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# AN INTEGER SEQUENCE WITH A DIVISIBILITY PROPERTY

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## Abstract

Modifying the divisibility condition, we introduce a new integer sequence and establish its representation in terms of Fibonacci numbers and the golden mean.
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
ai0 (mod n) for all nN.
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, . . . , zn1such that
Xn=
n
X
i=1
zi1 (mod n+ 1) for all nN.
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);
Fk11for n=Fk1(k > 4);
b cfor n=b 2c,n6=Fk(k > 2) and n6=Fk1(k > 4);
b 2cfor n=b c,n6=Fk(k > 2) and n6=Fk1(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=51
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 kN, the following identities are valid:
(i) bb cτc=b c+k1;
(ii) b(b c+ 1) τc=bkτc+k+ 1;
(iii) b(b c+k)τc= 2 bc+k;
(iv) b c=b cfor jNif 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 bc+ 1
⇒ bc<(b c+k) (τ1) <b c+ 1
⇒ bc<(b c+k)τ1<b c+ 1
⇒ bcτ < b c+k < (bkτc+ 1) τ
⇒ bc(τ1) < k < (bkτc+ 1) (τ1) + 1
⇒ bcτ1< k < (bkτc+ 1) τ1+ 1
⇒ bc< kτ < bkτ c+1+τ.
For (iv) suppose n=b c=bjτ c. Adding up relations nj τ < n + 1
and n1<≤ −n, we get 1<(jk)τ < 1. Multiplication
by τ1gives τ1< j k < τ1. This yields jk= 0 since 0 < τ1<1.
For a further reference, we adopt the convention
xyx+ 2 y.
Theorem 2.B. (Zeckendorf) Every positive integer nhas the unique
representation in the form
(∆) n=Fi1+Fi2+· · · +Fir
where i1i2≺ ·· · ≺ ir, and i12.
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:
nA1i1= 2 and i2is odd or n= 1;
nA2i1= 2 and i2is even;
nA3i1>2 and i1is even;
nA4i1>2 and i1is odd.
We also make use of the function e:NNdeﬁned by
e(n) = Fi11+Fi21+· · · +Fir1,
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={bc:nN}
and Sβ={bc:nN}form a disjoint decomposition of the set of
positive integers, i.e., SαSβ=Nand SαSβ=.
The next lemma provides some useful information about the sets A1,
A2,A3and A4and the function e.
Lemma 2.D. (i) If nA1A2and n= 1 + Fi2+· · · +Fir, then
b(1 + Fi2+· · · +Fir)τc= 1 + Fi2+1 +· · · +Fir+1 .
(ii) If nA3and n=Fi1+Fi2+· · ·+Fir, then b(Fi1+· · · +Fir)τc=
Fi1+1 +· · · +Fir+1 1.
(iii) If nA4and n=Fi1+Fi2+· · ·+Fir, then b(Fi1+· · · +Fir)τc=
Fi1+1 +· · · +Fir+1 .
(iv) If a(n) = bcand b(n) = b 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(Fj1) = Fj1and if jis even
and j > 2, then e(Fj1) = Fj11.
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)i21
τ+τ1
+· · · +τir+1 (τ1)ir1
τ+τ1
=τ2τ2
1
τ+τ1
+τi2+1 + (τ1)i2+1
τ+τ1
+· · · +τir+1 + (τ1)ir+1
τ+τ1
+1 + τ2
1(τ1)i2+1 (τ1)i21− · · · (τ1)ir+1 (τ1)ir1
τ+τ1
= 1 + Fi2+1 +· · · +Fir+1 +1 + τ2
1
τ+τ11(τ1)i21− · · · (τ1)ir1.
The above equality implies
(1 + Fi2+· · · +Fir)τ > 1 + Fi2+1 +· · · +Fir+1 +1 + τ2
1
τ+τ11τ2
1τ4
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 nA3and 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 A1A2,n+ 1 A3and n+ 1 A4.
(vi) If jis odd and j > 1, then e(Fj1) = e(F2+F4+· · · +Fj1) =
F1+F3+· · · +Fj2=Fj1. If jis even and j > 2, then e(Fj1) =
e(F1+F3+· · · +Fj11) = e(F3+· · · +Fj1) = F2+· · · +Fj2=
Fj11.
Lemma 2.E. Let S(τ) = {bc:nN}and S(τ2) = {b 2c:nN}.
Then
(i) S(τ)S(τ2) = Nand S(τ)S(τ2) = ;
(ii) S(τ2) = A4;
(iii) n1S(τ)⇒ bc=b(n1) τc+ 2 (n > 1);
AN INTEGER SEQUENCE WITH A DIVISIBILITY PROPERTY 5
(iv) n1S(τ2)⇒ bc=b(n1) τc+ 1 (n > 1).
Proof. (i) This follows immediately from Theorem 2.C since 1
τ+1
τ2= 1.
(ii) Let b(n) = b2cand n=Fi1+Fi2+· · · +Firbe the Fibonacci
representation of n.
If nA1A2, then b(n) = b2c=b c+n= 1 + Fi2+1 +· · · +
Fir+1 +1+Fi2+· · · +Fir=F3+Fi2+2 +· · · +Fir+2 A4.
If nA3, then b(n) = Fi1+2 +· · · +Fir+2 1 = F3+F5+· · · +
Fi1+1 +Fi2+2 +· · · +Fir+2 A4.
If nA4, then b(n) = Fi1+2 +· · · +Fir+2 A4.
The above formulas imply the inclusion S(τ2)A4.
Let us check the opposite inclusion.
Suppose nA4and n=Fi1+Fi2+· · · +Firbe the Fibonacci
representation of n. If i1>3, then Fi12+Fi22+· · · +Fir2belongs
to A4and b(Fi12+Fi22+· · · +Fir2)τ2c=Fi1+Fi2+· · · +Fir.
If i1= 3, then b(1 + Fi22+· · · +Fir2)τ2c=F3+Fi2+· · · +Fir.
Thus A4S(τ2).
(iii) The number of elements of S(τ) that are not larger than n1
equals n
τ. Indeed, let kbe the largest positive integer such that
b c ≤ n1. Then bkτc ≤ n1<b(k+ 1) τcwhich gives < n <
(k+ 1) τ, i.e., k=n
τ. If n1S(τ), then
n
τ= 1 + n1
τ⇒ b1c= 1 + b(n1) τ1c ⇐⇒ bc − n=
1 + b(n1) τc − (n1) ⇒ bc=b(n1) 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
. Lemma 2.A (iii) is possibly new. A proof of Zeckendorf ’s theorem
can be found in  (see also ). Lemma 2.D (i), (ii) and (iii) are
ours. Carlitz studied the function ein , 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]
3. Proof of Theorem 1.1
Let Mn=Pn1
n+1 for nN. 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=Mn1,Mn=Mn1if
Mn16=zk(k= 1, . . . , n 1) and zn=Mn1+n+ 1,Mn=Mn1+ 1
otherwise.
Proof. We shall proceed by induction. The claim is obviously true
for n= 3. Suppose that it holds for all 3 kn1. Then
6 MUHAREM AVDISPAHI ´
C AND FARUK ZEJNULAHI
{Mk}n1
k=1 is nondecreasing. Indeed, from Pk1=kMk1+ 1 and
Pk= (k+ 1) Mk+ 1 we get zk= (k+ 1) MkkMk1. Therefore,
Mk=Mk1if zk=Mk1, and Mk=Mk1+ 1 if zk=Mk1+k+ 1.
Hence, MkMk1for kn1.
The next observation we shall need is MkMk1+ 1 Mk2+ 2
· · · M1+k1 = k1< k.
Note also that zkMk1+k+ 1 for kn1.
Let us now derive the claim for zn. If Mn1does not appear among
z1, . . . , zn1, then we can take zn=Mn1since Pn=Pn1+Mn1=
nMn1+1+Mn11 (mod n+ 1) and Mn1< n 1. Note Mn=
Mn1< n 1 in this case.
If Mn1is equal to zkfor some kn1, then we can take zn=
Mn1+n+ 1 because
max
kn1zkmax {M1+ 3, . . . , Mn2+n}=Mn2+nMn1+n<Mn1+n+1,
and
Xn=Xn1+Mn1+n+1 = nMn1+1+Mn1+n+1 1 (mod n+ 1) .
Then
Mn=Pn1
n+ 1 =Mn1+ 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 nN.
(iii) znMn1+n+ 1 for nN.
(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 jNthere exists k2 such that j=Mk.
Then either j=zlfor some lkor 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=Fk1.
(iii) If n=b 2c,n6=Fj(j > 2),n6=Fj1 (j > 4), then zn=
b c=Mn.
(iv) If n=b c,n6=Fj(j > 2),n6=Fj1 (j > 4), then zn=
b 2c,Mn=k.
AN INTEGER SEQUENCE WITH A DIVISIBILITY PROPERTY 7
(v) If n=Fk1 (k > 4), then zn=Fk11 = Mn.
Proof. For n10, 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
F51
5
F5
6
b4τc
7
F61
8
F6
9
b6τc
10
4τ2
11
b7τc
12
F71
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 n10. Let us assume that
it holds for all 10 m < n and prove it is true for n.
(ii) If n=Fk, then n1 = Fk1. So, by the induction hypothesis,
zn1=Fk11 = Mn1. Then, Mn=Mn1+1 = Fk1,zn=Mn+n=
Fk1+Fk=Fk+1 according to the Proposition 3.1.
(iii) If n=b 2c,n6=Fj,n6=Fj1, then since b 2c=k+b c,
Lemma 2.A (i) gives n1 = k1 + bc=bbkτcτc. Obviously,
n16=Fj1 since n6=Fj. If n1 = bb cτc=Fs, then Lemma
2.D (iv) would yield bc=e(bbkτcτc) = e(Fs) = Fs1. By the same
reasoning b c=Fs1would imply k=Fs2. Then we would have
n=b 2c=k+b c=Fs2+Fs1=Fswhich is not the case. Thus,
Mn1=b c,zn1=bb cτ2c. Let us check whether Mn1appears
among z1, . . . , zn2. (We already have Mn16=zn1.)
(1) b ccannot take the form b 2c;
(2) b c 6=b cfor j6=kby Lemma 2.A (iv);
(3) b c=Fjwould yield k=e(bc) = e(Fj) = Fj1by Lemma
2.D (iv). Then we would have n=k+bkτc=Fj1+Fj=Fj+1
(4) Suppose b c=Fj1. For even j, one has Fj1 = Fj1+
· · ·+F3A4=S(τ2). However, bc ∈ S(τ2) is impossible. If
jis odd, then k=e(bkτc) = e(Fj1) = Fj1by Lemma 2.D
(vi). This would yield n=k+b c=Fj1 + Fj1=Fj+1 1,
contrary to the assumption n6=Fl1.
Hence, Mn1/∈ {z1, . . . , zn1}and zn=Mn1=b c=Mn.
(iv) If n=b c,n6=Fj,n6=Fj1, then obviously n16=Fj1.
We check the remaining three options:
(1) n1 = Fj: Then b c= 1 + Fjand k=e(b c) = 1 + Fj1.
The induction hypothesis gives us zn1=Fj+1, Mn1=Fj1.
Now, Fj1appears as zmfor m=Fj2< n 1. Hence, Mn=
Mn1+ 1 = Fj1+ 1 = kand zn=Mn+n=k+b c=bkτ2c;
8 MUHAREM AVDISPAHI ´
C AND FARUK ZEJNULAHI
(2) n1 = b 2cand n16=Fi: In this case, zn1=Mn1=bjτ c.
Thus, Mn=Mn1+ 1 = bjτ c+ 1 and zn=Mn+n. Then
b c=b 2c+ 1 = j+b c+ 1 = b(b c+ 1) τcby Lemma
2.A (ii). So, k=bjτ c+ 1. We end up with Mn=kand
zn=k+b c=b 2c;
(3) n1 = b cand n16=Fi: Then, bkτ c+k1 = bbkτcτc=
b(b c+ 1) τc=bjτc+j+ 1 = b c+j. Hence, k1 = j.
The induction hypothesis tells us that zn1=b 2c,Mn1=j.
Since b c=b(k1) τc+ 1, Lemma 2.E (iv) implies k1 =
jS(τ2). Thus, j=b 2c=b c+l.
Here, b c=Fsis not possible since we would have l=e(b c) =
Fs1. Then, Lemma 2.A (iii) would yield n1 = b(b c+l)τc=
2b c+l= 2Fs+Fs1=Fs+2 , in contradiction to n16=Fi.
Neither b c=Fs1. Namely, for sodd we would have
l=e(Fs1) = Fs1by Lemma 2.D (vi). Then n1 =
2b c+l= 2Fs2 + Fs1=Fs+2 2. This would imply
n=Fs+2 1, which is not the case. Analogously, for seven
l=e(Fs1) = Fs11. Then n1=2Fs2 + Fs11 =
Fs+2 3 = F5+· · · +Fs+1 A4=S(τ2) by Lemma 2.E (ii).
This is impossible because n1S(τ).
Thus, the induction hypothesis yields zbc=b 2c=j=Mn1.
Note that zbc∈ {z1, . . . , zn1}since b c< j =k1<
b c − 1 = n1. Therefore, Mn=Mn1+ 1 = j+ 1 = kand
zn=Mn+n=k+b c=b 2c.
(v) n=Fk1: Then n1 = Fk2. So, n16=Fjand n16=Fj1
for k > 5. We consider the remaining two cases:
(1) n1 = Fk2 cannot have the form b2c. Namely, for keven
we would have Fk2 = F2+F5+· · · +Fk1A1. Odd k
would lead us to Fk2 = F4+F6+· · · +Fk1A3. However,
(A1A3)S(τ2) = ;
(2) If n1 = bjτc, then zn1=bjτ 2c, Mn1=j. In this case, j=
Fk11. Indeed, we can see by induction that b(Fk11) τc=
Fk2 (= n1 = bjτ c). Let us check that j=Fk11/
{z1, . . . , zn1}:
2.1. Fk11 diﬀers from Fl1 (for l6=k1) and from Fl;
2.2. Suppose Fk11 = b 2c=zbc, where blτc 6=Fs, Fs
1. The case k1 being even would yield blτc=e(b 2c) =
e(Fk11) = Fk21, and this contradicts the assumption
b c 6=Fs1. The case k1 being odd would imply bc=
e(Fk11) = Fk2, contradicting bc 6=Fs;
AN INTEGER SEQUENCE WITH A DIVISIBILITY PROPERTY 9
2.3. Finally, suppose Fk11 = bc=zb2c, where blτ2c 6=
Fs, Fs1. If k1 is even, then Fk11S(τ2), which in
our case is impossible. For k1 odd, we get l=e(bc) =
e(Fk11) = Fk2. Then b 2c=l+b c=Fk2+Fk11 =
Fk1. This contradicts the assumption b 2c 6=Fs1.
Taking into account that Mn1=j=Fk11/∈ {z1, . . . , zn1}, we
get Mn=Mn1=Fk11 and zn=Fk11.
This completes the proof of the theorem.
Remark 3.1.The sequence {zn}
n=1 is initially considered in .
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
 S. Beatty, A. Ostrowski, J. Hyslop, and A. C. Aitken, Problems and Solutions:
Solutions: 3177, Amer. Math. Monthly, 34.3 (1927), 159–160.
 L. Carlitz, Fibonacci representations, The Fibonacci Quarterly, 6.4 (1968),
193–220.
10 MUHAREM AVDISPAHI ´
C AND FARUK ZEJNULAHI
 R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete mathematics. A