Evaluation of Some Centered Polygonal Numbers by Using the Division Algorithm

Vol. 10, 2017, 83-88

ISSN: 2349-0632 (P), 2349-0640 (online) Published 11 December 2017

www.researchmathsci.org

DOI: http://dx.doi.org/10.22457/jmi.v10a11

83

Journal of

Evaluation of Some Centered Polygonal Numbers by

Using the Division Algorithm

V. Pandichelvi1 and P. Sivakamasundari2

1

Department of Mathematics, Urumu Dhanalakshmi College Trichy, Tamilnadu, India

e-mail: [email protected]

2

Department of Mathematics, BDUCC, Lalgudi, Trichy, India. e-mail: [email protected]

Received 2 November 2017; accepted 7 December 2017

Abstract. In this communication, we find centered square number, centered octagonal number, centered dodecagonal number, centered hexadecagonal number and tetradecagonal number by using division algorithm.

Keywords: Division algorithm, centered polygonal numbers AMS Mathematics Subject Classification (2010): 11D99

1. Introduction

In [1] a function A: N →N is defined by A

( )

n =m where m is the smallest natural

number such that

∑

= + n k

k m divides n

1

2 _{. In [2] a function} ) (n

A such that A(n)=m

where mis the smallest natural number such that

∑

∑

= =

+

+ n

k n

k

k m and k m divides n

1 3

1

are given. In this communication, we obtain a function A(n) such that A(n)=m where mis some centered polygonal numbers, such that 2PP_n divides

) (acubical polynomial m+

Notations:

( )

n n

PP_n = 3+ 2 1

be a pentagonal pyramidal number of rank n

1 2 2 2− +

= n n

CS_n be a centered square number of rank

n

1 4

4 2 − +

= n n

84 1

6 6 2− +

= n n

CD_n be a centered dodecagonal number of rank n

1 7

7 2 − +

= n n

CT_n be a centered tetradecagonal number of rank n

1 8 8 2− +

= n n

CH_n be a centered Hexadecagonal number of rank

n

2. Method of analysis

SECTION A: Evaluation of centered square number

Let A: N →N be defined by A

( )

n =m where

m

is the smallest natural number such

that 2PP_n divides m+

(

n3−2n2 +3n−1

)

. If 2PP_ndivides

(

n3−2n2+3n−1

)

,

then A(n)=2n2 −2n+1, otherwise A(n)=

(

2n2−2n+1

)

−r where r is the least

non negative remainder when

(

n3−2n2+3n−1

)

is divided by 2PP_n. Hence Ais defined for all

n

. By division algorithm such remainder is given by

(

n3−2n2 +3n−1

)

−2qPPn where q is the quotient when

(

n3−2n2 +3n−1

)

is divided by 2PP_n and is given by the greatest integer function of

(

)

n PP

n n n

2

1 3 2 2 3_{− + −}

,

That is

   

  

 _{− + −}

=

n PP

n n n q

2

1 3 2 2 3

Hence,

(

) (

)

   

   

+ ×

   

   

+

− + − − − + − − +

= 3

3 2 3 2

3

3 _{2 3 1} 2 3 1 )

( n n

n n

n n n n

n n n n n A

(

)

3

3 2

3 3 2

3

3 2 2 1

1 3 2 )

( n n

n n

n n

n n

n n n

n n n n n

A × +

   

   

+ + − −

    

  

+ + + + − + − + =

(

)

3

3 2

2 2 2 1

1 1 2 2 )

( n n

n n

n n n

n n

A × +

   

   

+ + − − + + − =

0 1 2 2 )

(n = n2− n+ +

A _

  

  

< +

+ −

<2 2 1 1 0

3 2

n n

n n

∵

= + −

=2 2 1

)

(n n2 n

A centered square number.

SECTION B: Evaluation of centered octagonal number

Let A: N →N be defined by A

( )

n =m where

m

is the smallest natural number such

that 2PP_n divides m+

(

2n3−4n2 +6n−1

)

. If 2PP_ndivides

(

2n3−4n2+6n−1

)

,

85

negative remainder when

(

2n3−4n2 +6n−1

)

is divided by 2PP_n. Hence Ais defined for all

n

. By division algorithm such remainder is given by

(

2n3−4n2+6n−1

)

−2qPPn where q is the quotient when

(

2n3−4n2 +6n−1

)

is divided by 2PP_n and is given by the greatest integer function of

(

)

n PP

n n n

2

1 6 4

2 3− 2 + −

,

That is

   

  

 _{− + −}

=

n PP

n n n q

2

1 6 4

2 3 2

.

Hence ,

(

) (

)

   

   

+ ×

   

   

+

− + − −

− + − −

+

= 3

3 2 3 2

3

3 2 4 6 1

1 6 4 2 )

( n n

n n

n n n n

n n n

n n A

3 3

2 3

3 3 2

3

3 4 5 1

1 6 4 2 )

( n n

n n

n n n

n n

n n n

n n n n n

A × +

   

   

+

− + − +

    

  

+ + + + − + − + =

3 3

2 3 2

3 4 5 1

1 1 5 4 )

( n n

n n

n n n n

n n n

A × +

   

   

+ − + − + + + − + − =

3 2

3

1 5 4 )

(n n n n n n

A =− + − + + + _

  

  

< +

− + −

< 4 5 1 1

0

3 2 3

n n

n n n

∵

= + −

=4 4 1

)

(n n2 n

A centered octagonal number.

SECTION C : Evaluation of centered dodecagonal number

Let A: N →N be defined by A

( )

n =m where mis the least natural number such

that 2PP_ndivides m+

(

3n3−6n2 +9n−1

)

. If 2PP_ndivides

(

3n3−6n2+9n−1

)

,

then A(n)=6n2−6n+1, otherwise A(n)=

(

6n2−6n+1

)

−r where r is the least non

negative remainder when

(

3n3−6n2+9n−1

)

is divided by 2PP_n. Hence Ais defined

for all n. By division algorithm such residue is given by

(

3n3−6n2+9n−1

)

−2qPP_n

where q is the quotient when

(

3n3−6n2+9n−1

)

is divided by 2PP_n and is given by

the greatest integer function of

(

)

n

PP n n n

2

1 9 6

3 3 − 2 + −

,

That is

   

  

 _{− + −}

=

n PP

n n n q

2

1 9 6

3 3 2

.

86

(

) (

)

        + ×         + − + − − − + − − + = 3 3 2 3 2 3

3 3 6 9 1

1 9 6 3 )

( n n

n n n n n n n n n n n A

( )

3

3 2 3 3 3 2 3

3 _{3 6 9 1} 2 6 7 1 )

( n n

n n n n n n n n n n n n n n n

A × +

        + − + − +         + + + + − + − + = 3 3 2 3 2

3 _{6 8 1 2} 6 7 1 2

)

( n n

n n n n n n n n n

A × +

        + − + − + + + − + − = 3 2 3 2 2 1 8 6 2 )

(n n n n n n

A =− + − + + + _

      < + − + −

< 6 7 1 1

0 3 2 3 n n n n n ∵ = + −

=6 6 1

)

(n n2 n

A centered dodecagonal number.

Section D: Evaluation of tetradecagonal number

Let A: N →N be defined by A

( )

n =m where mis the smallest natural number such

that 2PP_ndivides m+

(

2n3−7n2 +9n−1

)

. If 2PP_ndivides

(

2n3−7n2 +9n−1

)

,

then A(n)=7n2−7n+1, otherwise A(n)=

(

7n2−7n+1

)

−r where r is the least non

negative remainder when

(

2n3−7n2 +9n−1

)

is divided by 2PP_n. Hence Ais defined for all

n

. By division algorithm such remainder is given by

(

2n3−7n2+9n−1

)

−2qPPn where q is the quotient when

(

2n3−7n2 +9n−1

)

is divided by 2PP_n and is given by the greatest integer function of

(

)

n PP n n n 2 1 9 7

2 3− 2+ −

, That is         _{− + −} = n PP n n n q 2 1 9 7

2 3 2

. Hence ,

(

) (

)

        + ×         + − + − − − + − − + = 3 3 2 3 2 3

3 2 7 9 1

1 9 7 2 )

( n n

n n n n n n n n n n n A

(

)

3

3 2 3 3 3 2 3

3 7 8 1

1 9 7 2 )

( n n

n n n n n n n n n n n n n n n

A × +

        + − + − +         + + + + − + − + =

(

)

3

3 2 3 2

3 _{7 8 1 1} 7 8 1 )

( n n

n n n n n n n n n

A × +

87 3

2 3

1 8 7 )

(n n n n n n

A =− + − + + +

(

)

_

  

  

< +

− + −

< 7 8 1 1

0

3 2 3

n n

n n n

∵

= + −

=7 7 1

)

(n n2 n

A centered tetradecagonal number.

SECTION E: Evaluation of centered hexadecagonal number

Let A: N →N be defined by A

( )

n =m where mis the lowest natural number such

that 2PP_ndivides m+

(

3n3−8n2 +11n−1

)

. If 2PP_ndivides

(

3n3 −8n2 +11n−1

)

,

then A(n)=8n2−8n+1, otherwise A(n)=

(

8n2−8n+1

)

−r where r is the least non

negative remainder when

(

3n3−8n2+11n−1

)

is divided by 2PP_n. Hence Ais defined for all

n

. By division algorithm such remainder is given by

(

3n3−8n2 +11n−1

)

−2qPPn where q is the quotient when

(

3n3−8n2+11n−1

)

is divided by 2PP_n and is given by the greatest integer function of

(

)

n PP

n n

n 2

1 11 8

3 3− 2 + −

,

That is

   

  

 _{− + −}

=

n PP

n n n q

2

1 11 8

3 3 2

.

Hence ,

(

) (

)

   

   

+ ×

   

   

+

− + − −

− + − −

+

= 3

3 2 3 2

3

3 3 8 11 1

1 11 8

3 )

( n n

n n

n n n n

n n n

n n A

( )

3

3 2 3

3 3 2

3

3 2 8 9 1

1 11 8

3 )

( n n

n n

n n n

n n

n n n

n n n n n

A × +

   

   

+ − + − +

    

  

+ + +

+ − + − + =

3 3

2 3 2

3 8 9 1

2 1 10 8

2 )

( n n

n n

n n n n

n n n

A × +

   

   

+ − + − + + + − + − =

3 2

3

2 2 1 10 8

2 )

(n n n n n n

A =− + − + + + _

  

  

< +

− + −

< 8 9 1 1

0

3 2 3

n n

n n n

∵

= + −

=8 8 1

)

(n n2 n

A centered hexadecagonal number.

3. Conclusion

In this communication, we find the centered polygonal numbers through the divisibility algorithm. In this manner, one can find the some special numbers through the divisibility algorithm.

REFERENCES

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2. M.A.Gopalan and V.Pandichelvi, Two remarkable functions on divisibility, Acta Ciencia Indica, XXXIV M (3) (2008) 1161.

3. van Nivan, Herbert , S.Zuckermann and L.Montgomery, An Introduction to Theory of Numbers, John Wiley and Sons Inc, New York 2004.

4. A.Choudhary, The general solution of a cubic Diophantine equation, Ganita, 50(1) (1999) 1-4.