Tn = n + [(n-1) ÷2] x [2 + (n – 2)]
Tn = (2n + n2 –n) ÷ 2
Tn = (n + n2) ÷ 2
Tn = (n ÷ 2) (n + 1)
Table 1: Triangular Numbers
From this table it is possible to see that each progressive units is equal to the sum of the previous unit plus n. Thus forming the simple formula of Tn = Tn-1 + n. In order to write this as a general formula, allowing for the calculation of any unit number without the prior knowledge of the previous sum it is necessary to use the formula Tn = 1 + [(n-1) ÷2] x [4 + (n – 2)] or Tn = (n ÷ 2) (n + 1)
Graph 1: Triangular Numbers- Number of dots in relation to the unit number
Stellar Numbers:
Diagram 2 : 6-Stellar Star
The Arithmetic Progression for the 6-point stellar star is that of twelve (12), or two (2) multiplied by the number of points of the star in such case being that of six (6). Therefore AP is equal to 2p.
Find the expression for the 6-stellar number at stage S7:
Sn = S1 + [(n-1) ÷2] x [2 x 2P + (n – 1 -1) 2P]
Sn = 1 + [(n-1) ÷2] x [2 x 12 + (n – 2) 12]
S7 = 1 + [(7-1) ÷2] x [2 x 12 + (7– 2) 12]
S7 = 1 + [(6) ÷2] x [24 + (5) 12]
S7 = 1 + [3] x [24 + 60]
S7 = 1 + 3 x 84
S7 = 1 + 252
S7 = 253
Find a general statement for the 6-stellar number at stage Sn in terms of n:
Sn = S1 + [(n-1) ÷2] x [2 x 2P + (n – 1 -1) 2P]
Sn = S1 + [(n-1) ÷2] x [2 x 12 + (n – 2) 12]
Sn = S1 + [(n-1) ÷2] x [24 + (n – 2) 12]
Test the general statement for the 6-stellar numbers for multiple stages:
General Statement: Sn = S1 + [(n-1) ÷2] x [24 + (n – 2) 12]
Stage S3:
S3 = S1 + [(3-1) ÷2] x [24 + (3 – 2) 12]
S3 = 1 + [2 ÷2] x [24 + (1) 12]
S3 = 1 + [1] x [24 + 12]
S3 = 1 + [36]
S3 = 37
Using the general statement it is possible to calculate that S3 of the 6- Stellar star has 37 dots in it. From the Diagram of the 6 point Stellar Star it is possible to show the formula is true for S3.
Stage S4:
S4 = S1 + [(4-1) ÷2] x [24 + (4 – 2) 12]
S4 = 1 + [(3) ÷2] x [24 + (2) 12]
S4 = 1 + [(3) ÷2] x [24 + 24]
S4 = 1 + [(3) ÷2] x [48]
S4 = 1 + 72
S4 = 73
The general statement makes it is possible to calculate that S4 of the 6- Stellar star has 73 dots in it. The Diagram of the 6 point Stellar Star it is possible to show the formula is true for S4.
Stage S5:
S5 = S1 + [(5-1) ÷2] x [24 + (5 – 2) 12]
S5 = 1 + [(4) ÷2] x [24 + (3) 12]
S5 = 1 + [2] x [24 + 36]
S5 = 1 + 2 x 60
S5 = 1 + 120
S5 = 121
Using the general statement makes it is possible to calculate that S5 of the 6- Stellar star has 121 dots. The Diagram of the 6 point Stellar Star it is possible to show the formula is true for S5.
Stage S6:
S6 = S1 + [(6-1) ÷2] x [24 + (6 – 2) 12]
S6 = 1 + [(5) ÷2] x [24 + (4) 12]
S6 = 1 + [(5) ÷2] x [24 + 48]
S6 = 1 + 180
S6 = 181
Using the general statement makes it is possible to calculate that S6 of the 6- Stellar star has 161 dots. The Diagram of the 6 point Stellar Star it is possible to show the formula is true for S6.
Table 2: 6- Stellar Numbers
From the progression it is possible to visualise the addition of the previous sum of units to two of number of points (in such case 6 x 2) multiple of n – 1.
Graph 2: 6-Stellar Star Numbers in relation to the unit number
In the case when p, the number of points on the star alters a new general statement must be used. This statement involves determining the new arithmetic progression of such series of numbers and then applying it to the new general statement. For example when only a 5-stellar number, the arithmetic progression changes to that of ten (10), which is double the number of points.
Diagram 3 : 5-Stellar Star
Diagram 4: 4-Stellar Star
Diagram 5: 7-Stellar Star
Find a general statement for any stellar star with different numbered points:
AP: Arithmetic Progression
Sn = S1 + [(n-1) ÷2] x [2 x AP + (n – 1 -1) AP]
Sn = S1 + [(n-1) ÷2] x [2 x AP + (n – 2) AP]
Testing the Validity of this statement
For a five point star:
AP: 10, or (2 multiplied by the number of points)
Sn = S1 + [(n-1) ÷2] x [2 x AP + (n – 2) AP]
Sn = S1 + [(n-1) ÷2] x [2 x 10 + (n – 2) 10]
Stage S2:
S2 = S1 + [(n-1) ÷2] x [2 x 10 + (n – 2) 10]
S2 = 1 + [(2-1) ÷2] x [20 + (2 – 2) 10]
S2 = 1 + [1 ÷2] x [20]
S2 = 1 + [10]
S2 = 11
Using the general statement it is possible to calculate that S2 of the 5- Stellar star has 11 dots in it. From the Diagram of the 5 point Stellar Star it is possible to show the formula is true for S2.
Stage S3:
S3 = S1 + [(n-1) ÷2] x [2 x 10 + (n – 2) 10]
S3 = 1 + [(3-1) ÷2] x [20 + (3 – 2) 10]
S3 = 1 + [(2) ÷2] x [20 + 10]
S3 = 1 + [30]
S3 = 31
Using the general statement it is possible to calculate that S3 of the 5- Stellar star has 31 dots in it. If the diagram of the 5 Stellar Star is continued it is possible to show that the general formula also applies to S3.
For a four point star:
AP: 8, or (2 multiplied by the number of points)
Sn = S1 + [(n-1) ÷2] x [2 x AP + (n – 2) AP]
Sn = S1 + [(n-1) ÷2] x [2 x 8 + (n – 2) 8]
Stage S2:
S2 = S1 + [(n-1) ÷2] x [2 x 8 + (n – 2) 8]
S2 = 1 + [(2-1) ÷2] x [16 + (2 – 2) 8]
S2 = 1 + [1 ÷2] x [16]
S2 = 1 + [8]
S2 = 9
Using the general statement it is possible to calculate that S2 of the 4- Stellar star has 9 dots in it. Diagram 4 shows that there are nine dots in S2.
Stage S3:
S3 = S1 + [(n-1) ÷2] x [2 x 8 + (n – 2) 8]
S3 = 1 + [(3-1) ÷2] x [16 + (3 – 2) 8]
S3 = 1 + [2 ÷2] x [24]
S3 = 25
Using the general statement it is possible to calculate that S3 of the 4- Stellar star has 25 dots in it. The Diagram of the four point star shows that in S3 has 25 dots in it.
For a seven point star:
AP: 14, or (2 multiplied by the number of points)
Sn = S1 + [(n-1) ÷2] x [2 x 14 + (n – 2) 14]
Sn = S1 + [(n-1) ÷2] x [28 + (n – 2) 14]
Stage S2:
S2 = S1 + [(n-1) ÷2] x [28 + (n – 2) 14]
S2 = 1 + [(2-1) ÷2] x [28 + (2 – 2) 14]
S2 = 1 + [1 ÷2] x [28]
S2 = 1 + [14]
S2 = 15
Using the general statement it is possible to calculate that S2 of the 7- Stellar star has 15 dots in it. Diagram 5 supports this result.
Stage S3:
S3 = S1 + [(n-1) ÷2] x [28 + (n – 2) 14]
S3 = 1 + [(3-1) ÷2] x [28 + (3 – 2) 14]
S3 = 1 + [2 ÷2] x [42]
S3 = 43
Using the general statement it is possible to calculate that S3 of the 7- Stellar star has 43 dots. By continuing diagram 5 it is possible to support this outcome of produced from the general statement.
Limitations:
There are many faults with this general statement. Although it applies to many different numbered points of Stellar Stars, for it to be applied it is necessary to determine that the star has the same formation as that of the star which the general statement was created for. In order to determine this it is necessary to determine that the arithmetic progression is equal to double that of the number of points of the star.
The number of points or P cannot be a negative number as this is not possible to form a negative star shape. The stellar shapes also only work when the number of point or vertices are equal to or greater than three (3). Thus it can be concluded that the formula only works for all possible stellar formation shapes.
This general statement also assumes that the first stellar unit is equal to that of one. Meaning that if the sequence of numbers started at any other numerical value the general formula may not apply. Therefore if your sequence did not include the first few layers it is possible the formula would not work as the S1.
