AN INTEGER SEQUENCE WITH A DIVISIBILITY PROPERTY

Abstract

Modifying the divisibility condition, we introduce a new integer sequence and establish its representation in terms of Fibonacci numbers and the golden mean.

Full text

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+τ11−(−τ1)i2−1− · · · − (−τ1)ir−1.
The above equality implies
(1 + Fi2+· · · +Fir)τ > 1 + Fi2+1 +· · · +Fir+1 +1 + τ2
1
τ+τ11−τ2
1−τ4
1− · · · 
>1 + Fi2+1 +· · · +Fir+1
and
(1 + Fi2+· · · +Fir)τ < 1 + Fi2+1 +· · · +Fir+1 +1 + τ2
1
τ+τ11 + 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
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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,
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[4] Th. Skolem, On certain distributions of integers in pairs with given differences,
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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