

A2,A3and A4and the function e.

Lemma 2.D. (i) If n∈A1∪A2and n= 1 + Fi2+· · · +Fir, then

b(1 + Fi2+· · · +Fir)τc= 1 + Fi2+1 +· · · +Fir+1 .

(ii) If n∈A3and n=Fi1+Fi2+· · ·+Fir, then b(Fi1+· · · +Fir)τc=

Fi1+1 +· · · +Fir+1 −1.

(iii) If n∈A4and n=Fi1+Fi2+· · ·+Fir, then b(Fi1+· · · +Fir)τc=

Fi1+1 +· · · +Fir+1 .

(iv) If a(n) = bnτcand b(n) = bnτ 2c, then e(a(n)) = nand

e(b(n)) = a(n).

(v) e(n) = b(n+ 1) τ1c.

(vi) If jis odd and j > 1, then e(Fj−1) = Fj−1and if jis even

and j > 2, then e(Fj−1) = Fj−1−1.

4 MUHAREM AVDISPAHI ´

C AND FARUK ZEJNULAHI

Proof. (i) One has

(1 + Fi2+· · · +Fir)τ= τ+τ1

τ+τ1

+τi2+ (−τ1)i2

τ+τ1

+· · · +τir+ (−τ1)ir

τ+τ1!τ

=τ2+ 1

τ+τ1

+τi2+1 −(−τ1)i2−1

τ+τ1

+· · · +τir+1 −(−τ1)ir−1

τ+τ1

=τ2−τ2

1

τ+τ1

+τi2+1 + (−τ1)i2+1

τ+τ1

+· · · +τir+1 + (−τ1)ir+1

τ+τ1

+1 + τ2

1−(−τ1)i2+1 −(−τ1)i2−1− · · · − (−τ1)ir+1 −(−τ1)ir−1

τ+τ1

= 1 + Fi2+1 +· · · +Fir+1 +1 + τ2

1

τ+τ11−(−τ1)i2−1− · · · − (−τ1)ir−1.

The above equality implies

(1 + Fi2+· · · +Fir)τ > 1 + Fi2+1 +· · · +Fir+1 +1 + τ2

1

τ+τ11−τ2

1−τ4

1− · · ·

>1 + Fi2+1 +· · · +Fir+1

and

(1 + Fi2+· · · +Fir)τ < 1 + Fi2+1 +· · · +Fir+1 +1 + τ2

1

τ+τ11 + τ2

1+τ4

1+· · ·

= 1 + Fi2+1 +· · · +Fir+1 + 1.

Combining these two estimates, we obtain (i).

(ii) and (iii) can be deduced in a similar way as (i).

(iv) follows easily from (i), (ii) and (iii). E.g., if n∈A3and n=

Fi1+· · · +Fir, then b(n) = Fi1+2 +· · · +Fir+2 −1 = F3+F5+· · · +

Fi1+1 +Fi2+2 +· · · +Fir+2 . So, we get e(b(n)) = F2+F4+· · · +Fi1+

Fi2+1 +· · · +Fir+1 =Fi1+1 −1 + Fi2+1 +· · · +Fir+1 =a(n).

(v) also follows from (i), (ii) and (iii). To see this, it is suﬃcient to

notice that b(n+ 1) τ1c=b(n+ 1) τc − (n+ 1) and then go through

the cases n+ 1 ∈A1∪A2,n+ 1 ∈A3and n+ 1 ∈A4.

(vi) If jis odd and j > 1, then e(Fj−1) = e(F2+F4+· · · +Fj−1) =

F1+F3+· · · +Fj−2=Fj−1. If jis even and j > 2, then e(Fj−1) =

e(F1+F3+· · · +Fj−1−1) = e(F3+· · · +Fj−1) = F2+· · · +Fj−2=

Fj−1−1.

Lemma 2.E. Let S(τ) = {bnτc:n∈N}and S(τ2) = {bnτ 2c:n∈N}.

Then

(i) S(τ)∪S(τ2) = Nand S(τ)∩S(τ2) = ∅;

(ii) S(τ2) = A4;

(iii) n−1∈S(τ)⇐⇒ bnτc=b(n−1) τc+ 2 (n > 1);

AN INTEGER SEQUENCE WITH A DIVISIBILITY PROPERTY 5

(iv) n−1∈S(τ2)⇐⇒ bnτc=b(n−1) τc+ 1 (n > 1).

Proof. (i) This follows immediately from Theorem 2.C since 1

τ+1

τ2= 1.

(ii) Let b(n) = bnτ2cand n=Fi1+Fi2+· · · +Firbe the Fibonacci

representation of n.

If n∈A1∪A2, then b(n) = bnτ2c=bnτ c+n= 1 + Fi2+1 +· · · +

Fir+1 +1+Fi2+· · · +Fir=F3+Fi2+2 +· · · +Fir+2 ∈A4.

If n∈A3, then b(n) = Fi1+2 +· · · +Fir+2 −1 = F3+F5+· · · +

Fi1+1 +Fi2+2 +· · · +Fir+2 ∈A4.

If n∈A4, then b(n) = Fi1+2 +· · · +Fir+2 ∈A4.

The above formulas imply the inclusion S(τ2)⊂A4.

Let us check the opposite inclusion.

Suppose n∈A4and n=Fi1+Fi2+· · · +Firbe the Fibonacci

representation of n. If i1>3, then Fi1−2+Fi2−2+· · · +Fir−2belongs

to A4and b(Fi1−2+Fi2−2+· · · +Fir−2)τ2c=Fi1+Fi2+· · · +Fir.

If i1= 3, then b(1 + Fi2−2+· · · +Fir−2)τ2c=F3+Fi2+· · · +Fir.

Thus A4⊂S(τ2).

(iii) The number of elements of S(τ) that are not larger than n−1

equals n

τ. Indeed, let kbe the largest positive integer such that

bkτ c ≤ n−1. Then bkτc ≤ n−1<b(k+ 1) τcwhich gives kτ < n <

(k+ 1) τ, i.e., k=n

τ. If n−1∈S(τ), then

n

τ= 1 + n−1

τ⇐⇒ bnτ1c= 1 + b(n−1) τ1c ⇐⇒ bnτc − n=

1 + b(n−1) τc − (n−1) ⇐⇒ bnτc=b(n−1) τc+ 2.

(iv) can be checked in an analogous way as (iii).

Remark 2.1.Lemma 2.A (i) and Lemma 2.A (ii) appear as a theorem in

[4]. Lemma 2.A (iii) is possibly new. A proof of Zeckendorf ’s theorem

can be found in [5] (see also [3]). Lemma 2.D (i), (ii) and (iii) are

ours. Carlitz studied the function ein [2], where Lemma 2.D (iv) can

be found. Lemma 2.D (v) is possibly new. Lemma 2.E (ii), (iii) and

(iv) are ours. Concerning the Beatty-Skolem theorem, we refer to [1, 4]

(see also [3]).

3. Proof of Theorem 1.1

Let Mn=Pn−1

n+1 for n∈N. One has z1= 1 and M1= 0; z2= 3

and M2= 1. We start by establishing the following proposition and its

corollary.

Proposition 3.1. For any n > 2, we have zn=Mn−1,Mn=Mn−1if

Mn−16=zk(k= 1, . . . , n −1) and zn=Mn−1+n+ 1,Mn=Mn−1+ 1

otherwise.

Proof. We shall proceed by induction. The claim is obviously true

for n= 3. Suppose that it holds for all 3 ≤k≤n−1. Then

6 MUHAREM AVDISPAHI ´

C AND FARUK ZEJNULAHI

{Mk}n−1

k=1 is nondecreasing. Indeed, from Pk−1=kMk−1+ 1 and

Pk= (k+ 1) Mk+ 1 we get zk= (k+ 1) Mk−kMk−1. Therefore,

Mk=Mk−1if zk=Mk−1, and Mk=Mk−1+ 1 if zk=Mk−1+k+ 1.

Hence, Mk≥Mk−1for k≤n−1.

The next observation we shall need is Mk≤Mk−1+ 1 ≤Mk−2+ 2 ≤

· · · ≤ M1+k−1 = k−1< k.

Note also that zk≤Mk−1+k+ 1 for k≤n−1.

Let us now derive the claim for zn. If Mn−1does not appear among

z1, . . . , zn−1, then we can take zn=Mn−1since Pn=Pn−1+Mn−1=

nMn−1+1+Mn−1≡1 (mod n+ 1) and Mn−1< n −1. Note Mn=

Mn−1< n −1 in this case.

If Mn−1is equal to zkfor some k≤n−1, then we can take zn=

Mn−1+n+ 1 because

max

k≤n−1zk≤max {M1+ 3, . . . , Mn−2+n}=Mn−2+n≤Mn−1+n<Mn−1+n+1,

and

Xn=Xn−1+Mn−1+n+1 = nMn−1+1+Mn−1+n+1 ≡1 (mod n+ 1) .

Then

Mn=Pn−1

n+ 1 =Mn−1+ 1 < n and zn=Mn+n.

Finally, let us also note that {Mk}∞

k=1 =N0.

Corollary 3.2. (i) The sequence {Mn}∞

n=1 is nondecreasing.

(ii) Mn< n for n∈N.

(iii) zn≤Mn−1+n+ 1 for n∈N.

(iv) {zn}∞

n=1 is a permutation of N.

Proof. Assertions (i), (ii) and (iii) follow easily from the proof of Propo-

sition 3.1. Let us have a look at (iv). The mapping n7→ znis obvi-

ously one-to-one since each zndiﬀers from all zl,l < n. Furthermore,

{zn}∞

n=1 =N. Indeed, for j∈Nthere exists k≥2 such that j=Mk.

Then either j=zlfor some l≤kor j=Mk=zk+1 by Proposition

3.1. Hence, n7→ znis a permutation of N.

We shall prove our theorem in the following apparently more general

form.

Theorem 3.3. (i) We have z1= 1 and M1= 0.

(ii) If n=Fk(k > 2), then zn=Fk+1 , Mn=Fk−1.

(iii) If n=bkτ 2c,n6=Fj(j > 2),n6=Fj−1 (j > 4), then zn=

bkτ c=Mn.

(iv) If n=bkτ c,n6=Fj(j > 2),n6=Fj−1 (j > 4), then zn=

bkτ 2c,Mn=k.

AN INTEGER SEQUENCE WITH A DIVISIBILITY PROPERTY 7

(v) If n=Fk−1 (k > 4), then zn=Fk−1−1 = Mn.

Proof. For n≤10, each of the cases (ii)-(v) appears at least once, as

one can see from the table below.

n1

F2

2

F3

3

F4

4

F5−1

5

F5

6

b4τc

7

F6−1

8

F6

9

b6τc

10

4τ2

11

b7τc

12

F7−1

13

F7

14

b9τc

15

6τ2

zn1 3 5 2 8 10 4 13 15 6 18 7 21 23 9

Mn0 1 2 2 3 4 4 5 6 6 7 7 8 9 9

Table 1. The ﬁrst 15 pairs (zn, Mn)

Thus, the assertion of theorem is valid for n≤10. Let us assume that

it holds for all 10 ≤m < n and prove it is true for n.

(ii) If n=Fk, then n−1 = Fk−1. So, by the induction hypothesis,

zn−1=Fk−1−1 = Mn−1. Then, Mn=Mn−1+1 = Fk−1,zn=Mn+n=

Fk−1+Fk=Fk+1 according to the Proposition 3.1.

(iii) If n=bkτ 2c,n6=Fj,n6=Fj−1, then since bkτ 2c=k+bkτ c,

Lemma 2.A (i) gives n−1 = k−1 + bkτc=bbkτcτc. Obviously,

n−16=Fj−1 since n6=Fj. If n−1 = bbkτ cτc=Fs, then Lemma

2.D (iv) would yield bkτc=e(bbkτcτc) = e(Fs) = Fs−1. By the same

reasoning bkτ c=Fs−1would imply k=Fs−2. Then we would have

n=bkτ 2c=k+bkτ c=Fs−2+Fs−1=Fswhich is not the case. Thus,

Mn−1=bkτ c,zn−1=bbkτ cτ2c. Let us check whether Mn−1appears

among z1, . . . , zn−2. (We already have Mn−16=zn−1.)

(1) bkτ ccannot take the form bjτ 2c;

(2) bkτ c 6=bjτ cfor j6=kby Lemma 2.A (iv);

(3) bkτ c=Fjwould yield k=e(bkτc) = e(Fj) = Fj−1by Lemma

2.D (iv). Then we would have n=k+bkτc=Fj−1+Fj=Fj+1

which contradicts the assumption n6=Fs;

(4) Suppose bkτ c=Fj−1. For even j, one has Fj−1 = Fj−1+

· · ·+F3∈A4=S(τ2). However, bkτc ∈ S(τ2) is impossible. If

jis odd, then k=e(bkτc) = e(Fj−1) = Fj−1by Lemma 2.D

(vi). This would yield n=k+bkτ c=Fj−1 + Fj−1=Fj+1 −1,

contrary to the assumption n6=Fl−1.

Hence, Mn−1/∈ {z1, . . . , zn−1}and zn=Mn−1=bkτ c=Mn.

(iv) If n=bkτ c,n6=Fj,n6=Fj−1, then obviously n−16=Fj−1.

We check the remaining three options:

(1) n−1 = Fj: Then bkτ c= 1 + Fjand k=e(bkτ c) = 1 + Fj−1.

The induction hypothesis gives us zn−1=Fj+1, Mn−1=Fj−1.

Now, Fj−1appears as zmfor m=Fj−2< n −1. Hence, Mn=

Mn−1+ 1 = Fj−1+ 1 = kand zn=Mn+n=k+bkτ c=bkτ2c;

8 MUHAREM AVDISPAHI ´

C AND FARUK ZEJNULAHI

(2) n−1 = bjτ 2cand n−16=Fi: In this case, zn−1=Mn−1=bjτ c.

Thus, Mn=Mn−1+ 1 = bjτ c+ 1 and zn=Mn+n. Then

bkτ c=bjτ 2c+ 1 = j+bjτ c+ 1 = b(bjτ c+ 1) τcby Lemma

2.A (ii). So, k=bjτ c+ 1. We end up with Mn=kand

zn=k+bkτ c=bkτ 2c;

(3) n−1 = bjτ cand n−16=Fi: Then, bkτ c+k−1 = bbkτcτc=

b(bjτ c+ 1) τc=bjτc+j+ 1 = bkτ c+j. Hence, k−1 = j.

The induction hypothesis tells us that zn−1=bjτ 2c,Mn−1=j.

Since bkτ c=b(k−1) τc+ 1, Lemma 2.E (iv) implies k−1 =

j∈S(τ2). Thus, j=blτ 2c=blτ c+l.

Here, blτ c=Fsis not possible since we would have l=e(blτ c) =

Fs−1. Then, Lemma 2.A (iii) would yield n−1 = b(blτ c+l)τc=

2blτ c+l= 2Fs+Fs−1=Fs+2 , in contradiction to n−16=Fi.

Neither blτ c=Fs−1. Namely, for sodd we would have

l=e(Fs−1) = Fs−1by Lemma 2.D (vi). Then n−1 =

2blτ c+l= 2Fs−2 + Fs−1=Fs+2 −2. This would imply

n=Fs+2 −1, which is not the case. Analogously, for seven

l=e(Fs−1) = Fs−1−1. Then n−1=2Fs−2 + Fs−1−1 =

Fs+2 −3 = F5+· · · +Fs+1 ∈A4=S(τ2) by Lemma 2.E (ii).

This is impossible because n−1∈S(τ).

Thus, the induction hypothesis yields zblτ c=blτ 2c=j=Mn−1.

Note that zblτ c∈ {z1, . . . , zn−1}since blτ c< j =k−1<

bkτ c − 1 = n−1. Therefore, Mn=Mn−1+ 1 = j+ 1 = kand

zn=Mn+n=k+bkτ c=bkτ 2c.

(v) n=Fk−1: Then n−1 = Fk−2. So, n−16=Fjand n−16=Fj−1

for k > 5. We consider the remaining two cases:

(1) n−1 = Fk−2 cannot have the form bjτ2c. Namely, for keven

we would have Fk−2 = F2+F5+· · · +Fk−1∈A1. Odd k

would lead us to Fk−2 = F4+F6+· · · +Fk−1∈A3. However,

(A1∪A3)∩S(τ2) = ∅;

(2) If n−1 = bjτc, then zn−1=bjτ 2c, Mn−1=j. In this case, j=

Fk−1−1. Indeed, we can see by induction that b(Fk−1−1) τc=

Fk−2 (= n−1 = bjτ c). Let us check that j=Fk−1−1/∈

{z1, . . . , zn−1}:

2.1. Fk−1−1 diﬀers from Fl−1 (for l6=k−1) and from Fl;

2.2. Suppose Fk−1−1 = blτ 2c=zblτ c, where blτc 6=Fs, Fs−

1. The case k−1 being even would yield blτc=e(blτ 2c) =

e(Fk−1−1) = Fk−2−1, and this contradicts the assumption

blτ c 6=Fs−1. The case k−1 being odd would imply blτc=

e(Fk−1−1) = Fk−2, contradicting blτc 6=Fs;

AN INTEGER SEQUENCE WITH A DIVISIBILITY PROPERTY 9

2.3. Finally, suppose Fk−1−1 = blτc=zblτ 2c, where blτ2c 6=

Fs, Fs−1. If k−1 is even, then Fk−1−1∈S(τ2), which in

our case is impossible. For k−1 odd, we get l=e(blτc) =

e(Fk−1−1) = Fk−2. Then blτ 2c=l+blτ c=Fk−2+Fk−1−1 =

Fk−1. This contradicts the assumption blτ 2c 6=Fs−1.

Taking into account that Mn−1=j=Fk−1−1/∈ {z1, . . . , zn−1}, we

get Mn=Mn−1=Fk−1−1 and zn=Fk−1−1.

This completes the proof of the theorem.

Remark 3.1.The sequence {zn}∞

n=1 is initially considered in [6].

Remark 3.2.If we are interested in listing, say, the ﬁrst 100000 mem-

bers of the sequence {zn}∞

n=1, we can easily obtain these by program-

ming our Proposition 3.1 in Phyton:

z list = [-1,1,3]

m list = [-1,0,1]

n = 2

for n in range(2,100000):

if m list[n] in z list:

m list.append(m list[n]+1)

z list.append(m list[n+1]+n+1)

else:

m list.append(m list[n])

z list.append(m list[n+1])

with open(’results,txt’, ’w’) as f:

f.write(’index, M, Z\n’)

for i in range(1,100000):

f.write(str(i) + ’,’+ str(m list[i]) +’,’+str(z list[i])+’\n’)

The ﬁrst one hundred members of {zn}∞

n=1 read: 1, 3, 5, 2, 8, 10, 4, 13,

15, 6, 18, 7, 21, 23, 9, 26, 28, 11, 31, 12, 34, 36, 14, 39, 41, 16, 44, 17,

47, 49, 19, 52, 20, 55, 57, 22, 60, 62, 24, 65, 25, 68, 70, 27, 73, 75, 29,

78, 30, 81, 83, 32, 86, 33, 89, 91, 35, 94, 96, 37, 99, 38, 102, 104, 40,

107, 109, 42, 112, 43, 115, 117, 45, 120, 46, 123, 125, 48, 128, 130, 50,

133, 51, 136, 138, 53, 141, 54, 144, 146, 56, 149, 151, 58, 154, 59, 157,

159, 61, 162.

References

[1] S. Beatty, A. Ostrowski, J. Hyslop, and A. C. Aitken, Problems and Solutions:

Solutions: 3177, Amer. Math. Monthly, 34.3 (1927), 159–160.

[2] L. Carlitz, Fibonacci representations, The Fibonacci Quarterly, 6.4 (1968),

193–220.

10 MUHAREM AVDISPAHI ´

C AND FARUK ZEJNULAHI

[3] R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete mathematics. A

foundation for computer science, Second Edition, Addison-Wesley, Reading,

MA, 1994.

[4] Th. Skolem, On certain distributions of integers in pairs with given diﬀerences,

Math. Scand., 5(1957), 57–68.

[5] E. Zeckendorf, Repr´esentation des nombres naturels par une somme de nombres

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[6] F. Zejnulahi, Doctoral Dissertation, University of Sarajevo, 2019.

University of Sarajevo, Department of Mathematics, Zmaja od Bosne

33-35, 71000 Sarajevo, Bosnia and Herzegovina

E-mail address, M. Avdispahi´c: mavdispa@pmf.unsa.ba

E-mail address, F. Zejnulahi: faruk.zejnulahi@gmail.com