Article [1] introduced the T₃ tree and showed a number of properties of tree, including divisibility, multiples and divisors and multiplications of the nodes. Looking through the other papers that are related with the article [1] , such as articles [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 important for finding N’s divisor. Accordingly, this article makes an investigation on the issue and presents the results.

Theorem 1. Let a > 1 be an integer and x > 0 be a real number; then it holds

and

Proof. Since a > 1 , the definition immediately yields

a ⌊ x ⌋ − 1 < a ⌊ x ⌋ ≤ a x < a ⌊ x ⌋ + 1

Since a and are integers, it yields by Lemma 2 (P13)

a ⌊ x ⌋ ≤ ⌊ a x ⌋ ≤ a ⌊ x ⌋ + 1

Considering, it knows a ⌊ x ⌋ − 1 < a ⌊ x ⌋ ; consequently

a ⌊ x ⌋ − 1 < a ⌊ x ⌋ ≤ ⌊ a x ⌋ ≤ a ⌊ x ⌋ + 1

Theorem 1. Let n be a positive integer and

b 0 = n 2 , b 1 = n 2 + 1 , ⋯ , b i = n 2 + i , ⋯ , b 2 n = n 2 + 2 n , b 2 n + 1 = n 2 + 2 n + 1

then

⌊ b i ⌋ ( i = 1 , ⋯ , 2 n ) = ⌊ b 0 ⌋ = n , ⌊ b 2 n + 1 ⌋ = n + 1 (3)

Proof. and ⌊ b 2 n + 1 ⌋ = n + 1 obviously hold. Now consider that, for i = 1 , 2 , ⋯ , 2 n , it holds

n = b 0 < b i = n 1 + i n 2 < b 2 n + 1 = n + 1

This is to say that,; since n is an integer, it is sure by definition of the floor function

⌊ b i ⌋ ( i = 1 , ⋯ , 2 n ) = n

Corollary 1. Let n and α be a positive integers and

b 0 = n 2 α , b 1 = n 2 α + 1 , ⋯ , b i = n 2 α + i , ⋯ , b 2 n α = n 2 α + 2 n α , b 2 n α + 1 = n 2 α + 2 n α + 1

then

Proof. (Omitted)

Example 1. Take b 0 = 2 8 , b 1 = 2 8 + 1 = 257 , ⋯ , b 2 × 2 4 = 2 8 + 2 × 2 4 = 288 , b 2 × 2 4 + 1 = 2 8 + 2 × 2 4 + 1 = 289 ; then ⌊ b 0 ⌋ = 2 4 , ⌊ b 1 ⌋ = 16 , ⋯ , ⌊ b 2 × 2 4 ⌋ = 16 , ⌊ b 2 × 2 4 + 1 ⌋ = 17 .

Theorem 2. Let n and α be a positive integers and

b 0 = n 2 α + 1 , b 1 = n 2 α + 1 + 1 , ⋯ , b i = n 2 α + 1 + i , ⋯ , b 2 n α = n 2 α + 1 + 2 n α n , b 2 n α + 1 = n 2 α + 1 + 2 n α n + 1

then

⌊ n α n ⌋ ≤ ⌊ b i ⌋ | ( i = 0 , 1 , 2 , ⋯ , 2 n α + 1 ) ≤ ⌊ n α n ⌋ + 1

Proof. See the following deductions.

1)

2) By Lemma 2(P13)

n 2 α + 1 = b 0 < b 1 = n 2 α + 1 + 1 < ⋯ < b 2 n α = n 2 α + 1 + 2 n α n < b 2 n α + 1 = ( n α n + 1 ) 2 ⇒ n α n = b 0 < b i | ( i = 1 , 2 , ⋯ , 2 n α ) < n α n + 1 ⇒ ⌊ n α n ⌋ = b 0 ≤ b i | ( i = 1 , 2 , ⋯ , 2 n α ) ≤ ⌊ n α n ⌋ + 1

Example 2. Take and b 2 n α + 1 by

, b 2 5 = 2 9 + 2 5 = 544 , b 2 5 + 1 = 2 9 + 2 5 + 1 = 545 then

⌊ b 2 4 + 1 ⌋ = ⌊ 2 9 + 17 ⌋ = ⌊ 529 ⌋ = 23 , ⋯ , ⌊ b 2 × 2 4 − 1 ⌋ = ⌊ 543 ⌋ = 23 , ⌊ b 2 × 2 4 ⌋ = ⌊ 544 ⌋ = 23 , ⌊ b 2 × 2 4 + 1 ⌋ = ⌊ 545 ⌋ = 23

Theorem 4 Suppose integer k satisfies k > 2 and N ( k , 0 ) be the leftmost node on level k of T₃; then is even if k is odd, whereas, it can be either odd or even if k is even.

Proof. Since N ( k , 0 ) = 2 k + 1 + 1 , it knows by Corollary 1 for an odd k. If k is even, let it be k = 2 α + 1 ; then by Theorem 2 ⌊ N ( k , 0 ) ⌋ = ⌊ 2 α 2 ⌋ or ⌊ N ( k , 0 ) ⌋ = ⌊ 2 α 2 ⌋ + 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.

N ( 7 , 0 ) = 2 7 + 1 + 1 = 257 ⇒ ⌊ N ( 7 , 0 ) ⌋ = 16 N ( 11 , 0 ) = 2 11 + 1 + 1 = 4097 ⇒ ⌊ N ( 11 , 0 ) ⌋ = 64 N ( 19 , 0 ) = 2 19 + 1 + 1 = 1048577 ⇒ ⌊ N ( 19 , 0 ) ⌋ = 1024

N ( 8 , 0 ) = 2 8 + 1 + 1 = 513 ⇒ ⌊ N ( 8 , 0 ) ⌋ = 22 N ( 10 , 0 ) = 2 10 + 1 + 1 = 1025 ⇒ ⌊ N ( 10 , 0 ) ⌋ = 45 N ( 16 , 0 ) = 2 16 + 1 + 1 = 131073 ⇒ ⌊ N ( 16 , 0 ) ⌋ = 362

Theorem 5. Suppose integers k and j satisfy k > 2 and 0 ≤ j ≤ 2 k − 1 ; let N ( k , j ) be the node at position j on level k of T₃; then it holds

Proof. Since 2 k + 1 + 1 ≤ N ( k , j ) ≤ 2 k + 2 − 1 , it yields 2 k + 1 < N ( k , j ) < 2 k + 2 ; hence it holds

2 k + 1 2 < N ( k , j ) < 2 k 2 + 1

By Lemma 2 (P13), it yields

⌊ 2 k + 1 2 ⌋ ≤ ⌊ N ( k , j ) ⌋ ≤ ⌊ 2 k 2 + 1 ⌋ (6)

By Theorem 1, it holds

2 ⌊ k + 1 2 ⌋ ≤ 2 k + 1 2

and

2 k 2 + 1 < 2 ⌊ k 2 ⌋ + 2

Hence it results in

2 ⌊ k + 1 2 ⌋ = ⌊ 2 ⌊ k + 1 2 ⌋ ⌋ ≤ ⌊ 2 k + 1 2 ⌋ ≤ ⌊ N ( k , j ) ⌋ ≤ ⌊ 2 k 2 + 1 ⌋ ≤ ⌊ 2 ⌊ k 2 ⌋ + 2 ⌋ = 2 ⌊ k 2 ⌋ + 2

That is

2 ⌊ k + 1 2 ⌋ ≤ ⌊ N ( k , j ) ⌋ ≤ 2 ⌊ k 2 ⌋ + 2 (7)

or equivalently

2 ⌊ k + 1 2 ⌋ − 1 < ⌊ N ( k , j ) ⌋ < 2 ⌊ k 2 ⌋ + 2 + 1 (8)

Corollary 2. ⌊ N ( k , j ) ⌋ is clamped in T₃ on level ⌊ k + 1 2 ⌋ − 1 and or level ⌊ k 2 ⌋ .

Proof. Since 2 ⌊ k + 1 2 ⌋ − 1 the biggest node on level ⌊ k + 1 2 ⌋ − 2 and 2 ⌊ k 2 ⌋ + 2 + 1 is the smallest node on level ⌊ k 2 ⌋ + 1 , it knows by (8) ⌊ N ( k , j ) ⌋ may be clamped on levels from ⌊ k + 1 2 ⌋ − 1 to ⌊ k 2 ⌋ , totally ⌊ k 2 ⌋ − ( ⌊ k + 1 2 ⌋ − 1 ) + 1 levels.

By Lemma 2 (P2) ⌊ k 2 ⌋ − ( ⌊ k + 1 2 ⌋ − 1 ) + 1 ≥ 2 + ⌊ k 2 − k + 1 2 ⌋ = 2 − 1 = 1

By Lemma 2 (P17 & P8) ⌊ k 2 ⌋ − ( ⌊ k + 1 2 ⌋ − 1 ) + 1 = 2 + ⌊ k 2 ⌋ − ( ⌊ k ⌋ − ⌊ k 2 ⌋ ) = 2 + 2 ⌊ k 2 ⌋ − ⌊ k ⌋ ≤ 2

Hence the corollary holds

Example 4. Taking the smallest nodes and the biggest nodes on level 7 and level 10 respectively, it can see that ⌊ N ( 7 , * ) ⌋ is clamped on 2 levels, whereas ⌊ N ( 10 , * ) ⌋ is clamped on 1 level.

N ( 7 , 0 ) = 2 7 + 1 + 1 = 257 ⇒ ⌊ N ( 7 , 0 ) ⌋ = 16 ≙ 2 N ( 7 , 2 7 − 1 ) = 2 7 + 2 − 1 = 511 ⇒ ⌊ N ( 7 , 2 7 − 1 ) ⌋ = 22 ≙ 3 N ( 10 , 0 ) = 2 11 + 1 = 2047 ⇒ ⌊ N ( 10 , 0 ) ⌋ = 45 ≙ 4 N ( 10 , 2 10 − 1 ) = 2 12 − 1 = 4095 ⇒ ⌊ N ( 10 , 2 10 − 1 ) ⌋ = 63 ≙ 4

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.

The research work is supported by the State Key Laboratory of Mathematical Engineering 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.

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