Stellar Numbers. In this folio task, we are going to determine difference geometric shapes, which lead to special numbers.

The results demonstrated from the calculator is y=0.5x2+0.5x+0

So,                  y= 12n2+ 12n=12n2+n

y= n2+n2                      ,    y=Un,       x=n

Un=(n2-n)/2

As the general term number can’t be zero or miners, the final general statement will be:

Un=(n2+n)÷2           for             n:n∈Z+   (n≥1)

Also, we can see a graph of the numbers by Select GRAPH,

Now we are able to see the graph which is plotted by the side no. and square no.

Curve is increasing.

6-STELLAR

Now another stellar shape (star) with (6) p vertices, leading to (6) p-stellar numbers will be involved.

Firstly, we are going to plot the stellar shape S5 and S6 by using the computer program “Stellar. Jar”.

 S5                                                                                                                 S6

After that, we can create another table for the stellar numbers,

                                                                        + 12(X)

One more layer will be adding on the privets stellar by increasing the general term number to 1. Thus, the different between Sn and Sn+1 will be the outer shell dots number of Sn. Consider to the diagrams, every new lay been added on the stellar, its outer shell number will increase 12 dots compare to the pervious shell.

Sn is the number of dots that the diagram contains.

When n=1, S1=1

n=2, S2=1+12×(2-1) =1+1×12

n=3, S3=1+12×(2-1)+12×(3-1)=1+3×12

n=4, S4=1+12×(2-1)+12×(3-1)+12×(4-1)=1+6×12

n=5, S5=1+12×(2-1)+12×(3-1)+12×(4-1)+12×(5-1)=1+10×12

n=6, S6=1+12×(2-1)+12×(3-1)+12×(4-1)+12×(5-1)+12×(6-1)=1+15×12

Therefore, we know that the number of dots in each stage above is,

Sn= 1+12(X)

We are now considering the triangular numbers at the first page, we can see the X values are same as the y values in the triangular diagrams.

As the order for X is same as the triangular numbers with general statement,

Un=(n2+n)/2       for       n:n∈Z+ (n≥1)

However, in this situation, When

n=2, X=1, which should be the triangular number when n=1.

n=3, X=3, which should be the triangular number when n=2

n=4, X=6, when should be the triangular number when n=3 etc.

Therefore,

 X =(n-12+n-1)/2

Let’s try the pattern for S7

X=U7=7-12+7-12=21

When n = 7, S7 = 1+12×(2-1)+12×(3-1)+12×(4-1)+12×(5-1)+12×(6-1)+12×(7-1) = 1+21×12

          S7 = 253

Now, subs X =(n-12+n-1)/2 into formula Sn=1+X×12,

The general formula for p(6) Stellar number will be

Sn=1+12n-12+n-1

=6n2-n+1

Again, test it by using n=5 (chose randomly),
Sn=6n2-n+1

=652-5+1

=121

The value of S5 is same as the S5 value in Table on the page 5.

Again, check the answer with the TI-84Puls calculator,

Follow the steps 1 – 3, the calculator should show,

The result is y=6x2-6x+1                 for                          y=Sn,       x=n

Thus, the formula can be shown as Sn=6n2-6n+1

This is same as result that worked out by hand

Sn=6(n2-n)+1

To see the graph, we can Select GRAPH again,

P (5)-STELLAR SHAPE

By using computer software “stellar”, we can plot the shapes below,

 

                            S5                                                         S6

Here are the first six stages of the P (5)-STELLAR SHAPE, and we can now create a table to demonstrate the number of dots in each of the stage,

One more layer will be adding on the privets stellar by increasing the general term number to 1. Thus, the different between F1 and Fn+1 will be the outer shell dots number (Fn). Every new lay been added on the stellar, its outer shell number will have 10 more dots.

Fn is the number of dots that the diagram contains.

When n=1, F1=1

n=2, F2=1+10×(2-1) =1+1×10

n=3, F3=1+10×(2-1)+10×(3-1)=1+3×10

n=4, F4=1+10×(2-1)+10×(3-1)+10*(4-1)=1+6×10

n=5, F5=1+10×(2-1)+10×(3-1)+10×(4-1)+10×(5-1)=1+10×10

n=6, F6=1+10×(2-1)+10×(3-1)+10×(4-1)+10×(5-1)+10×(6-1)=1+15×10 etc.

When n=n, Fn=1+X×10

X (the number that multiple to 10), has the following relationship with n when p=6

It also follows the general formula X =(n-12+n-1)/2

Try the pattern F7,

X=U7=7-12+7-12=21

When n = 7, F7 = 1+10×(2-1)+10×(3-1)+10×(4-1)+10×(5-1)+10×(6-1)+10×(7-1) = 1+21×10

          F7 = 210

Substitute X =(n-12+n-1)/2 into formula Fn=1+X×10,

Hence the general formula for Stellar number will be

Fn=1+10((n-1)2+n-1)2

=5n2-n+1

Now, test the formula by using n=4,
Fn=5n2-n+1

=542-4+1

=61

The F4 value is also same as the F4 value in Table on the page 7.

Again, check the answer by using the TI-84Puls calculator with the same method which is used for the triangular numbers.

The result shown is y=5x2-5x+1                           for                                   y=Sn,       x=n

So, the formula can be shown as Fn=5n2-5n+1

This is same as result that worked out by hand,

Fn=5(n2-n)+1

The graph will be,

P (7)-STELLAR SHAPE

By using computer software “stellar”, we can plot the shapes below,

 

                                S5                                                                 S6

On the pervious page, there are the first six stages of the P (7)-STELLAR SHAPE, and we can now create a table to demonstrate the number of dots in each of the stage,

One more layer will be adding on the privets stellar by increasing the general term number to 1. Thus, the different between K1 and Kn+1 will be the outer shell dots number (Kn). Every new lay been added on the stellar, its outer shell number will have more dots.

Kn is the number of dots that the diagram contains. When,

n=1, K1=1

n=2, K2=1+14×(2-1) =1+1×14

n=3, K3=1+14×(2-1)+14*(3-1)=1+3×14

n=4, K4=1+14×(2-1)+14×(3-1)+14*(4-1)=1+6×14

n=5, K5=1+14×(2-1)+14×(3-1)+14×(4-1)+14×(5-1)=1+10×14

n=6, K6=1+14×(2-1)+14×(3-1)+14×(4-1)+14×(5-1)+14×(6-1)=1+15×14 etc.

When n=n, Kn=1+X×14

X (the number that multiple to 14), has the following relationship with n when p=5

It also follows the general formula X =(n-12+n-1)/2

Try this pattern for K7.

X=U7=7-12+7-12=21

When n = 7, K7 = 1+14×(2-1)+14×(3-1)+14×(4-1)+14×(5-1)+14×(6-1)+14×(7-1) = 1+21×14

           E7 = 294

Subs X =(n-12+n-1)/2 into formula Kn=1+X×14,

Thus, the general formula for stellar number will be,

Kn=1+14((n-1)2+n-1)2

=7n2-n+1

Test formula by using n=2,
Kn=7n2-n+1

=722-2+1 =15

The K2 now is same as the K2 value in the table before.

Again, check the answer by using the TI-84Puls calculator with the same method that use for the triangular numbers.

The result is y=7x2-7x+1                   for                          y=Kn,       x=n

Thus, the formula can write down as Kn=7n2-7n+1

This is same as result that worked out by hand,

Kn=7(n2-n)+1

P(8) stellar number.

        

Consider the stellar number diagram up to M6,

One more layer will be adding on the privets stellar by increasing the general term number to 1. Thus, the different between M1and Mn+1 will be the outer shell dots number (Mn). Every new lay been added on the stellar, its outer shell number will have 16 more dots.

Mn is the number of dots that the diagram contains. When,

n=1, M1=1

n=2, M2=1+16×(2-1) =1+1×16

n=3, M3=1+16×(2-1)+16×(3-1)=1+3×16

n=4, M4=1+16×(2-1)+16×(3-1)+16×(4-1)=1+6×16

n=5, M5=1+16×(2-1)+16×(3-1)+16×(4-1)+16×(5-1)=1+10×16

n=6, M6=1+16×(2-1)+16×(3-1)+16×(4-1)+16×(5-1)+16×(6-1)=1+15×16 etc.

When n=n, Mn=1+X×16

X (the number that multiple to 16), has the following relationship with n when p=7

It also follows the general formula X =(n-12+n-1)/2

Try the pattern M7,

X=U7=7-12+7-12=21

When n = 7, M7 = 1+16×(2-1)+16×(3-1)+16×(4-1)+16×(5-1)+16×(6-1)+16×(7-1) = 1+21×16

          M7 = 337

Subs X =(n-12+n-1)/2 into formula Mn=1+X×16,

Thus, the general formula for Stellar number in this case will be

Mn=1+16((n-1)2+n-1)2

=8n2-n+1

Let’s test the formula by using n=4,
Mn=8n2-n+1

=842-4+1 

=97

The M4 now is same as the M4 value in the table before.

Again, check the answer by using the TI-84Puls calculator with the same method that use for the triangular numbers.

The calculator will show,

The result shown in the calculator is y=8x2-8x+1            for                 y=Mn,       x=n

So, the formula can be shown as Mn=8n2-8n+1

And this is same as result that worked out by hand,

Mn=8(n2-n)+1

Then, analysis the 5-STELLAR, 6-STELLAR and 6-STELLAR, hence, produce a general statement for all of them. The statement will be worked out in terms of p and n, it is aiming to demonstrate the sequence of p –stellar numbers for any value of p at stage Sn.

Firstly, we have to test the general statements for each of the STELLARS is working certainly and we are going to take n=3 for the test,

P (5)-STELLAR

Equal to each other (=31), therefore, it is valid.

Sn=5n2-5n+1S3=5(3)2-53+1=31

P (6)-STELLAR

Equal to each other (=37), therefore, it is valid.

Sn=6n2-6n+1Sn=6(3)2-63+1=37

 P (7)-STELLAR

Equal to each other (=43), therefore, it is valid.

Sn=7n2-7n+1Sn=7(3)2-73+1=43

After that, we are able to determine that, when

P (5) - stellar numbers, p=5,   Fn=5(n2-n)+1

P (6) - stellar numbers, p=6,   Sn=6(n2-n)+1

P (7) - stellar numbers, p=7,   On=7(n2-n)+1

P (8) - stellar numbers, p=8,   Mn=8(n2-n)+1

As the result, all the general formulas have a constant statement of Yn=x(n2-n)+1,

And the value of x is same as the p value of the stellar numbers.

∴for a p-stellar number,

Yn=p(n2-n)+1

Because of that, we are now testing the general formula with p(9) – stellar numbers and p(10) – stellar numbers.

P(9) Stellar number

Refer to the stellar number diagrams up to S6, a table can be list as the following,

Find the general formula by using the TI-84Puls calculator with the same method on the previous pages,

After finished steps from 1 to 3, the calculator should get the answer as diagram shown,

The result shown is y=9x2-9x+1                       for                      y=Y'n,       x=n

Therefore, the formula can be shown as Y'n=9n2-9n+1

And again, it is same as result which had been proved,

Y'n=9(n2-n)+1

P(10) – Stellar number

Refer to the stellar number diagrams up to Y”6, another table can be created below,

Again, test the general formula by using the TI-84Puls calculator with the same method that from the pervious pages.

Followed the steps from 1 to 3, the calculator should get the answer as diagram below,.

The result shown is y=10x2-10x+1              for                y=Y"n,       x=n

So, the formula can be shown as Y"n=10n2-10n+1

And it is same as last result

Y"n=10(n2-n)+1

Scope and Limitation

This general formula only can be applied to p – stellar when p≥3 (p∈Z ).

If we place 3 to be the stellar number, the diagram will be shown like this and it will be still a stellar diagram.

3-stellar

 However, it we place a (p) number which is smaller than 3, using 2 as an example, it cannot form a geometric shape.

And also, a non-integer number (p) cannot form a geometric shape as well. Furthermore, the general statement can also use for any polygon not only the stellar shapes. They can also contain same number of dots which is same as the stellar number.

For example, p( 5 )-stellar shape as an example.

As the diagram show, although we had changed the density of the p (5)-stellar shape, they also contain 5 vertexes and actually contain the same number of dots.

The result is reliable, as the general statement is supported by the equations generated in the cases of a p (5), p (6), p (7) and p (8) -stellar shape. The numerical coefficients in each equation are equal to the corresponding stellar number.

However, all the general term number (n) must be greater than 1 (n≥1，n∈Z+). As a general term number, it can’t be 0, miner’s term and even decimal term.

Sn=pn2-pn+1

(p) Must be greater than 3 and applied to all p-stellar shapes given that p ϵ R.

 

        

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P.T.O. For Appendix.

The methods of use TI-84 Plus calculator to find out the relationship between general term number n and the triangular number (number of dots) in a general statement.

1. Press STAT then ENTER and choose “1.Edit”, (P.T.O to see) From the calculator screen,
2. Type in General term number (n)into “L1” and Triangular number (Un) into “L2” in the order as the diagram shown (P.T.O to see) below. And then press2ND and MODE to choose “QUIT”.

3. Press STAT then to “CALC”, and chose “5. QuadReg”. Finally, press “ENTER” to get the result in diagram in the right hand side.

 

• This method can also use on finding the general formula of the special number other geometric shapes Sn, Kn, Fn or Mn for L2 instead of Un.