ISSN Online: 2160-0384 ISSN Print: 2160-0368
DOI: 10.4236/apm.2018.87039 Jul. 20, 2018 666 Advances in Pure Mathematics
Brief Investigation on Square Root
of a Node of
T
3 Tree
Guihong Chen
1,2, Jianhui Li
1,31Department of Computer Science and Technology, Neusoft Institute Guangdong, Foshan, China 2School of Electronics and Information Technology, Sun Yat-sen University, Guangzhou, China 3State Key Laboratory of Mathematical Engineering and Advanced Computing, Wuxi, China
Abstract
The article investigates some properties of square root of T3 tree’s nodes. It
first proves several inequalities that are helpful to estimate the square root of a node, and then proves several theorems to describe the distribution of the square root of the nodes on T3 tree.
Keywords
Square Root, Node, Valuated BinaryTree, Inequality
1. Introduction
Article [1] introduced the T3 tree and showed a number of properties of tree,
in-cluding divisibility, multiples and divisors and multiplications of the nodes. Looking through the other papers that are related with the article [1], such as ar-ticles [2]-[7], one can see that the T3 tree is really a new attempt to study integers.
However, one can also see that, there has not been an article that concerns the square root of a node in the T3 tree. As is known, a divisor of integer N must be
no bigger than N . Hence the location where N lies in the T3 tree is
im-portant for finding N’s divisor. Accordingly, this article makes an investigation on the issue and presents the results.
2. Preliminaries
2.1. Symbols and Notations
Symbol T3 is the T3 tree that was introduced in [1] and [2] and symbol N( )k j, is by default the node at position j on level k of T3, where k≥0 and 0≤ ≤j 2 1k− .
An integer X is said to be clamped on level k of T3 if 2k+1+ ≤1 X ≤2k+2−1 and
How to cite this paper: Chen, G.H. and Li, J.H. (2018) Brief Investigation on Square
Root of a Node of T3 Tree. Advances in
Pure Mathematics, 8, 666-671.
https://doi.org/10.4236/apm.2018.87039 Received: June 22, 2018
Accepted: July 17, 2018 Published: July 20, 2018 Copyright © 2018 by authors and Scientific Research Publishing Inc. This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).
DOI: 10.4236/apm.2018.87039 667 Advances in Pure Mathematics symbol X k indicates X is clamped on level k. An odd interval
[ ]
a b, is a set of consecutive odd numbers that take a as lower bound and b as upper bound, for example,[ ]
3,11 ={
3,5,7,9,11}
. Intervals in this whole article are by default the odd ones unless particularly mentioned. Symbol x is the floor function, an integer function of real number x that satisfies inequality x− <1 x ≤x, or equivalently x ≤ <x x +1. Symbol A⇒B means conclusion B can be de-rived from condition A.2.2. Lemmas
Lemma 1 (See in [1]). Let N( )k j, be the node at the jth position on the kth lev-el of T3 with k≥0 and 0≤ ≤j 2 1k− ; then 2k+1+ ≤1 N( )k j, ≤2k+2−1.
Lemma 2 (See in [8]). For real numbers x and y, it holds (P2) x − y − ≤1 x y− ≤ x − y < x − y +1
(P8) n x ≤ nx with n being a positive integer (P13) x y≤ ⇒ x ≤ y
(P17) 1 2
2
x +x+ = x
, 2x + x2+1= x
3. Main Results and Proofs
Theorem 1. Let a>1 be an integer and x>0 be a real number; then it holds
1 1
x x x x
a − <a ≤a <a + (1) and
1 1
x x x x
a − <a ≤ a ≤a +
(2)
Proof. Since a>1, the definition x− <1 x ≤ <x x +1 immediately yields
1 1
x x x x
a − <a ≤a <a +
Since a and x are integers, it yields by Lemma 2 (P13) 1
x x x
a ≤ a ≤a +
Considering axx1 a 1
a
−
= > , it knows 1
x x
a − <a ; consequently
1 1
x x x x
a − <a ≤ a ≤a +
Theorem 1. Let n be a positive integer and
2 2 2 2 2
0 , 1 1, , i , , 2n 2 , 2 1n 2 1
b =n b n= + b n i= + b =n + n b + =n + n+ then
( 1, ,2 ) 0
i i n
b = b n
= =
, b2 1n+ = + n 1 (3)
DOI: 10.4236/apm.2018.87039 668 Advances in Pure Mathematics for i=1,2, ,2 n, it holds
0 i 1 i2 2 1n 1
n b b n b n
n +
= < = + < = +
This is to say that, i i( 1,2, ,2 )
(
, 1)
nb = ∈ n n+
; since n is an integer, it is sure by
definition of the floor function
( 1, ,2 )
i i n
b = n
=
Corollary 1. Let n and
α
be a positive integers and2 2 2 2 2
0 , 1 1, , i , , 2n 2 , 2n 1 2 1
b n b n= α = α+ b n= α+i b α =nα+ n bα α+ =n α+ nα+
then
(
0,1,2, ,2)
2 1| , 1
i i n n
b α n b α n
α α
+
= = = + (4)
Proof. (Omitted)
Example 1. Take b0 =2 ,8 b1=2 1 257, ,8+ = b2 2×4 =2 2 28+ × 4 =288, 4 8 4
2 2 1 2 2 2 1 289
b× + = + × + = ; then
4 4
4
0 2 , 1 16, , 2 2 16, 2 2 1 17
b b b× b× +
= = = =
.
Theorem 2. Let n and
α
be a positive integers and2 1 2 1 2 1 2 1
0 1 2
2 1
2 1
, 1, , , , 2 ,
2 1
i n
n
b n b n b n i b n n n
b n n n
α
α
α α α α α
α α
+ + + +
+ +
= = + = + = +
= + +
then
(
0,1,2, ,2 1)
1i
i n
n nα b α n nα
= +
≤ ≤ +
Proof. See the following deductions.
1)
(
)
2 2 1
2 1 2 1
2 1
2 1 1 1
1
n n
n
b n n n n n b n n
b n n
α α
α
α α α α
α +
+ +
+
= + + = + ⇒ = +
⇒ = +
2) By Lemma 2(P13)
(
)
(
)
(
)
2
2 1 2 1 2 1
0 1 2 2 1
0 1,2, ,2
0 1,2, ,2
1 2 1
1
1
n n
i i n
i i n
n b b n b n n n b n n
n n b b n n
n n b b n n
α α
α
α
α α α α α
α α
α α
+ + +
+
=
=
= < = + < < = + < = +
⇒ = < < +
⇒ = ≤ ≤ +
Example 2. Take b b0, , ,1 b2nα and b2nα+1 by
5
9 9 9 5
0 2 , 1 2 1 513, , 2 1 2 2 1 543
DOI: 10.4236/apm.2018.87039 669 Advances in Pure Mathematics
5 9 5
2 1 2 2 1 545
b + = + + = then
4
0
1
2
9 4
2
512 22,
513 22,
514 22,
,
2 2 528 22,
b b b
b
= =
= =
= =
= + = =
4
4 4 4
9 2 1
2 2 1
2 2
2 2 1
2 17 529 23, ,
543 23,
544 23,
545 23
b
b
b
b
+
× −
×
× +
= + = =
= =
= =
= =
Theorem 4 Suppose integer k satisfies k>2 and N( )k,0 be the leftmost node on level k of T3; then N( )k,0 is even if k is odd, whereas, it can be ei-ther odd or even if k is even.
Proof. Since N( )k,0 =2k+1+1, it knows by Corollary 1 ( )
1 2
,0 2
k k
N +
=
for an
odd k. If k is even, let it be k=2α+1; then by Theorem 2 N( )k,0 = 2 2α or N( )k,0 =2 2α +1, which indicates N( )k,0 can be either odd or even.
Example 3. Taking N( )7,0,N(11,0),N(19,0),N( )8,0 ,N(10,0) and N(16,0) as examples results in the following results.
( ) ( )
( ) ( )
( ) ( )
7 1
7,0 7,0
11 1
11,0 11,0
19 1
19,0 19,0
2 1 257 16
2 1 4097 64
2 1 1048577 1024
N N
N N
N N
+
+
+
= + = ⇒ =
= + = ⇒ =
= + = ⇒ =
( ) ( )
( ) ( )
( ) ( )
8 1
8,0 8,0
10 1
10,0 10,0
16 1
16,0 16,0
2 1 513 22
2 1 1025 45
2 1 131073 362
N N
N N
N N
+
+
+
= + = ⇒ =
= + = ⇒ =
= + = ⇒ =
Theorem 5. Suppose integers k and j satisfy k>2 and 0≤ ≤j 2 1k− ; let
( )k j,
N be the node at position j on level k of T3; then it holds
( )
1 1 2 2
2 2 2 2
,
2 1 2 2 2 1
k k k k
k j
N
+ +
+ +
− < ≤ ≤ < +
(5)
DOI: 10.4236/apm.2018.87039 670 Advances in Pure Mathematics holds
( )
1 1
2 2
,
2k Nk j 2k
+ +
< <
By Lemma 2 (P13), it yields
( )
1 1
2 2
,
2k Nk j 2k
+ +
≤ ≤
(6)
By Theorem 1, it holds
1 1
2 2
2 2
k+ k
+
≤
and
2
1 2
2
2 2
k k+ +
< Hence it results in
( )
1 1 1 1 2 2
2 2 2 2 2 2
,
2 2 2 2 2 2
k k k k k k
k j
N
+ +
+ + + +
= ≤ ≤ ≤ ≤ =
That is
( )
1 2
2 2
,
2 2
k k
k j
N
+
+
≤ ≤
(7)
or equivalently
( )
1 2
2 2
,
2 1 2 1
k k
k j
N
+
+
− < < +
(8)
Corollary 2. N( )k j, is clamped in T3 on level 1 1 2
k+
−
and or level
2
k
.
Proof. Since 1 2
2 1
k+
− the biggest node on level 1 2
2
k+
−
and
2 2
2 1
k
+
+
is the smallest node on level 1 2
k
+
, it knows by (8) N( )k j, may be clamped on levels from 1 1
2
k+
−
to 2
k
, totally 2 21 1 1
k k+
− − +
le-vels.
By Lemma 2 (P2) 1 1 1 2 1 2 1 1
2 2 2 2
k k+ k k+
− − + ≥ + − = − =
By Lemma 2 (P17 & P8)
1 1 1 2 2 2 2
2 2 2 2 2
k k+ k k k k k
− − + = + − − = + − ≤
Hence the corollary holds
DOI: 10.4236/apm.2018.87039 671 Advances in Pure Mathematics level 10 respectively, it can see that N( )7,* is clamped on 2 levels, whereas
(10,*)
N
is clamped on 1 level.
( ) ( )
(
)
(
)
( ) ( )
(
)
(
)
7 7
10 10
7 1
7,0 7,0
7 2
7,2 1 7,2 1
11
10,0 10,0
12
10,2 1 10,2 1
2 1 257 16 2
2 1 511 22 3
2 1 2047 45 4
2 1 4095 63 4
N N
N N
N N
N N
+
+
− −
− −
= + = ⇒ =
= − = ⇒ =
= + = ⇒ =
= − = ⇒ =
4. Conclusion
Elementary number theory shows that an integer must have a divisor smaller than the square root of the integer itself. Hence the square root is undoubtedly an important issue of an integer. Since T3 tree is considered to be a new tool to study
integers, the square root of a node is certainly helpful to know the distribution of the node’s divisors. The properties proved in this article are sure to provide a know-about the square root of the nodes. We hope it will be useful in the future.
Acknowledgements
The research work is supported by the State Key Laboratory of Mathematical En-gineering and Advanced Computing under Open Project Program No. 2017A01, the Youth Innovative Talents Project (Natural Science) of Education Department of Guangdong Province under grant 2016KQNCX192, 2017KQNCX230. The authors sincerely present thanks to them all.
References
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