SOME PROPERTIES OF GENERALIZED
TETRANACCI SEQUENCE MODULO m
Mansi N. Zaveri1, Dr. Jayant K. Patel2
Asst. Prof., A.S.H. Dept., S.N.P.I.T & RC, GTU, Umrakh, Bardoli, Gujarat, India 1
Asso. Prof., Mathematics Dept., Navyug Science College, V.N.S.G.U., Surat, Gujarat, India 2
mansizaveri@ymail.com1 , jrustpatel@gmail.com 2
ABSTRACT: Every Tribonacci Sequence starting with an arbitrary triple of integers is
periodic modulo m for any modulus m > 1. Our research was encouraged by the relations
between the primitive periods of sequences obtained by reducing Tribonacci sequence by a
given prime modulus p and by its powers pt , which were deduced by M. E. Waddill. In this
paper we derive some results for the case of a generalized Tetranacci sequence that Start with
an arbitrary quadruple of integers.
Keywords: Modular periodicity, Periodic sequence, Tetranacci sequence, Tribonacci
sequence.
1. INTRODUCTION
Let (𝑡𝑛 )∞
𝑛=0 be a Tetranacci sequence 0, 0, 0, 1, 1, 2, 4, 8. 15, 29, 56, 108, 208, 401,
773,……..defined by the recurrence relation 𝑡𝑛+4 = 𝑡𝑛+3 + 𝑡𝑛+2 + 𝑡𝑛+1 + 𝑡𝑛 and with
initial values [𝑡0 , 𝑡1 , 𝑡2 , 𝑡3 ] = [0, 0, 0, 1] and let (𝑇𝑛 )∞
𝑛=0 be a Tetranacci sequence
defined by the recurrence relation 𝑇𝑛+4 = 𝑇𝑛+3 + 𝑇𝑛+2 + 𝑇𝑛+1 + 𝑇𝑛 with an arbitrary
quadruple of integers [𝑎, 𝑏, 𝑐, 𝑑]. It is well known that the sequences (𝑡𝑛 𝑚𝑜𝑑 𝑚)∞
𝑛=0 and
(𝑇𝑛 𝑚𝑜𝑑 𝑚)∞
𝑛=0 are periodic sequences for any arbitrary modulus m > 1. We denote by
ℎ(𝑚) and ℎ(𝑚)[𝑎, 𝑏, 𝑐, 𝑑] the primitive periods of these sequences respectively. Here, we
derive relationship between the numbers ℎ(𝑝)[𝑎, 𝑏, 𝑐, 𝑑] and ℎ(𝑝𝑒 )[𝑎, 𝑏, 𝑐, 𝑑], where 𝑝 is an
arbitrary prime and 𝑒 ∈ 𝑁.
2. TETRANACCI SEQUENCE – CONCEPTUAL FRAMEWORK
The linear recurrence relation 𝑡𝑛+4 = 𝑡𝑛+3 + 𝑡𝑛+2 + 𝑡𝑛+1 + 𝑡𝑛 is defined as a mapping of
ordered four tuples (𝑡𝑛 , 𝑡𝑛+1 , 𝑡𝑛+2 , 𝑡𝑛+3 ) onto (𝑡𝑛+1 , 𝑡𝑛+2 , 𝑡𝑛+3 , 𝑡𝑛+4 ), see [3]. This
mapping is represented by 𝑇(𝑡𝑛 , 𝑡𝑛+1 , 𝑡𝑛+2 , 𝑡𝑛+3 ) = (𝑡𝑛+1 , 𝑡𝑛+2 , 𝑡𝑛+3 , 𝑡𝑛+4 ). Where the
Tetranacci matrix T can be defined by
(2.1)
0
𝑇 = [0
0
1
1
0
0
1
0
1
0
1
0
0]
1
1
and
𝑡𝑛−1
𝑡
𝑇 𝑛 = [𝑡 𝑛
𝑛+1
𝑡𝑛+2
𝑠𝑛−1
𝑠𝑛
𝑠𝑛+1
𝑠𝑛+2
𝑚𝑛−1
𝑚𝑛
𝑚𝑛+1
𝑚𝑛+2
𝑡𝑛
𝑡𝑛+1
𝑡𝑛+2 ] for n > 1
𝑡𝑛+3
The Tetranacci sequence {𝑡𝑛 } is defined by:
𝑡𝑛+4 = 𝑡𝑛+3 + 𝑡𝑛+2 + 𝑡𝑛+1 + 𝑡𝑛
𝑡0 = 𝑡1 = 𝑡2 = 0, 𝑡3 = 1
Two companion sequences {𝑠𝑛 } 𝑎𝑛𝑑 {𝑚𝑛 } are defined as:
𝑠𝑛+4 = 𝑠𝑛+3 + 𝑠𝑛+2 + 𝑠𝑛+1 + 𝑠𝑛
𝑠0 = 𝑠3 = 1, 𝑠1 = 𝑠2 = 0
𝑚𝑛+4 = 𝑚𝑛+3 + 𝑚𝑛+2 + 𝑚𝑛+1 + 𝑚𝑛
𝑚0 = 𝑚2 = 0, 𝑚1 = 𝑚3 = 1
For any arbitrary 𝑛 ∈ 𝑁 and an arbitrary modulus m, 𝑇 𝑛 assumes a unique form
𝑇 𝑛 = 𝐵 + 𝑚𝐴 , where 𝐴 = [𝑎𝑖𝑗 ] and 𝐵 = [𝑏𝑖𝑗 ] are integer matrices such that 0  bij  m  1
and [𝑎𝑖𝑗 ] are nonnegative integers.
If 𝑛 = ℎ(𝑚), then 𝑇ℎ(𝑚) ≡ 𝐸 (𝑚𝑜𝑑 𝑚), where E is an identity matrix. Therefore,
we can express 𝑇ℎ(𝑚) as 𝑇ℎ(𝑚) = 𝐸 + 𝑚𝐴. We will use this fact for the alternative proof
of Theorem 2.1 published by M. E. Waddill in 1978 (see [5]). The alternative proof of this
Theorem is based on matrix algebra. Its modification can also be used for the proof of
generalized linear recurrence of order k. It is particularly applied to the case of Fibonacci
sequences (see [2], page 527).
Theorem 2.1. let 𝑝 be an arbitrary prime and ℎ(𝑝) ≠ ℎ(𝑝𝑒 ) then
(2.2)
ℎ(𝑝𝑒 ) = 𝑝𝑒−1 ℎ(𝑝),
for all 𝑒 ∈ 𝑁
𝑒
𝑒
Proof: The matrix 𝑇ℎ(𝑝 ) can be written as 𝑇ℎ(𝑝 ) = 𝐸 + 𝑝𝑒 𝐴. By using binomial expansion,
we can write,
𝑒
𝑇 𝑝ℎ(𝑝 ) = (𝐸 + 𝑝𝑒 𝐴)𝑝 = 𝐸 𝑝 + (
𝑝
𝑝
) 𝐸 𝑝−1 ( 𝑝𝑒 𝐴 ) + ( ) 𝐸 𝑝−2 (𝑝𝑒 𝐴)2 + ⋯ + (𝑝𝑒 𝐴)𝑝
1
2
By using congruence, the above result can be written as,
𝑒
𝑇 𝑝ℎ(𝑝 ) ≡ 𝐸( 𝑚𝑜𝑑 𝑝𝑒+1 )
Since ℎ(𝑝𝑒+1 ) is the primitive period, we have ℎ(𝑝𝑒+1 ) | 𝑝ℎ(𝑝𝑒 ).
And it is obvious that ℎ(𝑝𝑒 ) | ℎ(𝑝𝑒+1 ) and as a result one of the following is true.
ℎ(𝑝𝑒+1 ) = ℎ(𝑝𝑒 ) or ℎ(𝑝𝑒+1 ) = 𝑝ℎ(𝑝𝑒 )
(2.3)
Now we use induction by 𝑒. For e = 1 the assertion is evident and for e = 2 it follows from
the assumption. Assume that ℎ(𝑝𝑒 ) = 𝑝ℎ(𝑝𝑒−1 ) = 𝑝𝑒−1 ℎ(𝑝) holds for 𝑒 ≥ 1.
We will prove this equation for 𝑒 + 1. The induction assumption ℎ(𝑝𝑒−1 ) ≠ ℎ(𝑝𝑒 ) implies
that 𝑇ℎ(𝑝
𝑒−1 )
= 𝐸 + 𝑝𝑒−1 𝐴. where 𝑝 ∤ 𝐴 . Thus we have
𝑝
𝑇
𝑝ℎ(𝑝𝑒−1 )
𝑝
= (𝐸 + 𝑝𝑒−1 𝐴)𝑝 = ∑ ( ) 𝐸 𝑝−𝑖 (𝑝𝑒−1 𝐴)𝑖
𝑖
𝑖=0
𝑒
𝑒−1
Hence 𝑇ℎ(𝑝 ) = 𝑇 𝑝ℎ(𝑝 ) ≢ 𝐸(𝑚𝑜𝑑 𝑝𝑒+1) and therefore, ℎ(𝑝𝑒+1 ) ∤ ℎ(𝑝𝑒 ) and hence
ℎ(𝑝𝑒+1 ) ≠ ℎ(𝑝𝑒 ). Thus from equation (2.3) we can say that ℎ(𝑝𝑒+1 ) = 𝑝ℎ(𝑝𝑒 ) and
ℎ(𝑝𝑒+1 ) = 𝑝𝑒 ℎ(𝑝).
Theorem 2.2. Let 𝑒 ∈ 𝑁 satisfies ℎ(𝑝) = ℎ(𝑝2 ) =. . . . . = ℎ(𝑝𝑒 ) ≠ ℎ(𝑝𝑒+1 ). Then for
arbitrary 𝑡 ≥ 𝑒, we have ℎ(𝑝𝑒 ) = 𝑝𝑡−𝑒 ℎ(𝑝)
Proof. We proceed with the analogy of Theorem 2.1.
Problem 2.3. Whether the supposition ℎ(𝑝) ≠ ℎ(𝑝2 ) is essential or whether the equality
ℎ(𝑝) = ℎ(𝑝2 ) never occurs is open. No instance of ℎ(𝑝) = ℎ(𝑝2 ) has been found up to
present, neither it is proved that such type of equality can never be happen. However, for
sequences defined by a general linear recurrence of order four, the condition ℎ(𝑝) ≠ ℎ(𝑝2 )
need not be satisfied. If (𝑡𝑛 )∞
𝑛=0 be a Tetranacci sequence defined by the linear recurrence
𝑡𝑛+4 = 3𝑡𝑛+3 − 𝑡𝑛+2 − 𝑡𝑛+1 − 𝑡𝑛 and with initial quadruple [0, 0, 0, 1] then
ℎ(2) = ℎ(22 ) = 5. In [6], it is proved the positive answer to the question whether
ℎ(𝑝) ≠ ℎ(𝑝2 ) holds for all primes implies the validity of the first case of Fermat’s last
theorem. However, questions related to the validity of the equation ℎ(𝑝) = ℎ(𝑝2 ) are not
evaluated here. In the continuation we will always assume ℎ(𝑝) ≠ ℎ(𝑝2 ).
3. ELEMENTARY
SEQUENCE
OBSERVATIONS
OF
GENERALIZED
TETRANACCI
Clearly the relation ℎ(𝑝𝑒 )[𝑎, 𝑏, 𝑐, 𝑑] = 𝑝𝑒−1 ℎ(𝑝)[𝑎, 𝑏, 𝑐, 𝑑] is generally not valid.
We have, for instance, ℎ(𝑝)[0, 0, 0, 0] = ℎ(𝑝𝑒 )[0, 0, 0, 0] = 1 for arbitrary 𝑝, 𝑒. Put
𝑥0 = [𝑎, 𝑏, 𝑐, 𝑑]𝑇 and 𝑥𝑛 = [𝑇𝑛 , 𝑇𝑛+1 , 𝑇𝑛+2 , 𝑇𝑛+3 ]𝑇 , where T is the transposition. Then 𝑥𝑛
can be expressed in terms of 𝑥0 using the equation 𝑥𝑛 = 𝑇 𝑛 𝑥0 . If a Tetranacci sequence is
determined by the quadruple [0, 0, 0, 1] then ℎ(𝑚) is the smallest number h for which 𝑇 ℎ ≡
𝐸(𝑚𝑜𝑑 𝑚). But by the following example we can see that, this result is not true for arbitrary
Tetranacci sequence with quadruple [𝑎, 𝑏, 𝑐, 𝑑].
Example 3.1.
Let 𝑥0 = [1, 1, 1, 1] and 𝑚 = 6 then we can easily verify that
5
𝑇 ≢ 𝐸(𝑚𝑜𝑑 6). Also 𝑇 5 𝑥0 ≡ 𝑥0 (𝑚𝑜𝑑 6) holds. We have ℎ(3)[1, 1, 1, 1] = 1 and one can
assume using Theorem 2.1 that ℎ(32 )[1, 1, 1, 1] = 3 but ℎ(32 )[1, 1, 1, 1] = 78.
The relationship between ℎ(𝑝)[𝑎, 𝑏, 𝑐, 𝑑] = ℎ(𝑝𝑒 )[𝑎, 𝑏, 𝑐, 𝑑] seem to be more complex and
needs closer study. First we will prove two important lemmas.
Lemma 3.2. Let 𝑝 be an arbitrary prime. Then for every 𝑒 ∈ 𝑁 and 1 ≤ 𝑖 ≤ 𝑒, we have,
ℎ(𝑝𝑒 )[𝑝𝑒−𝑖 𝑎, 𝑝𝑒−𝑖 𝑏, 𝑝𝑒−𝑖 𝑐, 𝑝𝑒−𝑖 𝑑] = ℎ(𝑝𝑖 )[𝑎, 𝑏, 𝑐, 𝑑].
Proof: let (𝑇𝑛 )∞
𝑛=0 be a Tetranacci sequence defined by the arbitrary quadruple [𝑎, 𝑏, 𝑐, 𝑑]
and let 𝑝 be any prime number. Then,
𝑒−𝑖
( 𝑝𝑒−𝑖 𝑇𝑛 𝑚𝑜𝑑 𝑝𝑒 )∞
( 𝑇𝑛 𝑚𝑜𝑑 𝑝𝑖 )∞
𝑛=0 = 𝑝
𝑛=0
Therefore, we can write,
(3.1)
ℎ(𝑝𝑒 )[𝑝𝑒−𝑖 𝑎, 𝑝𝑒−𝑖 𝑏, 𝑝𝑒−𝑖 𝑐, 𝑝𝑒−𝑖 𝑑] = ℎ(𝑝𝑖 )[𝑎, 𝑏, 𝑐, 𝑑].
Particularly, for 𝑖 = 1 the above result gives
ℎ(𝑝𝑒 )[𝑝𝑒−1 𝑎, 𝑝𝑒−1 𝑏, 𝑝𝑒−1 𝑐, 𝑝𝑒−1 𝑑] = ℎ(𝑝)[𝑎, 𝑏, 𝑐, 𝑑].
Remark 3.2.1 By lemma (3.2), the investigation of the periods for arbitrary quadruples
[𝑎, 𝑏, 𝑐, 𝑑] can be reduced to the case with [𝑎, 𝑏, 𝑐, 𝑑] ≢ [0, 0, 0, 0](𝑚𝑜𝑑𝑝).
Lemma 3.3. Let 𝑝 be an arbitrary prime. For every arbitrary integers [𝑎, 𝑏, 𝑐, 𝑑] and every
𝑒, 𝑡 ∈ 𝑁, where 𝑒 ≤ 𝑡, we have, ℎ(𝑝𝑒 )[𝑎, 𝑏, 𝑐, 𝑑] | ℎ(𝑝𝑡 )[𝑎, 𝑏, 𝑐, 𝑑]. In particular,
we
have
(3.2)
ℎ(𝑝)[𝑎, 𝑏, 𝑐, 𝑑] | ℎ(𝑝𝑡 )[𝑎, 𝑏, 𝑐, 𝑑].
Proof: Given that 𝑝 be any arbitrary prime number. And let 𝑇𝑛 be a Tetranacci sequence
defined by the arbitrary initial quadruple [𝑎, 𝑏, 𝑐, 𝑑].
Let 𝑥0 = [𝑎, 𝑏, 𝑐, 𝑑]𝑇 , ℎ = ℎ(𝑝𝑒 )[𝑎, 𝑏, 𝑐, 𝑑] and 𝑘 = ℎ(𝑝𝑡 )[𝑎, 𝑏, 𝑐, 𝑑]. Then we have,
𝑇 𝑘 𝑥0 ≡ 𝑥0 (𝑚𝑜𝑑 𝑝𝑡 ). It follows that 𝑇 𝑘 𝑥0 ≡ 𝑥0 (𝑚𝑜𝑑 𝑝𝑒 ). This means that k is the period
of a Tetranacci sequence with quadruple [𝑎, 𝑏, 𝑐, 𝑑], reduced by the modulus 𝑝𝑒 . Therefore a
primitive period divides an arbitrary period, we have ℎ | 𝑘.
𝑒
𝑒
Remark 3.3.1 Moreover, 𝑇ℎ(𝑝 ) ≡ 𝐸 (𝑚𝑜𝑑 𝑝𝑒 ) implies 𝑇ℎ(𝑝 ) 𝑥0 ≡ 𝑥0 (𝑚𝑜𝑑 𝑝𝑒 ). For any
𝑥0 = [𝑎, 𝑏, 𝑐, 𝑑]𝑇 and 𝑒 ∈ 𝑁 and therefore 𝑥ℎ(𝑝𝑒) ≡ 𝑥0 (𝑚𝑜𝑑 𝑝𝑒 ). So that we have,
ℎ(𝑝𝑒 )[𝑎, 𝑏, 𝑐, 𝑑] | ℎ(𝑝𝑒 )[0, 0, 0, 1].
Theorem 3.4. If 𝑛 | 𝑚, then for a given Tetranacci sequence,
ℎ(𝑛)[𝑎, 𝑏, 𝑐, 𝑑] | ℎ(𝑚)[𝑎, 𝑏, 𝑐, 𝑑].
Proof: Let T = (𝑇𝑛 )∞
𝑛=1 be a generalised Tetranacci sequence defined by an arbitrary integers
[𝑎, 𝑏, 𝑐, 𝑑]. Let ℎ = ℎ(𝑚)[𝑎, 𝑏, 𝑐, 𝑑]. We need to show that 𝑇(𝑚𝑜𝑑 𝑛) repeats in blocks of
length h. We can prove this by showing that 𝑇𝑖 ≡ 𝑇𝑖+ℎ (𝑚𝑜𝑑 𝑛) regardless of our choice of i.
Certainly we know that 𝑇𝑖 ≡ 𝑇𝑖+ℎ (𝑚𝑜𝑑 𝑚). For some 0 ≤ 𝑏 < 𝑚, we have
𝑇𝑖 = 𝑏 + 𝑚𝑥 and 𝑇𝑖+ℎ = 𝑏 + 𝑚𝑦.
Now, we have, 𝑛 | 𝑚 and let 𝑚 = 𝑛𝑟 and substitute this value in the previous equations, we
get,
𝑇𝑖 = 𝑏 + 𝑛𝑟𝑥 and 𝑇𝑖+ℎ = 𝑏 + 𝑛𝑟𝑦.
We can say that 𝑏 = 𝑏 ′ + 𝑛𝑤 (0 ≤ 𝑏 ′ < 𝑛) and substitute for b in the previous equations.
We have,
𝑇𝑖 = 𝑏 ′ + 𝑛(𝑤 + 𝑟𝑥) and 𝑇𝑖+ℎ = 𝑏 ′ + 𝑛(𝑤 + 𝑟𝑦).
This implies that 𝑇𝑖 ≡ 𝑇𝑖+ℎ (𝑚𝑜𝑑 𝑛) which we want to be shown.
We can make this theorem more exact by expressing ℎ(𝑚) in terms of ℎ(𝑝𝑖 𝑒𝑖 ), where 𝑚 have
the prime factorization 𝑚 = ∏ 𝑝𝑖 𝑒𝑖 .
Theorem 3.5. Let m have the prime factorization 𝑚 = ∏ 𝑝𝑖 𝑒𝑖 , then
ℎ(𝑚)[𝑎, 𝑏, 𝑐, 𝑑] = 𝑙𝑐𝑚 {ℎ(𝑝𝑖 𝑒𝑖 )[𝑎, 𝑏, 𝑐, 𝑑]},
the least common multiple of ℎ(𝑝𝑖 𝑒𝑖 )[𝑎, 𝑏, 𝑐, 𝑑].
Proof: Here, 𝑚 have a prime factorization 𝑚 = ∏ 𝑝𝑖 𝑒𝑖 . Therefore 𝑝𝑖 𝑒𝑖 | 𝑚. By Theorem 2.4.
we have
ℎ(𝑝𝑖 𝑒𝑖 )[𝑎, 𝑏, 𝑐, 𝑑]| ℎ(𝑚)[𝑎, 𝑏, 𝑐, 𝑑], for all 𝑖.
It follows that
(3.3)
𝑙𝑐𝑚 {ℎ(𝑝𝑖 𝑒𝑖 )[𝑎, 𝑏, 𝑐, 𝑑]} | ℎ(𝑚)[𝑎, 𝑏, 𝑐, 𝑑]
Since ℎ(𝑝𝑖 𝑒𝑖 )[𝑎, 𝑏, 𝑐, 𝑑] | 𝑙𝑐𝑚 {ℎ(𝑝𝑖 𝑒𝑖 )[𝑎, 𝑏, 𝑐, 𝑑]}, We know that 𝑇(𝑚𝑜𝑑 𝑝𝑖 𝑒𝑖 ) repeats in
blocks of length 𝑙𝑐𝑚 {ℎ(𝑝𝑖 𝑒𝑖 )[𝑎, 𝑏, 𝑐, 𝑑]}.
Hence, for all 𝑖,
𝑇𝑙𝑐𝑚 {ℎ(𝑝𝑖 𝑒𝑖 )[𝑎,𝑏,𝑐,𝑑]} ≡ 𝑇0 (𝑚𝑜𝑑 𝑝𝑖 𝑒𝑖 ),
𝑇𝑙𝑐𝑚 {ℎ(𝑝𝑖 𝑒𝑖 )[𝑎,𝑏,𝑐,𝑑]}+1 ≡ 𝑇1 (𝑚𝑜𝑑 𝑝𝑖 𝑒𝑖 ),
𝑇𝑙𝑐𝑚 {ℎ(𝑝𝑖 𝑒𝑖 )[𝑎,𝑏,𝑐,𝑑]} + 2 ≡ 𝑇2 (𝑚𝑜𝑑 𝑝𝑖 𝑒𝑖 ),
𝑇𝑙𝑐𝑚 {ℎ(𝑝𝑖 𝑒𝑖 )[𝑎,𝑏,𝑐,𝑑]} + 3 ≡ 𝑇3 (𝑚𝑜𝑑 𝑝𝑖 𝑒𝑖 ).
Here all 𝑝𝑖 𝑒𝑖 are relatively prime and therefore by Chinese Remainder Theorem, we can
write,
𝑇𝑙𝑐𝑚 {ℎ(𝑝𝑖 𝑒𝑖 )[𝑎,𝑏,𝑐,𝑑]} ≡ 𝑇0 (𝑚𝑜𝑑 𝑚),
𝑇𝑙𝑐𝑚 {ℎ(𝑝𝑖 𝑒𝑖 )[𝑎,𝑏,𝑐,𝑑]} +1 ≡ 𝑇1 (𝑚𝑜𝑑 𝑚).
𝑇𝑙𝑐𝑚 {ℎ(𝑝𝑖 𝑒𝑖 )[𝑎,𝑏,𝑐,𝑑]} + 2 ≡ 𝑇2 (𝑚𝑜𝑑 𝑚),
𝑇𝑙𝑐𝑚 {ℎ(𝑝𝑖 𝑒𝑖 )[𝑎,𝑏,𝑐,𝑑]} + 3 ≡ 𝑇3 (𝑚𝑜𝑑 𝑚).
Thus 𝑇(𝑚𝑜𝑑 𝑚) repeats in blocks of length 𝑙𝑐𝑚{ℎ(𝑝𝑖 𝑒𝑖 )] and we can say that
ℎ(𝑚)[𝑎, 𝑏, 𝑐, 𝑑] | 𝑙𝑐𝑚 {ℎ(𝑝𝑖 𝑒𝑖 )[𝑎, 𝑏, 𝑐, 𝑑]}
(3.4)
Since from (3.3) and (3.4) we can write,
ℎ(𝑚)[𝑎, 𝑏, 𝑐, 𝑑] = 𝑙𝑐𝑚 {ℎ(𝑝𝑖 𝑒𝑖 )[𝑎, 𝑏, 𝑐, 𝑑]}.
This completes the proof.
Theorem 3.6. For any Tetranacci sequence with arbitrary quadruple [𝑎, 𝑏, 𝑐, 𝑑], we have
ℎ([𝑚, 𝑛])[𝑎, 𝑏, 𝑐, 𝑑] = [ℎ(𝑚)[𝑎, 𝑏, 𝑐, 𝑑], ℎ(𝑛)[𝑎, 𝑏, 𝑐, 𝑑]]. Where [ ] denotes the least
common multiple.
Proof: For the simplicity of this Theorem let 𝑘(𝑚) = ℎ(𝑚)[𝑎, 𝑏, 𝑐, 𝑑] be the period of
generalized Tetranacci sequence with arbitrary quadruple [𝑎, 𝑏, 𝑐, 𝑑]. We know that,
𝑚 | [𝑚, 𝑛] and 𝑛 | [𝑚, 𝑛]. By Theorem (3.4) we can write,
𝑘(𝑚) | 𝑘([𝑚, 𝑛]) and 𝑘(𝑛) | 𝑘([𝑚, 𝑛]).
It follows that
[𝑘(𝑚), 𝑘(𝑛)] | 𝑘 ([𝑚, 𝑛])
We have the prime factorization [𝑚, 𝑛] = 𝑝1 𝑒1 . 𝑝2 𝑒2 … … 𝑝𝑟 𝑒𝑟 , for all 𝑟. Then,
𝑘([𝑚, 𝑛]) = 𝑘(𝑝1 𝑒1 . 𝑝2 𝑒2 … … 𝑝𝑟 𝑒𝑟 )
= [𝑘(𝑝1 𝑒1 ), 𝑘(𝑝2 𝑒2 ), … … , 𝑘(𝑝𝑟 𝑒𝑟 )].
Since 𝑝𝑟 𝑒𝑟 divides either 𝑚 or 𝑛, for all 𝑟. Therefore either
𝑘(𝑝𝑟 𝑒𝑟 ) | 𝑘(𝑚) or 𝑘(𝑝𝑟 𝑒𝑟 ) | 𝑘(𝑛), for all 𝑟.
Thus,
[𝑘(𝑝1 𝑒1 ), … … , 𝑘(𝑝𝑟 𝑒𝑟 )] | [𝑘(𝑚), 𝑘(𝑛)]
Therefore,
𝑘([𝑚, 𝑛])| [𝑘(𝑚), 𝑘(𝑛)]
Hence, we have,
𝑘([𝑚, 𝑛]) = [𝑘(𝑚), 𝑘(𝑛)].
The above Theorems restricts the form of the numbers ℎ(𝑝𝑒 )[𝑎, 𝑏, 𝑐, 𝑑]. As we will see here
that the relation between ℎ(𝑝𝑒 )[𝑎, 𝑏, 𝑐, 𝑑] and ℎ(𝑝)[𝑎, 𝑏, 𝑐, 𝑑] also depends on the form of
the factorization of polynomial 𝑓(𝑥) over the field 𝔽𝑝 .
4. CONCLUSION
In this paper we have introduced the Tetranacci matrix and by matrix algebra proved some
theorems. We proved some results for the Tetranacci sequence with quadruple [0, 0, 0, 1] and
introduced generalized Tetranacci sequence with arbitrary quadruple [𝑎, 𝑏, 𝑐, 𝑑] and verify
∞
some properties that is true for (𝑡𝑛 )∞
𝑛=0 but not applicable for the (𝑇𝑛 )𝑛=0 . We proved some
theorems to define period of Tetranacci sequence (𝑇𝑛 )∞
𝑛=0 . Defined relationship between the
𝑡 )[𝑎,
numbers ℎ(𝑝)[𝑎, 𝑏, 𝑐, 𝑑] and ℎ(𝑝
𝑏, 𝑐, 𝑑]. Where 𝑝 is an arbitrary prime and 𝑡 ∈ 𝑁.
5. REFERENCES
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