Stellar Numbers 2d5j1w

Marcus Silliman IB DP Math

Stellar Numbers Aim: In this investigation, I have examined geometric shapes which lead to special numbers. The simplest example of these are square numbers, 1, 4, 9, 16, which can be represented by squares of side 1, 2, 3 and 4. The following diagrams show a triangular pattern of evenly spaced dots. The number of dots in each diagram are example of triangular numbers (1, 3, 6,…).

Complete the triangular numbers sequence with three more . In order to find the next three in this sequence we must first find a pattern in the increase of the . To do this we must first arrange our data in a table. Term

1 1

2 3

3 6

4 10

5 15

Upon examination of the data, we can see a pattern that is very consistent with the data. Each term is increasing in numerical order: Term Interval

1 1

2 3 +2

Term Interval

2 3

3 6 +3

Marcus Silliman IB DP Math

Term

3 6

Term

4 10

Interval

4 10

Interval

+4

5 15 +5

Assuming that the sequence follows the same pattern, the next three would be: Term

5 15

6 21

Interval

Term

+6 Term Interval

Find a general statement that represents the

6 21

Interval 7 28

7 28 +7

8 36 +8 triangular number in of n.

Using the data above, we are able to further develop the pattern into an equation that can be used to find the term of the triangular numbers.  =1 Recognizing that the value from the previous term  = 3 = 1+2 ( ) plus the term we are looking for ( ), we 

= 6 = 3+3



= 10 = 6+4



= 15 = 10+5

can conclude that the general statement that represents the term in of n is:

Consider stellar (star) shapes with p vertices, leading to p-stellar number. The first four representations for a star with six vertices are shown in the four stages - below. The 6stellar number at each stage is the total of sots in the diagram.

Marcus Silliman IB DP Math

Find the number of dots (i.e. the stellar number) in each stage up to that you can recognize and describe any patterns. Term Stellar #

1

13

37

+24

+12

. Organize the data so

73

+48

+36

+12

+12 +12

Find an expression for the 6-stellar number at stage

121

181

+60

+12

.

Based on the above information, we can conclude that to find the 6-stellar number at stage ( ) we can use the following expression: Find a general statement for the 6-stellar number at stage

in of n.

Based on the information above, we can conclude that a general statement for the 6-stellar number at stage in of n is: ( ) Now repeat the steps above for other values of p.  If the P-stellar number is 8, then the general statement is: The stellar numbers up to stage are: 1, 17, 49, 97, 161, 241

(

)

,

Marcus Silliman IB DP Math

Term Stellar #

1

17

49

+32

+16

97

161

+64

+48

+16

+80

+16

+16

+16

The general statement for the stage

( )

is:

(

 If the P-stellar number is 5, then the general statement is: The stellar numbers up to stage are: 1, 11, 31, 61, 101, 151 Term Stellar # 1 11 31 61

+20

+10

+10

+16

+10

(

+8 +8

( )

81

+32

+24

+8

is:

+50

( )

is:

 If the P-stellar number is 4, then the general statement is: The stellar numbers up to stage are: 1, 9, 25, 49, 81, 121 Term Stellar # 1 9 25 49

The general statement for the stage

151

+40

+10

+8

)

101

+30

+10

The general statement for the stage

241

121

+40

+8

)

Marcus Silliman IB DP Math

Hence, produce the general statement, in of p and n, that generates the sequence of pstellar numbers for any value of p at stage . Based on the examples above, it can be concluded that the general statement that generates the sequence of p-stellar numbers for any value of p at stage is: ( ) Test the validity of the general statement. 

For p-stellar number 4,



For p-stellar number 5,

( ) ( ) ( ) ( )

Discuss the scope or limitations of the general statement. The first term, , is always equal to one because there can only be one center point in a star. The second term can always be found using the equation . In order for the general statement to be correct, . Explain how you arrived at the general statement. The general statement was found with the realization that the stellar number of each inner circle was counted as part of the stellar number of the next term in the series. It is accumulative, like the sum of an arithmetic sequence. In comparison with the precedent term, the sequence increased by a factor of 2 dots. The only exception is that this pattern does not work for the relationship between the 1st and 2nd term of the stellar sequence. Therefore, it could not be a geometric sum. Since the 2nd term could always be found using 2p+1 and the increasing difference between each term was always a factor 2p, then 2p would replace d in the arithmetic sum formula which is

( )(

(

) ). That way, it becomes a constant

even though the difference between the stellar numbers of each term does not seem to be constant when counting them. Word Count: 960