I am going to investigate how changing the number of tiles at the centre of a pattern, will affect the number of border tiles I
Contents
Page 1 ~ Introduction
Page 2 ~ Patterns for 2 centre tiles
Page 5 ~ Patterns for 3 centre tiles
Page 8 ~ Patterns for 4 centre tiles
Page 11 ~ Patterns for 5 centre tiles
Page 14 ~ Summary for patterns with a single row of centre tiles
Page 16 ~ Patterns for 4 centre tiles
Page 19 ~ Patterns for 6 centre tiles
Page 22 ~ Patterns for 8 centre tiles
Page 25 ~ Patterns for 10 centre tiles
Page 28 ~ Summary for pattern with a double row of centre tiles
Page 30 ~ Summary for single and double rows of tiles
Page 31 ~ Patterns for 6 centre tiles
Page 34 ~ Patterns for 9 centre tiles
Page 37 ~ Patterns for 12 centre tiles
Page 40 ~ Patterns for 15 centre tiles
Page 43 ~ Summary for patterns with a triple row of centre tiles
Page 44 ~ Conclusion
Borders Coursework
Introduction
For my experiment I am going to investigate how changing the number of tiles at the centre of a pattern, will affect the number of border tiles I will need. I will do this to find patterns and a formula, to link back to each set of patterns. Each formula will be tested by using a larger border, but with the same number of centre tiles, this will ensure my formula is correct. I will then try to find a general formula, that will enable me to predict the border for any size centre tiles. I will also do the same for the total tiles in the pattern.
Key
N ~ Pattern
B ~ Outer border tiles
T ~ Total Tiles
C ~ Centre Tiles
D ~ First difference
H ~ Height
W ~ Width
/ ~ x(y
~ Prediction
For two centre tiles
Table 1:
Pattern: N
2
3
4
5
Border Tiles: B
6
0
4
8
22
+4 +4 +4 +4
B= 4N + 2
To get 4N at the start of the formula, you have to put the added outer borders into a table ( see above ). Each pattern number is counted as N and the squares in that outer most border is counted as B. These numbers change as each outer border gets bigger. Every pattern number with only centre tiles has no border, so it has not been included in the sequence to make my formulae. You notice you add 4 each time so that 4 is the multiplier, so you have 4 ??N which equals 4N. To get the +2 you subtract the difference of 4 from the original number of tiles you started with, 6. This will give you the equation, 6 - 4, which equals +2 and this is then added to give the pattern formula, 4N + 2.
Testing my Pattern formula.
I will test my formula to see if it is producing the same values as in Table 1.
N=1, B= 4(1)+ 2 = 6.
N=2, B= 4(2)+ 2 = 10.
N=3, B= 4(3)+ 2 = 14.
I have proved that this formula gives me the same sequence of outer border tiles as I have counted from my diagrams.
Prediction
I can use my pattern formula to predict how many border tiles I would need for any pattern number, with 2 centre tiles. To test this hypothesis I will draw a larger diagram and count the outer border tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
B= 4(7) + 2
= 28 +2
= 30
So I expect that pattern number 7 will have 30 outer border tiles.
I have drawn pattern number 7 and counted 30 border tiles, which confirms that my formula is correct.
Total Tiles
I am also going to investigate whether I can find a formula which give the total amount of tiles in my pattern, T, provided I have the pattern number.
Table 2
Pattern: N
2
3
4
5
Total Tiles: T
8
8
32
50
72
+10 +14 +18 +22
+4 +4 +4
Table two shows the pattern number and total amount of tiles in that particular pattern. There is no constant first difference until the second difference (which is constant). This tells you the formula is based on N². The second difference is 4 and you divide this by 2, to get 2. This is the multiplier of N², so you should now have 2N². To get the rest of the formula you first find the multiplier of the 2N( when you subtract it from the original total tiles number, 8 - 4 = 4, which you multiply by, N, to get 4N. You then take this away from the original number to give you the final number to add, 2. So the formulae for total tiles, T, for patterns with two centre tiles is:
T= 2N² + 4N + 2
Testing my formula.
I will test my formula to see if it is produces the same values as in Table 2.
N=1, T = 2(1)² + 4(1) + 2 = 8.
N=2, T = 2(2)² + 4(2) + 2 = 18.
N=3, T = 2(3)² + 4(3) + 3 = 32.
I have proved that this formula gives me the same sequence of total tiles as I have counted from my diagrams.
Prediction
I can use my Total Tiles formula to predict how many tiles I would need for any pattern number, with 2 centre tiles. To test this hypothesis I will draw a larger diagram and count the total amount of tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
T= 2N² + 4N + 2
= 2(7)² + 4(7) + 2
= 2(49) + 30
= 128
So I expect that pattern number 7 will have 128 tiles.
I have drawn pattern 8 and counted 98 tiles, this confirms that my formula is correct.
For three centre tiles
Table 1
Pattern: N
2
3
4
5
Border Tiles: B
8
2
6
20
24
+4 +4 +4 +4
B= 4n + 4
To get 4N at the start of the formula, you have to put the added outer borders into a table ( see above ). Each pattern number is counted as N and the squares in that outer most border is counted as B. These numbers change as each outer border gets bigger. Every pattern number with only centre tiles has no border, so it has not been included in the sequence to make my formulae. You notice you add 4 each time so that 4 is the multiplier, so you have 4 ??N which equals 4N. To get the +4 you subtract the difference of 4 from the original number of tiles you started with, 8. This will give you the equation, 8 - 4, which equals +4 and this is then added to give the pattern formula, 4N + 4.
Testing my Pattern formula.
I will test my formula to see if it is producing the same values as in Table 1.
N=1, B= 4(1) + 4 = 8.
N=2, B= 4(2) + 4 = 12.
N=3, B= 4(3) + 4 = 16.
I have proved that this formula gives me the same sequence of outer border tiles as I have counted from my diagrams.
Prediction
I can use my pattern formula to predict how many border tiles I would need for any pattern number, with 3 centre tiles. To test this hypothesis I will draw a larger diagram and count the outer border tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
B= 4(7) + 4
= 32
So I expect that pattern number 7 will have 32 outer border tiles.
I have drawn pattern number 7 and counted 32 border tiles, which confirms that my formula is correct.
Total Tiles
I am also going to investigate whether I can find a formula which give the total amount of tiles in my pattern, T, provided I have the pattern number.
Table 2
Pattern: N
2
3
4
5
Total Tiles: T
1
23
39
59
83
+12 +16 +20 +24
+4 +4 +4
Table two shows the pattern number and total amount of tiles in that particular pattern. There is no constant first difference until the second difference (which is constant). This tells you the formula is based on N². The second difference is 4 and you divide this by 2, to get 2. This is the multiplier of N², so you should now have 2N². You will then need to find the second part of the equation, this will be the difference between 2N( subtracted from original number and 2(2)( subtracted from the original number. The final part of the equation is to add on the number of centre tiles. So the formula for total tiles, T, for patterns with 3 centre tiles is:
T= 2N² + 6N + 3
Testing my formula.
I will test my formula to see if it is produces the same values as in Table 2.
N=1, T = 2(1)² + 6(1) + 3 = 11.
N=2, T = 2(2)² + 6(2) + 3 = 23.
N=3, T = 2(3)² + 6(1) + 3 = 39.
This formula gives me the same sequence of total tiles as I counted from my diagrams.
Prediction
I can use my Total Tiles formula to predict how many tiles I would need for any pattern number, with 3 centre tiles. To test this hypothesis I will draw a larger diagram and count the total amount of tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
T= 2N² + 6N + 3
= 2(7)² + 6(7) + 3
= 2(49) + 45
= 143
So I expect that pattern number 7 will have 143 tiles.
I have drawn pattern 7 and counted 143 tiles, this confirms that my formula is correct.
For four centre tiles
Table 1
Pattern: N
2
3
4
5
Border Tiles: B
0
4
8
22
26
+4 +4 +4 +4
B= 4N + 6
To get 4N at the start of the formula, you have to put the added outer borders into a table ( see above ). Each pattern number is counted as N and the squares in that outer most border is counted as B. These numbers change as each outer border gets bigger. Every pattern number with only centre tiles has no border, so it has not been included in the sequence to make my formulae. You notice you add 4 each time so that 4 is the multiplier, so you have 4 ??N which equals 4N. To get the +6 you subtract the difference of 4 from the original number of tiles you started with, 10. This will give you the equation, 10 - 4, which equals +6 and this is then added to give the pattern formula, 4N + 6.
Testing my Pattern formula.
I will test my formula to see if it is producing the same values as in Table 1.
N=1, B= 4(1) + 6 = 10.
N=2, B= 4(2) + 6 = 14.
N=3, B= 4(3) + 6 = 18.
I have proved that this formula gives me the same sequence of outer border tiles as I have counted from my diagrams.
Prediction
I can use my pattern formula to predict how many border tiles I would need for any pattern number, with 4 centre tiles. To test this hypothesis I will draw a larger diagram and count the outer border tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
B= 4(7) + 6
= 28 +6
= 34
So I expect that pattern number 7 will have 34 outer border tiles.
I have drawn pattern number 7 and counted 30 border tiles, which confirms that my formula is correct.
Total Tiles
I am also going to investigate whether I can find a formula which give the total amount of tiles in my pattern, T, provided I have the pattern number.
Table 2
Pattern: N
2
3
4
5
Total Tiles: T
4
28
46
68
94
+14 +18 +22 +26
+4 +4 +4
Table two shows the pattern number and total amount of tiles in that particular pattern. There is no constant first difference until the second difference (which is constant). This tells you the formula is based on N². The second difference is 4 and you divide this by 2, to get 2. This is the multiplier of N², so you should now have 2N². You will then need to find the second part of the equation, this will be the difference between 2N( subtracted from original number and 2(2)( subtracted from the original number. The final part of the equation is to add on the number of centre tiles. So the formulae for total tiles, T, for patterns with 4 centre tiles is:
T= 2N² + 8N + 4
Testing my formula.
I will test my formula to see if it is produces the same values as in Table 2.
N=1, T = 2(1)² + 4(1) + 6 = 14.
N=2, T = 2(2)² + 4(2) + 6 = 28.
N=3, T = 2(3)² + 4(3) + 6 = 46.
I have proved that this formula gives me the same sequence of total tiles as I have counted from my diagrams.
Prediction
I can use my Total Tiles formula to predict how many tiles I would need for any pattern number, with 4 centre tiles. To test this hypothesis I will draw a larger diagram and count the total amount of tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
T= 2N²
= 2(7)²
= 2(49)
= 98
So I expect ...
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I have proved that this formula gives me the same sequence of total tiles as I have counted from my diagrams.
Prediction
I can use my Total Tiles formula to predict how many tiles I would need for any pattern number, with 4 centre tiles. To test this hypothesis I will draw a larger diagram and count the total amount of tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
T= 2N²
= 2(7)²
= 2(49)
= 98
So I expect that pattern number 7 will have 98 tiles.
I have drawn pattern 8 and counted 98 tiles, this confirms that my formula is correct.
For five centre tiles
Table 1
Pattern: N
2
3
4
5
Border Tiles: B
2
6
20
24
28
+4 +4 +4 +4
B= 4N + 8
To get 4N at the start of the formula, you have to put the added outer borders into a table ( see above ). Each pattern number is counted as N and the squares in that outer most border is counted as B. These numbers change as each outer border gets bigger. Every pattern number with only centre tiles has no border, so it has not been included in the sequence to make my formulae. You notice you add 4 each time so that 4 is the multiplier, so you have 4 ??N which equals 4N. To get the +8 you subtract the difference of 4 from the original number of tiles you started with, 12. This will give you the equation, 12 - 4, which equals +8 and this is then added to give the pattern formula, 4N + 8.
Testing my Pattern formula.
I will test my formula to see if it is producing the same values as in Table 1.
N=1, B= 4(1) + 8 = 12.
N=2, B= 4(2) + 8 = 16.
N=3, B= 4(3) + 8 = 20.
I have proved that this formula gives me the same sequence of outer border tiles as I have counted from my diagrams.
Prediction
I can use my pattern formula to predict how many border tiles I would need for any pattern number, with 4 centre tiles. To test this hypothesis I will draw a larger diagram and count the outer border tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
B= 4(7) + 8
= 28 +8
= 36
So I expect that pattern number 7 will have 36 outer border tiles.
I have drawn pattern number 7 and counted 36 border tiles, which confirms that my formula is correct.
Total Tiles
I am also going to investigate whether I can find a formula which give the total amount of tiles in my pattern, T, provided I have the pattern number.
Table 2
Pattern: N
2
3
4
5
Total Tiles: T
7
33
53
75
05
+16 +20 +24 +28
+4 +4 +4
Table two shows the pattern number and total amount of tiles in that particular pattern. There is no constant first difference until the second difference (which is constant). This tells you the formula is based on N². The second difference is 4 and you divide this by 2, to get 2. This is the multiplier of N², so you should now have 2N². You will then need to find the second part of the equation, this will be the difference between 2N( subtracted from original number and 2(2)( subtracted from the original number. The final part of the equation is to add on the number of centre tiles. So the formulae for total tiles, T, for patterns with 5 centre tiles is:
T= 2N² + 10N + 5
Testing my formula.
I will test my formula to see if it is produces the same values as in Table 2.
N=1, T = 2(1)² + 10(1) + 5 = 17.
N=2, T = 2(2)² + 10(2) + 5 = 33.
N=3, T = 2(3)² + 10(3) + 5 = 53.
I have proved that this formula gives me the same sequence of total tiles as I have counted from my diagrams.
Prediction
I can use my Total Tiles formula to predict how many tiles I would need for any pattern number, with 5 centre tiles. To test this hypothesis I will draw a larger diagram and count the total amount of tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
T= 2N² + 10N + 5
= 2(7)² + 10(7) + 5
= 2(49) + 75
= 173
So I expect that pattern number 7 will have 173 tiles.
I have drawn pattern 7 and counted 173 tiles, this confirms that my formula is correct.
Summary for patterns with a single row of centre tiles
Border Tiles
I have finished my formula for, B, to make it easier to find hidden patterns which might help me. The factorised expressions form another pattern, as I thought, and I can express this new algebraic sequence for border tiles, in terms of both, N, the pattern number and, C, the number of centre tiles. If I factorise 2 out, I am left with brackets that all have 2N as the first term and a number added which is one less than 'C', i.e. (C - 1). I can now form a complete formula to predict the number of border tiles required.
The complete formula is:-
B = 2(2N + C - 1)
Test
I will now test my complete formula using my previous data, to check that it works correctly.
If C = 5 and N = 3 If C = 3 and N = 5
Then B = 2(2 x 5 + 3 - 1) Then B = 2(2 x 5 + 3 - 1)
B = 20 B = 24
If C = 2 and N = 4
Then B = 2(2 x 4 + 2 - 1)
B = 18
These three checks are correct so I think my formula is working. I will now use it to make predictions for diagram number, N = 8.
Prediction
If C = 8 and N = 2. If C = 8 and N = 5
B = 2(2 x 5 + 8 - 1) B = 2(2 x 2 + 8 - 1)
B = 22 B = 34
I will now find the formula to predict the number of tiles in the first border. Every pattern with only centre tiles has no border, so it has not been included in the sequence to make my formulae.
+2 +2 +2 +2
D = 2C + 2
This formula allows me to predict the very first difference in my sequences, which I have not included in my formulae. The very first difference is always different to all the other differences and would, therefore, prevent me from developing my formulae. So this formula allows me to calculate the number of border tiles in the very first border of any given pattern with, C, centre tiles.
Total Tiles
I have finished my formula for, T to make it easier to find hidden patterns which might help me. The quadratic expressions form a pattern, as I thought, 2N( is recurrent, the last number which is added on, is equal to the number of centre tiles in the pattern and the number times by, N, is double that number. I can express this new algebraic sequence for total tiles, in terms of both, N, the pattern number and, C, the number of centre tiles. I can now form a complete formula to predict the amount of total tiles required.
The complete formula is:-
T = 2N( + 2CN + C
Test
I will now test my complete formula using my previous data, to check that it works correctly.
If C = 2 and N = 3 If C = 3 and N = 5
Then T = 2(3)( + 2(2)(3) + 2 Then T = 2(5)( + 2(3)(5) + 3 T = 32 T = 83
If C = 5 and N = 5
Then T = 2(5)( + 2(5)(5) + 5
T = 105
These three checks are correct so I think my formula is working. I will now use it to make predictions for diagram number, N = 7.
Prediction
If C = 5 and N = 7.
Then T = 2(7)( + 2(5)(7) + 5
T = 173
I have referred back to my diagrams with 5 centre squares and this once again proves my formula is correct.
For four centre tiles in a square
Table 1
Pattern: N
2
3
4
5
Border Tiles: B
8
2
6
20
24
+4 +4 +4 +4
B= 4N + 4
To get 4N at the start of the formula, you have to put the added outer borders into a table ( see above ). Each pattern number is counted as N and the squares in that outer most border is counted as B. These numbers change as each outer border gets bigger. Every pattern number with only centre tiles has no border, so it has not been included in the sequence to make my formulae. You notice you add 4 each time so that 4 is the multiplier, so you have 4 ??N which equals 4N. To get the +4 you subtract the difference of 4 from the original number of tiles you started with, 8. This will give you the equation, 8 - 4, which equals +4 and this is then added to give the pattern formula, 4N + 4.
Testing my Pattern formula.
I will test my formula to see if it is producing the same values as in Table 1.
N=1, B= 4(1) + 4 = 8.
N=2, B= 4(2) + 4 = 12.
N=3, B= 4(3) + 4 = 16.
I have proved that this formula gives me the same sequence of outer border tiles as I have counted from my diagrams.
Prediction
I can use my pattern formula to predict how many border tiles I would need for any pattern number, with 4 centre tiles. To test this hypothesis I will draw a larger diagram and count the outer border tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
B= 4(7) + 4
= 28 +4
= 32
So I expect that pattern number 7 will have 32 outer border tiles.
I have drawn pattern number 7 and counted 32 border tiles, which confirms that my formula is correct.
Total Tiles
I am also going to investigate whether I can find a formula which give the total amount of tiles in my pattern, T, provided I have the pattern number.
Table 2
Pattern: N
2
3
4
5
Total Tiles: T
2
24
40
60
84
+12 +16 +20 +24
+4 +4 +4
Table two shows the pattern number and total amount of tiles in that particular pattern. There is no constant first difference until the second difference (which is constant). This tells you the formula is based on N². The second difference is 4 and you divide this by 2, to get 2. This is the multiplier of N², so you should now have 2N². You will then need to find the second part of the equation, this will be the difference between 2N( subtracted from original number and 2(2)( subtracted from the original number. The final part of the equation is to add on the number of centre tiles. So the formulae for total tiles, T, for patterns with four centre tiles is:
T= 2N² + 6N + 4
Testing my formula.
I will test my formula to see if it is produces the same values as in Table 2.
N=1, T = 2(1)² + 6(1) + 4 = 10.
N=2, T = 2(2)² + 6(2) + 4 = 24.
N=3, T = 2(3)² + 6(3) + 4 = 40.
I have proved that this formula gives me the same sequence of total tiles as I have counted from my diagrams.
Prediction
I can use my Total Tiles formula to predict how many tiles I would need for any pattern number, with 4 centre tiles. To test this hypothesis I will draw a larger diagram and count the total amount of tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
T= 2N² + 6N + 4
= 2(7)² + 6(7) + 4
= 2(49) + 46
= 144
So I expect that pattern number 7 will have 144 tiles.
I have drawn pattern 7 and counted 144 tiles, this confirms that my formula is correct.
For six centre tiles in a square
Table 1
Pattern: N
2
3
4
5
Border Tiles: B
0
4
8
22
26
+4 +4 +4 +4
B= 4N + 6
To get 4N at the start of the formula, you have to put the added outer borders into a table ( see above ). Each pattern number is counted as N and the squares in that outer most border is counted as B. These numbers change as each outer border gets bigger. Every pattern number with only centre tiles has no border, so it has not been included in the sequence to make my formulae. You notice you add 4 each time so that 4 is the multiplier, so you have 4 ??N which equals 4N. To get the +6 you subtract the difference of 4 from the original number of tiles you started with, 10. This will give you the equation, 10 - 4, which equals +6 and this is then added to give the pattern formula, 4N + 6.
Testing my Pattern formula.
I will test my formula to see if it is producing the same values as in Table 1.
N=1, B= 4(1) + 6 = 10.
N=2, B= 4(2) + 6 = 14.
N=3, B= 4(3) + 6 = 18.
I have proved that this formula gives me the same sequence of outer border tiles as I have counted from my diagrams.
Prediction
I can use my pattern formula to predict how many border tiles I would need for any pattern number, with 4 centre tiles. To test this hypothesis I will draw a larger diagram and count the outer border tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
B= 4(7) + 6
= 28 +6
= 34
So I expect that pattern number 7 will have 34 outer border tiles.
I have drawn pattern number 7 and counted 34 border tiles, which confirms that my formula is correct.
Total Tiles
I am also going to investigate whether I can find a formula which give the total amount of tiles in my pattern, T, provided I have the pattern number.
Table 2
Pattern: N
2
3
4
5
Total Tiles: T
6
30
48
70
96
+14 +18 +22 +26
+4 +4 +4
Table two shows the pattern number and total amount of tiles in that particular pattern. There is no constant first difference until the second difference (which is constant). This tells you the formula is based on N². The second difference is 4 and you divide this by 2, to get 2. This is the multiplier of N², so you should now have 2N². You will then need to find the second part of the equation, this will be the difference between 2N( subtracted from original number and 2(2)( subtracted from the original number. The final part of the equation is to add on the number of centre tiles. So the formulae for total tiles, T, for patterns with six centre tiles is:
T= 2N² + 8N + 6
Testing my formula.
I will test my formula to see if it is produces the same values as in Table 2.
N=1, T = 2(1)² + 8(1) + 6 = 16.
N=2, T = 2(2)² + 8(2) + 6 = 30.
N=3, T = 2(3)² + 8(3) + 6 = 48.
I have proved that this formula gives me the same sequence of total tiles as I have counted from my diagrams.
Prediction
I can use my Total Tiles formula to predict how many tiles I would need for any pattern number, with 6 centre tiles. To test this hypothesis I will draw a larger diagram and count the total amount of tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
T= 2N² + 8N + 6
= 2(7)² + 8(7) + 6
= 2(49) + 62
= 160
So I expect that pattern number 7 will have 160 tiles.
I have drawn pattern 7 and counted 160 tiles, this confirms that my formula is correct.
For eight centre tiles in a square
Table 1
Pattern: N
2
3
4
5
Border Tiles: B
2
6
20
24
28
+4 +4 +4 +4
B= 4N + 8
To get 4N at the start of the formula, you have to put the added outer borders into a table ( see above ). Each pattern number is counted as N and the squares in that outer most border is counted as B. These numbers change as each outer border gets bigger. Every pattern number with only centre tiles has no border, so it has not been included in the sequence to make my formulae. You notice you add 4 each time so that 4 is the multiplier, so you have 4 ??N which equals 4N. To get the +8 you subtract the difference of 4 from the original number of tiles you started with, 10. This will give you the equation, 12 - 4, which equals +8 and this is then added to give the pattern formula, 4N + 8.
Testing my Pattern formula.
I will test my formula to see if it is producing the same values as in Table 1.
N=1, B= 4(1) + 8 = 12.
N=2, B= 4(2) + 8 = 16.
N=3, B= 4(3) + 8 = 20.
I have proved that this formula gives me the same sequence of outer border tiles as I have counted from my diagrams.
Prediction
I can use my pattern formula to predict how many border tiles I would need for any pattern number, with 8 centre tiles. To test this hypothesis I will draw a larger diagram and count the outer border tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
B= 4(7) + 8
= 28 +8
= 36
So I expect that pattern number 7 will have 36 outer border tiles.
I have drawn pattern number 7 and counted 36 border tiles, which confirms that my formula is correct.
Total Tiles
I am also going to investigate whether I can find a formula which give the total amount of tiles in my pattern, T, provided I have the pattern number.
Table 2
Pattern: N
2
3
4
5
Total Tiles: T
20
36
56
80
08
+16 +20 +24 +28
+4 +4 +4
Table two shows the pattern number and total amount of tiles in that particular pattern. There is no constant first difference until the second difference (which is constant). This tells you the formula is based on N². The second difference is 4 and you divide this by 2, to get 2. This is the multiplier of N², so you should now have 2N². You will then need to find the second part of the equation, this will be the difference between 2N( subtracted from original number and 2(2)( subtracted from the original number. The final part of the equation is to add on the number of centre tiles. So the formulae for total tiles, T, for patterns with eight centre tiles is:
T= 2N² + 10N + 8
Testing my formula.
I will test my formula to see if it is produces the same values as in Table 2.
N=1, T = 2(1)² + 10(1) + 8 = 20.
N=2, T = 2(2)² + 10(2) + 8 = 36.
N=3, T = 2(3)² + 10(3) + 8 = 56.
I have proved that this formula gives me the same sequence of total tiles as I have counted from my diagrams.
Prediction
I can use my Total Tiles formula to predict how many tiles I would need for any pattern number, with 8 centre tiles. To test this hypothesis I will draw a larger diagram and count the total amount of tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
T= 2N² + 10N + 8
= 2(7)² + 10(7) + 8
= 2(49) + 78
= 176
So I expect that pattern number 7 will have 177 tiles.
I have drawn pattern 7 and counted 177 tiles, this confirms that my formula is correct.
For ten centre tiles in a square
Table 1
Pattern: N
2
3
4
5
Border Tiles: B
4
8
22
26
30
+4 +4 +4 +4
B= 4N + 10
To get 4N at the start of the formula, you have to put the added outer borders into a table ( see above ). Each pattern number is counted as N and the squares in that outer most border is counted as B. These numbers change as each outer border gets bigger. Every pattern number with only centre tiles has no border, so it has not been included in the sequence to make my formulae. You notice you add 4 each time so that 4 is the multiplier, so you have 4 ??N which equals 4N. To get the +10 you subtract the difference of 4 from the original number of tiles you started with, 14. This will give you the equation, 14 - 4, which equals +10 and this is then added to give the pattern formula, 4N + 10.
Testing my Pattern formula.
I will test my formula to see if it is producing the same values as in Table 1.
N=1, B= 4(1) + 10 = 14.
N=2, B= 4(2) + 10 = 18.
N=3, B= 4(3) + 10 = 22.
I have proved that this formula gives me the same sequence of outer border tiles as I have counted from my diagrams.
Prediction
I can use my pattern formula to predict how many border tiles I would need for any pattern number, with 8 centre tiles. To test this hypothesis I will draw a larger diagram and count the outer border tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
B= 4(7) + 10
= 28 +10
= 38
So I expect that pattern number 7 will have 38 outer border tiles.
I have drawn pattern number 7 and counted 38 border tiles, which confirms that my formula is correct.
Total Tiles
I am also going to investigate whether I can find a formula which give the total amount of tiles in my pattern, T, provided I have the pattern number.
Table 2
Pattern: N
2
3
4
5
Total Tiles: T
20
36
56
80
08
+16 +20 +24 +28
+4 +4 +4
Table two shows the pattern number and total amount of tiles in that particular pattern. There is no constant first difference until the second difference (which is constant). This tells you the formula is based on N². The second difference is 4 and you divide this by 2, to get 2. This is the multiplier of N², so you should now have 2N². You will then need to find the second part of the equation, this will be the difference between 2N( subtracted from original number and 2(2)( subtracted from the original number. The final part of the equation is to add on the number of centre tiles. So the formulae for total tiles, T, for patterns with two centre tiles is:
T= 2N² + 12N + 10
Testing my formula.
I will test my formula to see if it is produces the same values as in Table 2.
N=1, T = 2(1)² + 12(1) + 10 = 20.
N=2, T = 2(2)² + 12(2) + 10 = 36.
N=3, T = 2(3)² + 12(3) + 10 = 56.
I have proved that this formula gives me the same sequence of total tiles as I have counted from my diagrams.
Prediction
I can use my Total Tiles formula to predict how many tiles I would need for any pattern number, with 10 centre tiles. To test this hypothesis I will draw a larger diagram and count the total amount of tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
T= 2N² + 12N + 10
= 2(7)² + 12(7) + 10
= 2(49) + 94
= 192
So I expect that pattern number 7 will have 192 tiles.
I have drawn pattern 7 and counted 192 tiles, this confirms that my formula is correct.
Summary for patterns with a double row of centre tiles
Border Tiles
I have finished my formula for, B, to make it easier to find hidden patterns which might help me. The factorised expressions form another pattern, as I thought, and I can express this new algebraic sequence for border tiles, in terms of both, N, the pattern number and, C, the number of centre tiles. I have not drawn a separate diagram for 2x1, as it is the same as 1x2, which I have already drawn in the first section of data. From this summary table I can see a pattern that is recurring: 2 is a factor, then each bracket is made up of (2N + C/2) I can now form a complete formula to predict the number of border tiles required.
The complete formula is:-
B = 2(2N + C/2 )
This can be times by 2 and simplified:-
B = 4N + C
Test
I will now test my complete formula using my previous data, to check that it works correctly.
If C = 6 and N = 5 If C = 4 and N = 3
Then B = 4(5) + 6 Then B = 4(3) + 4
B = 26 B = 16
If C = 10 and N = 2
Then B = 4(2) + 10
B = 18
These three checks are correct so I think my formula is working. I will now use it to make predictions for diagram number, N = 8.
Prediction
If C = 12 and N = 8 If C = 4 and N = 8
B = 4(8) + 12 B = 4(8) + 4
B = 44 B = 36
I will now find the formula to predict the number of tiles in the first border. Every pattern with only centre tiles has no border, so it has not been included in the sequence to make my formulae.
+2 +2 +2 +2
D = 2C + 2
This formula allows me to predict the very first difference in my sequences, which I have not included in my formulae. The very first difference is always different to all the other differences and would, therefore, prevent me from developing my formulae. So this formula allows me to calculate the number of border tiles in the very first border of any given pattern with, C, centre tiles.
Total Tiles
I have finished my formula for, T, to make it easier to find hidden patterns which might help me. The quadratic expressions form a pattern, as I thought, 2N( is recurrent, the last number which is added on, is equal to the number of centre tiles in the pattern and the number times by, N, is two more than ,C. I can express this new algebraic sequence for total tiles, in terms of both, N, the pattern number and, C, the number of centre tiles. I can now form a complete formula to predict the amount of total tiles required.
The complete formula is:-
T = 2N( + (C + 2)N + C
Test
I will now test my complete formula using my previous data, to check that it works correctly.
If C = 2 and N = 3 If C = 6 and N = 5
Then T = 2(3)( + (2 + 2)(3) + 2 Then T = 2(5)( + (6 + 2)(5) + 6 T = 32 T = 96
If C = 12 and N = 5
Then T = 2(5)( + (12 + 2)(5) + 12
T = 132
These three checks are correct so I think my formula is working. I will now use it to make predictions for diagram number, N = 8.
Prediction
If C = 4 and N = 8.
Then T = 2(8)( + (4 + 2)(8) + 4
T = 180
I have referred back to my diagrams with 4 centre squares and this once again proves my formula is correct.
Summary of results for single row of tiles and double rows of tiles.
Border Tiles
From my first two sets of results for design types, 1 x W and 2 x W, I can easily see a pattern. I will predict that for design type, 3 x W, the formula will be:
B = 2(2N + C + 1)
This is because the C term is increasing by one each time, so I will just add one to C to get C + 1. Using H x W as the abbreviation for design type, then I can write a final linear formula so as I could predict the border tiles in any pattern, with any amount of centre tiles.
Final Formula
B = 2(2N + C + H - 2)
The 'H - 2' term can be seen above in my table since: 1 - 2 = -1
2 - 2 = 0
3 - 2 = +1
This I will prove in my, '3 x W', row of tiles patterns.
Total Tiles
From my first two sets of results for design types, 1 x W and 2 x W, I can easily see a pattern. I will predict that for design type, 3 x W, the formula will be:
T = 2N( + (2/3C + 4)N + C
This is because the end numbers are C, the C term in the bracket is a multiple of 3. There is a gap of 2 so the common difference is 2/3. If I work backwards from 6 and subtract 2 I get 4, which is added on to 2/3C, giving me (2/3C + 4). This is the number before N. I can write a final Quadratic formula so as I could predict the total tiles in any pattern, with any amount of centre tiles.
Final Formula
T = 2N( + 2(W + H - 1)N + C
I got this formula from:-
T = 2N( + (2C / H + 2H - 2)N + C
T = 2N( + (2(HxW) + 2H - 2)N +C
H
= 2N( + (2W + 2H -2)N + C
T = 2N( + 2(W + H -1)N + C
This I will prove in my, '3 x W', row of tiles patterns.
For six centre tiles in a rectangle
Table 1
Pattern: N
2
3
4
5
Border Tiles: B
0
4
8
22
26
+4 +4 +4 +4
B= 4N + 6
To get 4N at the start of the formula, you have to put the added outer borders into a table ( see above ). Each pattern number is counted as N and the squares in that outer most border is counted as B. These numbers change as each outer border gets bigger. Every pattern number with only centre tiles has no border, so it has not been included in the sequence to make my formulae. You notice you add 4 each time so that 4 is the multiplier, so you have 4 ??N which equals 4N. To get the +6 you subtract the difference of 4 from the original number of tiles you started with, 10. This will give you the equation, 10 - 4, which equals +6 and this is then added to give the pattern formula, 4N + 6.
Testing my Pattern formula.
I will test my formula to see if it is producing the same values as in Table 1.
N=1, B= 4(1) + 6 = 10.
N=2, B= 4(2) + 6 = 14.
N=3, B= 4(3) + 6 = 18.
I have proved that this formula gives me the same sequence of outer border tiles as I have counted from my diagrams.
Prediction
I can use my pattern formula to predict how many border tiles I would need for any pattern number, with 6 centre tiles. To test this hypothesis I will draw a larger diagram and count the outer border tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
B= 4(7) + 6
= 28 +6
= 34
So I expect that pattern number 7 will have 34 outer border tiles.
I have drawn pattern number 7 and counted 34 border tiles, which confirms that my formula is correct.
Total Tiles
I am also going to investigate whether I can find a formula which give the total amount of tiles in my pattern, T, provided I have the pattern number.
Table 2
Pattern: N
2
3
4
5
Total Tiles: T
6
30
48
70
96
+14 +18 +22 +26
+4 +4 +4
Table two shows the pattern number and total amount of tiles in that particular pattern. There is no constant first difference until the second difference (which is constant). This tells you the formula is based on N². The second difference is 4 and you divide this by 2, to get 2. This is the multiplier of N², so you should now have 2N². You will then need to find the second part of the equation, this will be the difference between 2N( subtracted from original number and 2(2)( subtracted from the original number. The final part of the equation is to add on the number of centre tiles. So the formulae for total tiles, T, for patterns with two centre tiles is:
T= 2N² + 8N + 6
Testing my formula.
I will test my formula to see if it is produces the same values as in Table 2.
N=1, T = 2(1)² + 8(1) + 6 = 16.
N=2, T = 2(2)² + 8(2) + 6 = 30.
N=3, T = 2(3)² + 8(3) + 6 = 48.
I have proved that this formula gives me the same sequence of total tiles as I have counted from my diagrams.
Prediction
I can use my Total Tiles formula to predict how many tiles I would need for any pattern number, with 6 centre tiles. To test this hypothesis I will draw a larger diagram and count the total amount of tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
T= 2N² + 8N + 6
= 2(7)² + 8(7) + 6
= 2(49) + 62
= 160
So I expect that pattern number 7 will have 160 tiles.
I have drawn pattern 7 and counted 160 tiles, this confirms that my formula is correct.
For nine centre tiles in a rectangle
Table 1
Pattern: N
2
3
4
5
Border Tiles: B
2
6
20
24
28
+4 +4 +4 +4
B= 4N + 8
To get 4N at the start of the formula, you have to put the added outer borders into a table ( see above ). Each pattern number is counted as N and the squares in that outer most border is counted as B. These numbers change as each outer border gets bigger. Every pattern number with only centre tiles has no border, so it has not been included in the sequence to make my formulae. You notice you add 4 each time so that 4 is the multiplier, so you have 4 ??N which equals 4N. To get the +8 you subtract the difference of 4 from the original number of tiles you started with, 12. This will give you the equation, 12 - 4, which equals +8 and this is then added to give the pattern formula, 4N + 8.
Testing my Pattern formula.
I will test my formula to see if it is producing the same values as in Table 1.
N=1, B= 4(1) + 8 = 12.
N=2, B= 4(2) + 8 = 16.
N=3, B= 4(3) + 8 = 20.
I have proved that this formula gives me the same sequence of outer border tiles as I have counted from my diagrams.
Prediction
I can use my pattern formula to predict how many border tiles I would need for any pattern number, with 9 centre tiles. To test this hypothesis I will draw a larger diagram and count the outer border tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
B= 4(7) + 8
= 28 +8
= 36
So I expect that pattern number 7 will have 36 outer border tiles.
I have drawn pattern number 7 and counted 36 border tiles, which confirms that my formula is correct.
Total Tiles
I am also going to investigate whether I can find a formula which give the total amount of tiles in my pattern, T, provided I have the pattern number.
Table 2
Pattern: N
2
3
4
5
Total Tiles: T
21
37
57
81
09
+16 +20 +24 +28
+4 +4 +4
Table two shows the pattern number and total amount of tiles in that particular pattern. There is no constant first difference until the second difference (which is constant). This tells you the formula is based on N². The second difference is 4 and you divide this by 2, to get 2. This is the multiplier of N², so you should now have 2N². You will then need to find the second part of the equation, this will be the difference between 2N( subtracted from original number and 2(2)( subtracted from the original number. The final part of the equation is to add on the number of centre tiles. So the formulae for total tiles, T, for patterns with two centre tiles is:
T= 2N² + 10N + 9
Testing my formula.
I will test my formula to see if it is produces the same values as in Table 2.
N=1, T = 2(1)² + 10(1) + 9 = 21.
N=2, T = 2(2)² + 10(2) + 9 = 37.
N=3, T = 2(3)² + 10(3) + 9 = 57.
I have proved that this formula gives me the same sequence of total tiles as I have counted from my diagrams.
Prediction
I can use my Total Tiles formula to predict how many tiles I would need for any pattern number, with 9 centre tiles. To test this hypothesis I will draw a larger diagram and count the total amount of tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
T= 2N² + 10N + 9
= 2(7)² + 10(7) + 9
= 2(49) + 79
= 177
So I expect that pattern number 7 will have 177 tiles.
I have drawn pattern 7 and counted 177 tiles, this confirms that my formula is correct.
For twelve centre tiles in a rectangle
Table 1
Pattern: N
2
3
4
5
Border Tiles: B
4
8
22
26
30
+4 +4 +4 +4
B= 4N + 10
To get 4N at the start of the formula, you have to put the added outer borders into a table ( see above ). Each pattern number is counted as N and the squares in that outer most border is counted as B. These numbers change as each outer border gets bigger. Every pattern number with only centre tiles has no border, so it has not been included in the sequence to make my formulae. You notice you add 4 each time so that 4 is the multiplier, so you have 4 ??N which equals 4N. To get the +10 you subtract the difference of 4 from the original number of tiles you started with, 14. This will give you the equation, 14 - 4, which equals +10 and this is then added to give the pattern formula, 4N + 10.
Testing my Pattern formula.
I will test my formula to see if it is producing the same values as in Table 1.
N=1, B= 4(1) + 10 = 14.
N=2, B= 4(2) + 10 = 18.
N=3, B= 4(3) + 10 = 22.
I have proved that this formula gives me the same sequence of outer border tiles as I have counted from my diagrams.
Prediction
I can use my pattern formula to predict how many border tiles I would need for any pattern number, with 12 centre tiles. To test this hypothesis I will draw a larger diagram and count the outer border tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
B= 4(7) + 10
= 28 +10
= 38
So I expect that pattern number 7 will have 38 outer border tiles.
I have drawn pattern number 7 and counted 38 border tiles, which confirms that my formula is correct.
Total Tiles
I am also going to investigate whether I can find a formula which give the total amount of tiles in my pattern, T, provided I have the pattern number.
Table 2
Pattern: N
2
3
4
5
Total Tiles: T
26
44
66
92
24
+18 +22 +26 +32
+4 +4 +4
Table two shows the pattern number and total amount of tiles in that particular pattern. There is no constant first difference until the second difference (which is constant). This tells you the formula is based on N². The second difference is 4 and you divide this by 2, to get 2. This is the multiplier of N², so you should now have 2N². You will then need to find the second part of the equation, this will be the difference between 2N( subtracted from original number and 2(2)( subtracted from the original number. The final part of the equation is to add on the number of centre tiles. So the formulae for total tiles, T, for patterns with two centre tiles is:
T= 2N² + 12N + 12
Testing my formula.
I will test my formula to see if it is produces the same values as in Table 2.
N=1, T = 2(1)² + 12(1) + 12 = 26.
N=2, T = 2(2)² + 12(2) + 12 = 44.
N=3, T = 2(3)² + 12(3) + 12 = 66.
I have proved that this formula gives me the same sequence of total tiles as I have counted from my diagrams.
Prediction
I can use my Total Tiles formula to predict how many tiles I would need for any pattern number, with 12 centre tiles. To test this hypothesis I will draw a larger diagram and count the total amount of tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
T= 2N² + 12N + 12
= 2(7)² + 12(7) + 12
= 2(49) + 96
= 194
So I expect that pattern number 7 will have 194 tiles.
I have drawn pattern 7 and counted 194 tiles, this confirms that my formula is correct.
For fifteen centre tiles in a rectangle
Table 1
Pattern: N
2
3
4
5
Border Tiles: B
6
20
24
28
32
+4 +4 +4 +4
B= 4N + 12
To get 4N at the start of the formula, you have to put the added outer borders into a table ( see above ). Each pattern number is counted as N and the squares in that outer most border is counted as B. These numbers change as each outer border gets bigger. Every pattern number with only centre tiles has no border, so it has not been included in the sequence to make my formulae. You notice you add 4 each time so that 4 is the multiplier, so you have 4 ??N which equals 4N. To get the +12 you subtract the difference of 4 from the original number of tiles you started with, 16. This will give you the equation, 16 - 4, which equals +12 and this is then added to give the pattern formula, 4N + 12.
Testing my Pattern formula.
I will test my formula to see if it is producing the same values as in Table 1.
N=1, B= 4(1) + 12 = 16.
N=2, B= 4(2) + 12 = 20.
N=3, B= 4(3) + 12 = 24.
I have proved that this formula gives me the same sequence of outer border tiles as I have counted from my diagrams.
Prediction
I can use my pattern formula to predict how many border tiles I would need for any pattern number, with 15 centre tiles. To test this hypothesis I will draw a larger diagram and count the outer border tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
B= 4(7) + 12
= 28 +12
= 40
So I expect that pattern number 7 will have 40 outer border tiles.
I have drawn pattern number 7 and counted 40 border tiles, which confirms that my formula is correct.
Total Tiles
I am also going to investigate whether I can find a formula which give the total amount of tiles in my pattern, T, provided I have the pattern number.
Table 2
Pattern: N
2
3
4
5
Total Tiles: T
31
51
75
03
35
+20 +24 +28 +32
+4 +4 +4
Table two shows the pattern number and total amount of tiles in that particular pattern. There is no constant first difference until the second difference (which is constant). This tells you the formula is based on N². The second difference is 4 and you divide this by 2, to get 2. This is the multiplier of N², so you should now have 2N². You will then need to find the second part of the equation, this will be the difference between 2N( subtracted from original number and 2(2)( subtracted from the original number. The final part of the equation is to add on the number of centre tiles. So the formulae for total tiles, T, for patterns with two centre tiles is:
T= 2N² + 14N + 15
Testing my formula.
I will test my formula to see if it is produces the same values as in Table 2.
N=1, T = 2(1)² + 14(1) + 15 = 31.
N=2, T = 2(2)² + 14(2) + 15 = 51.
N=3, T = 2(3)² + 14(3) + 15 = 75.
I have proved that this formula gives me the same sequence of total tiles as I have counted from my diagrams.
Prediction
I can use my Total Tiles formula to predict how many tiles I would need for any pattern number, with 12 centre tiles. To test this hypothesis I will draw a larger diagram and count the total amount of tiles to check my formula. If I was to let the pattern number be N= 7, (See below) then:
T= 2N² + 14N + 15
= 2(7)² + 14(7) + 15
= 2(49) + 113
= 211
So I expect that pattern number 7 will have 211 tiles.
I have drawn pattern 7 and counted 211 tiles, this confirms that my formula is correct.
Summary for patterns with a triple row of centre tiles
Border Tiles
I have finished my formula for, B, to make it easier to find hidden patterns which might help me. The factorised expressions form another pattern, as I thought, and I can express this new algebraic sequence for border tiles, in terms of both, N, the pattern number and, C, the number of centre tiles. From this summary table I can see a pattern that is recurring: 2 is a factor, then each bracket is made up of (2N + C + 1). I can now form a complete formula to predict the number of border tiles required.
The complete formula is :-
B = 2(2N + C +1)
Test
I will now test my complete formula using my previous data, to check that it works correctly.
If C = 6 and N = 3 If C = 15 and N = 5
Then B = 2(2 x 3 + 6 + 1) Then B = 2(2 x 15 + 5 + 1)
B = 18 B = 32
If C = 12 and N = 4
Then B = 2(2 x 4 + 12 +1)
B = 26
These three checks are correct so I know my formula is working. I will now use it to make predictions for diagram number, N = 8.
Prediction
If C = 15 and N = 7. If C = 15 and N = 8
B = 2(2 x 7 + 15 + 1) B = 2(2 x 8 + 15 + 1)
B = 40 B = 44
I will now find the formula to predict the number of tiles in the first border. Every pattern with only centre tiles has no border, so it has not been included in the sequence to make my formulae.
+2 +2 +2 +2
D = 2C + 2
This formula allows me to predict the very first difference in my sequences, which I have not included in my formulae. The very first difference is always different to all the other differences and would, therefore, prevent me from developing my formulae. So this formula allows me to calculate the number of border tiles in the very first border of any given pattern with, C, centre tiles.
Total Tiles
I have finished my formula for, T to make it easier to find hidden patterns which might help me. The quadratic expressions form a pattern, as I thought, 2N( is recurrent, the last number which is added on, is equal to the number of centre tiles in the pattern and the number times by, N, is double that number. I can express this new algebraic sequence for total tiles, in terms of both, N, the pattern number and, C, the number of centre tiles. I can now form a complete formula to predict the amount of total tiles required.
The complete formula is:-
T = 2N( + (2 C + 4)N + C
3 .
Test
I will now test my complete formula using my previous data, to check that it works correctly.
If C = 6 and N = 3 If C = 15 and N = 5
Then T = 2(3)( + (2 (6) + 4)(3) + 6 Then T = 2(5)( + (2(15) + 4)(5) +15
3 3
T = 48 T = 135
If C = 12 and N = 4
Then T = 2(4)( + (2(12) +4)(4) + 12
3
T = 92
These three checks are correct so I know my formula is working. I will now use it to make predictions for diagram number, N = 7.
Prediction
If C = 15 and N = 7.
Then T = 2(7)( + (2(15) + 4)(7) + 15
3
T = 211
This once again proves my formula is correct.
Conclusion
I have proved that my formulas work and that there is a pattern between them. This proves that my prediction for border tiles B = 2(2N + C + H - 2) and my prediction for total tiles T = 2N( + 2(W + H - 1)N + C was correct, therefore I could predict the total tiles and border tiles for any pattern with any number of centre tiles.
E.g.
Border Tiles
Border Tiles in the third pattern of a diagram with nine centre tiles.
If the pattern number was N=3, then:
B = 2(2N + C + H - 2)
B = 2(2(3) + 9 + 3 - 2)
B = 32
I have checked this against my previous data and have concluded that the data matches. I can therefore conclude that I can predict any border tile pattern for any number of centre tiles.
Total Tiles
Total tiles in the fifth pattern of a diagram with 9 centre tiles.
If N=5, C=9, W=3, H=3
T = 2N( + 2(W + H - 1)N + C
T = 2(3)( + 2(5 + 3 - 1)5 + 9
T = 50 + 50 + 9
T = 109
I have checked this against my previous data and have concluded that the data matches. I can therefore conclude that I can predict any amount of total tiles pattern for any number of centre tiles.