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Extending the Golden Ratio and
the Binet-de Moivr´e Formula
Felix P. Muga II
Mathematics Department
School of Science and Engineering
Ateneo de Manila University
Quezon City, Philippines
fmuga@ateneo.edu
Abstract
Whitford generalized the Fibonacci numbers using the Binet’s formula of the second
order recurrence an=an−1+tan−2where t∈Z+. Horadam presented a higher form
of generalization of the Binet’s formula with the recurrence an=kan−1+tan−2where
k, t ∈Z. In this paper, we shall discuss and present some results on recurrence of the
form an=kan−1+tan−2where k, t ∈Z+.
1 Introduction
The golden ratio denoted by φis defined to be the ratio of two positive integers aand b,
where a>b, such that
φ=a
b=a+b
a=1 + √5
2≈1.6180339887498949 . . .
Let φk,t, where k, t ∈Z+, be a generalization of the golden ratio such that
φk,t =a
b=ka +tb
a.
Thus,
φk,t =k+t
φk,t
φ2
k,t −kφk,t −t= 0
1
The roots of this quadratic equation are:
φk,t =k+√k2+ 4t
2and φk,t =k−√k2+ 4t
2.
Clearly φ1,1=φ.
Whitford [3] introduced a generalization of this ratio with k= 1 and extended the Binet’s
formula for the recurrence an=an−1+tan−2for n > 1, with a0= 0, a1= 1, and t∈Z+
such that
an=φn
1,t −φn
1,t
√1+4t.
Horadam [1] presented a higher form of generalization of the Binet’s formula with the
recurrence an=kan−1+tan−2where k, t ∈Z.
In this paper we shall examine the case where kand tare positive integers with
an=φn
k,t −φn
k,t
√k2+ 4t.
We shall determine the values of kand tfor φk,t to be an integer multiple of φand for φk,t
to have an integer value.
2 Extending the Golden Ratio
Suppose that φk,t =mφ where mis a positive integer. Then
k+√k2+ 4t
2=m+m√5
2
k+√k2+ 4t=m+m√5
k−m+√k2+ 4t−m√5 = 0 (1)
If the root √k2+ 4tis an integer, then left hand side of Equation 1cannot be zero since
m > 0.
Thus, we have (k−m) + √k2+ 4t−m√5= 0 where the first and the second terms of
the left hand side are respectively equal to zero.
Consequently, k=mand
√k2+ 4t−m√5 = 0
√k2+ 4t=m√5
k2+ 4t= 5m2
4t= 4m2
t=m2.
2
Note that these are the only integer values of kand tfor which φk,t is an integer multiple
of φ.
Therefore, we have the following theorem.
Theorem 1. Let m, k, t ∈Z+such that k=mand t=m2. Then φk,t is an integer multiple
of the golden ratio, i.e.,φm,m2=mφ. This is the only φk,t configuration to obtain an integer
multiple of the golden ratio.
This theorem implies that the golden ratio is obtained only from φk,t where k=t= 1.
For the integer sequence A085449 published electronically in The On-Line Encyclopedia
of Integer Sequences [4], [5] at http://oeis.org, the ratio is φ2,4= 2φ. For the integer
sequence A099012, the ratio is φ3,9= 3φand for the integer sequence A099133, the ratio is
φ4,16 = 4φ.
3 The Ratio φk,t with Integer Values
In this section, we shall determine the integer values of kand tso that the ratio φk,t is also
an integer value.
We have the following theorems.
Theorem 2. Let m, k, t ∈Z+. Then k∈Z?
mand t=m2−mk if and only if φk,t =m.
Proof. 1. Suppose that k∈Z?
mand t=m2−mk. Then k < m and
k+√k2+ 4t
2=k+√4m2−4mk +k2
2
=2m
2=m
2. Suppose that φk,t =mwhere mis a positive integer. Then
k+√k2+ 4t
2=m⇐⇒ √k2+ 4t= 2m−k
k2+ 4t= 4m2−4mk +k2
t=m2−mk.
Since k, t ∈Z+, it follows that k= 1,2, . . . , m −1 and m > k.
Note that these are the only integer values for kand tfor which φk,t is a positive integer.
Table 1enumerates the only integer sequences for m= 2,3,4,5,6,7.
Note that the sequence of Jacobsthal numbers [2], [6] is the only integer sequence from
the second order linear recurrence of the form an=kan−1+tan−2, with a0= 0, a1= 1 and
k, t ∈Z+where the ratio is equal to 2.
3
Table 1: The Only Integer Sequences Derived from the Second Order Homogeneous Recur-
rences an=kan−1+tan−2with a0= 0, a1= 1, and k, t ∈Z+Having φk,t with Integer Values
m.mInteger Sequence k t
2A001045 1 2
3A015441 1 6
3A015518 2 3
4A053404 1 12
4A003683 2 8
4A015521 3 4
5A053428 1 20
5A079773 2 15
5A015528 3 10
5A015531 4 5
6A053430 1 30
6A051958 2 24
6A080424 3 18
6A053524 4 12
6A015540 5 6
7 1 42
7 2 35
7 3 28
7 4 21
7A053573 5 14
7A015552 6 7
4
Theorem 3. The values m(m+k)where mand kare positive integers are the only integer
values of tfor which the Binet-de Moivr´e formula on the recurrence bn=kbn−1+tbn−2for
n > 1with b0= 0 and b1= 1 has a root which is a square. In particular, √k2+ 4t= 2m+k
which is a positive integer.
Thus, the Binet-de Moivr´e formula is given by bn=1
2m+k(m+k)n−(−m)n.
Proof. The Binet-de Moivr´e formula of the recurrence bn=kbn−1+tbn−2for n > 1 with
b0= 0, b1= 1 and k, t ∈Z+is given by
bn=1
√k2+ 4t k+√k2+ 4t
2!n
− k−√k2+ 4t
2!n!.
Thus, if t=m(m+k), then √k2+ 4t=p(2m+k)2= 2m+k.
Hence, the characteristic roots are r1=m+kand r2=−m.
Therefore, the Binet-de Moivr´e formula is given by bn=1
2m+k(m+k)n−(−m)n.
Table 2presents 31 integer sequences published in the Online Encyclopedia of Integer
Sequences where their respective terms amsatisfy t=am=m(m+k) for positive integers
kand m.
Example 1. Note that the second and the third terms of the integer sequence A002378 are
2 and 6 respectively, i.e., A002378(1) = 2 and A002378(2) = 6. Thus, the corresponding
integer sequences are A001045 with Binet-de Moivr´e formula bn= (2n−(−1)n)/3 and
A015441 with bn= (3n−(−1)n)/4.
Also, A005563(1) = 3 and A005564(2) = 8. Thus, the corresponding integer sequences
are A015518 with bn= (3n−(−2)n)/5 and A003683 with bn= (4n−(−2)n)/6.
4 On the Convergence of an+1
an
to φk,t
For positive integers kand t, we consider the recurrence relation an=kan−1+tan−2for
n= 2,3, . . . such that a0= 0 and a1= 1.
The characteristic equation of this recurrence relation is equal to r2−kr −t= 0 where
the roots are
φk,t =k+√k2+ 4t
2and φk,t =k−√k2+ 4t
2
Hence it can be shown that the solution to the recurrence relation is a sequence ann≥0
where an=1
√k2+ 4tφn
k,t −φn
k,t.
Theorem 4. Let ann≥0be a sequence with an=1
√k2+ 4tφn
k,t −φn
k,t.
Then lim
n→∞
an+1
an
=φk,t.
5
Table 2: Binet-de Moivr´e Formula of the Integer Sequences from bn=kbn−1+tbn−2for n > 1
with b0= 0, b1= 1 and t, k ∈Z+where t=m(m+k) for m∈Z+and k= 1,2,...,31.
Integer Sequence amBinet-de Moivr´e Formula of
ksuch that t=ambn=kbn−1+tbn−2
1A002378 (m+ 1)n−(−m)n/(2m+ 1)
2A005563 (m+ 2)n−(−m)n/(2m+ 2)
3A028552 (m+ 3)n−(−m)n/(2m+ 3)
4A028347 (m+ 4)n−(−m)n/(2m+ 4)
5A028557 (m+ 5)n−(−m)n/(2m+ 5)
6A028560 (m+ 6)n−(−m)n/(2m+ 6)
7A028563 (m+ 7)n−(−m)n/(2m+ 7)
8A028566 (m+ 8)n−(−m)n/(2m+ 8)
9A028569 (m+ 9)n−(−m)n/(2m+ 9)
10 A098603 (m+ 10)n−(−m)n/(2m+ 10)
11 A119412 (m+ 11)n−(−m)n/(2m+ 11)
12 A098847 (m+ 12)n−(−m)n/(2m+ 12)
13 A132759 (m+ 13)n−(−m)n/(2m+ 13)
14 A098848 (m+ 14)n−(−m)n/(2m+ 14)
15 A132760 (m+ 15)n−(−m)n/(2m+ 15)
16 A098849 (m+ 16)n−(−m)n/(2m+ 16)
17 A132761 (m+ 17)n−(−m)n/(2m+ 17)
18 A098850 (m+ 18)n−(−m)n/(2m+ 18)
19 A132762 (m+ 19)n−(−m)n/(2m+ 19)
20 A120071 (m+ 20)n−(−m)n/(2m+ 20)
21 A132763 (m+ 21)n−(−m)n/(2m+ 21)
22 A132764 (m+ 22)n−(−m)n/(2m+ 22)
23 A132765 (m+ 23)n−(−m)n/(2m+ 23)
24 A132766 (m+ 24)n−(−m)n/(2m+ 24)
25 A132767 (m+ 25)n−(−m)n/(2m+ 25)
26 A132768 (m+ 26)n−(−m)n/(2m+ 26)
27 A132769 (m+ 27)n−(−m)n/(2m+ 27)
28 A132770 (m+ 28)n−(−m)n/(2m+ 28)
29 A132771 (m+ 29)n−(−m)n/(2m+ 29)
30 A132772 (m+ 30)n−(−m)n/(2m+ 30)
31 A132773 (m+ 31)n−(−m)n/(2m+ 31)
6
Proof. Since k > 0 and t > 0, we have k+√k2+ 4t > 0 and k2< k2+ 4t. Thus,
k < √k2+ 4t
k−√k2+ 4t < 0
k−√k2+ 4t
k+√k2+ 4t<0
φk,t
φk,t
<0.
−k−√k2+ 4t < k −√k2+ 4t < k +√k2+ 4t
−1<k−√k2+ 4t
k+√k2+ 4t<1
−1<φk,t
φk,t
<1
φk,t
φk,t
<1
Thus, lim
n→∞
φk,t
φk,t
n
= 0. This implies that lim
n→∞ φk,t
φk,t n
= 0.
Since
an+1
an
=φn+1
k,t −φn+1
k,t
φn
k,t −φn
k,t
=φk,t +φn
k,t φk,t −φk,t
φn
k,t −φn
k,t
an+1
an
=φk,t +φk,t
φk,t n
φk,t −φk,t
1−φk,t
φk,t n
Therefore, lim
n→∞
an+1
an
=φk,t.
The following corollaries follow immediately from the previous theorem.
Corollary 1. The ratio between two consecutive terms of the sequence of natural numbers
0,1, m, 2m2,3m3, . . . , an, . . .
where an=man−1+m2an−2for n= 2,3, . . . converges to mφ.
7
Corollary 2. Let m, k, t ∈Z+. The ratio between two consecutive terms of the sequence of
natural numbers
0,1, k, k2+t, k3+ 2t, . . . , an, . . .
where an=kan−1+tan−2,k= 1,2, . . . , m −1and t=m2−mk for n= 2,3, . . . converges
to m.
Hence, we have the following results.
Theorem 5. If n > −rln 10 −ln √k2+ 4t−10−r
ln k2+ 2t−k√k2+ 4t−ln 2 −ln t, then
an+1
an−φk,t≤10−rwhere
r∈Z+.
Proof. For a given positive integer rwe find the integer value of nsuch that
an+1
an−φk,t≤10−r.
an+1
an
=φn+1
k,t −φn+1
k,t
φn
k,t −φn
k,t
=φk,t +φn
k,t φk,t −φk,t
φn
k,t −φn
k,t
an+1
an
=φk,t +φk,t
φk,t n
φk,t −φk,t
1−φk,t
φk,t n
an+1
an−φk,t =φk,t
φk,t n
φk,t −φk,t
1−φk,t
φk,t n
φk,t
φk,t
=k−√k2+ 4t
k+√k2+ 4t
=k−√k2+ 4t
k+√k2+ 4t·k−√k2+ 4t
k−√k2+ 4t
=2k2+ 4t−2k√k2+ 4t
−4t
φk,t
φk,t
=k2+ 2t−k√k2+ 4t
−2t
8
Since φk,t −φk,t =√k2+ 4t, we have
an+1
an−φk,t = k2+ 2t−k√k2+ 4t
−2t!n√k2+ 4t
1− k2+ 2t−k√k2+ 4t
−2t!n
an+1
an−φk,t
= k2+ 2t−k√k2+ 4t
2t!n√k2+ 4t
1− k2+ 2t−k√k2+ 4t
−2t!n
<10−r
Thus,
k2+ 2t−k√k2+ 4t
2t!n√k2+ 4t < 10−r
1− k2+ 2t−k√k2+ 4t
−2t!n
<10−r 1 + k2+ 2t−k√k2+ 4t
−2t!n!
k2+ 2t−k√k2+ 4t
2t!n√k2+ 4t < 10−r 1 +
k2+ 2t−k√k2+ 4t
−2t
n!
Since kand tare positive integers, we have k4+ 4k2t+ 4t2> k4+ 4k2t > 0. Thus,
k2+ 2t>k√k2+ 4t
k2+ 2t−k√k2+ 4t > 0
k2+ 2t−k√k2+ 4t
−2t<0
Hence,
k2+ 2t−k√k2+ 4t
2t!n√k2+ 4t < 10−r+ 10−r k2+ 2t−k√k2+ 4t
2t!n
k2+ 2t−k√k2+ 4t
2t!n
√k2+ 4t−10−r<10−r
9
Since k, t and rare positive integers, we have √k2+ 4t−10−r>0. Thus,
k2+ 2t−k√k2+ 4t
2t!n
√k2+ 4t−10−r<10−r
k2+ 2t−k√k2+ 4t
2t!n
<10−r
√k2+ 4t−10−r
The logarithmic function is an increasing function. Thus, we have
ln k2+ 2t−k√k2+ 4t
2t!n
<ln 10−r
√k2+ 4t−10−r
nln k2+ 2t−k√k2+ 4t
2t!<−rln 10 −ln √k2+ 4t−10−r
Since kand tare positive integers and the square root function is increasing in its domain,
we have
k2< k2+ 4t
k < √k2+ 4t.
Thus,
k2< k√k2+ 4t
k2−k√k2+ 4t < 0
k2+ 2t−k√k2+ 4t < 2t
k2+ 2t−k√k2+ 4t
2t<1
Hence, 0 <k2+ 2t−k√k2+ 4t
2t<1. This implies that ln k2+ 2t−k√k2+ 4t
2t!<0.
10
This means that
nln k2+ 2t−k√k2+ 4t
2t!<−rln 10 −ln √k2+ 4t−10−r
n > −rln 10 −ln √k2+ 4t−10−r
ln k2+ 2t−k√k2+ 4t
2t!
n > −rln 10 −ln √k2+ 4t−10−r
ln k2+ 2t−k√k2+ 4t−ln(2t)
n > −rln 10 −ln √k2+ 4t−10−r
ln k2+ 2t−k√k2+ 4t−ln 2 −ln t
The following corollaries follow immediately.
Corollary 3. In the infinite sequence of Fibonacci numbers (Fn)n≥0,
if n > −rln 10 −ln √5−10−r
ln 3−√5−ln 2 then
Fn+1
Fn−φ≤10−rwhere r∈Z+.
Corollary 4. If k=m, t =m2where m∈Z+and
if n > −rln 10 −ln m√5−10−r
ln 3−√5−ln 2 , then
an+1
an−mφ≤10−rwhere r∈Z+.
Corollary 5. If k= 1,2, . . . , m −1, t =m2−km where k, t, m ∈Z+and
if n > −rln 10 −ln (2m−k−10−r)
ln (m−k)−ln m, then
an+1
an−m≤10−rwhere r∈Z+.
Example 2. For the sequence of Fibonacci numbers Fnpublished electronically in The On-
Line Encyclopedia of Integer Sequences at http://oeis.org as A000045,
F41
F40 −φ
<10−16.
For A099133,
a42
a41 −4φ
<10−16.
For A015540,
a23
a22 −6
<10−16.
5 Acknowledgements
We would like to thank the editors of The On-Line Encyclopedia of Integer Sequences for
their valuable comments to my submissions to the Encyclopedia which are derived from this
paper.
This work is being supported by a grant from the Dean’s Office, School of Science and
Engineering, Ateneo de Manila University.
11
References
[1] A. F. Horadam, Basic properties of a certain generalised sequence of numbers, Fibonacci
Quart. 3(1965), 161–176.
[2] A. F. Horadam, Jacobsthal representations numbers. Fibonacci Quart. 34 (1996) 40–54.
[3] A. K. Whitford, Binet’s formula generalized, Fibonacci Quart. 15 (1979), 21,24,29.
[4] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, Published electroni-
cally at http://oeis.org.
[5] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, Notices Amer. Math.
Soc. 50 (2003), 912–915.
[6] E. W. Weisstein, Jacobsthal Number, MathWorld–A Wolfram Web Resource. Published
electronically at http://mathworld.wolfram.com/JacobsthalNumber.html.
2010 Mathematics Subject Classification: Primary 11B37; Secondary 11B39.
Keywords: golden ratio, Binet-de Moivr´e formula, Fibonacci number, convergence. recur-
rence.
(Concerned with sequences A000045,A001045,A002378,A003683,A005563,A015441,A015518,
A015521,A015528,A015531,A015540,A015552,A028347,A028552,A028557,A028560,
A028563,A028566,A028569,A051958,A053404,A053428,A053430,A053524,A053573,
A079773,A080424,A085449,A098603,A098847,A098848,A098849,A098850,A099012,
A099133,A119412,A120071,A132759,A132760,A132761,A132762,A132763,A132764,
A132765,A132766,A132767,A132768,A132769,A132770,A132771,A132772, and A132773.
)
12
