Trios and Extensions of Trios

Aim:

My aim is to study the topic of trios and work on from that, to discover patterns and links. I will study trios using permutations as I believe these to have the most regularity and a much easier sequence to work with.

I am going to investigate how many trios there are for the number 5. Once I have worked this out, I will work out how many trios there are for the number 6; then for the number 7 and so on. I will then try to make a connection between the above numbers and come up with a formula which can be used to work out a trio.

I will then explore further into this question, by investigating ‘quartets’, ‘quintets’, ‘sextet’ etc.

What is a trio?

A trio is a set of 3 numbers greater than zero, that when added together make another number. For example, 1 + 2 + 2 = 5, also 2 + 1 + 2 = 5.

TRIOS:

I will, first of all, investigate the number of trios there are for the number 5:

Trios for 5:

1        2        2

2        1        2

2        2        1

1        1        3

1        3        1

3        1        1

Trios for 6:

2        2        2

3        2        1

3        1        2

2        1        3

2        3        1

4        1        1

1        4        1

1        1        4

1        2        3

1        3        2

Trios for 7:

3        2        2                        5        1        1

2        3        2                        1        5        1

2        2        3                        1        1        5

1        2        4                        1        3        3

1        4        2                        3        1        3

2        4        1                        3        3        1

2        1        4                        

4        2        1                        

4        1        2                        

The total number of cans in this model is 15; hence it is a triangular number. The next triangular number is 21 which is achieved by adding 6 cans to the bottom of the above model, and the next triangular number is made by adding 7 onto that.

The formula for a triangular number is:

Since the number of trios for a certain number is always triangular, I can adapt the formula for triangular numbers to come up with a formula for trios. That formula is:

(n-1)(n-2)

2

I will now test this formula:

How many trios for the number 7?

(7-1)(7-2)

6 x 5 = 30                30 / 2 = 15

This proves that my formula works! I know this because I know that 15 is a triangular number and I also know that it is true that there are 15 trios for 7 because I have worked it out on page 1.

Now that I have proven my formula for trios, I can progress and work out the formula for ‘quartets’.

QUARTETS:

I will first of all investigate how many quartets there are for the number 5:

Quartets for 5:

2        1        1        1

1        2        1        1

1        1        2        1

1        1        1        2

Quartet for 6:

2        2        1        1

2        1        2        1

2        1        1        2

1        2        2        1

1        2        1        2

1        1        2        2

3        1        1        1

1        3        1        1

1        1        3        1

1        1        1        3

Quartets for 7:

1        1        1        4

1        1        4        1

1        4        1        1

4        1        1        1

1        1        2        3

1        1        3        2

1        2        3        1

1        3        2        1

1        3        1        2

1        2        1        3

2        2        2        1

2        2        1        2

2        1        2        2

1        2        2        2

3        2        1        1

3        1        2        1

3        1        1        2

2        3        1        1

2        1        3        1

2        1        1        3

Number of quartets:

5        →        4

6        →        10

7        →        20

Difference between numbers of quartets:

4        →        10        =        6

10        →        20        =        10

6 and 10 are triangular numbers. This suggests to me that there are connections between the formula for trios, which is relative to the formula for triangular numbers, and the formula for quartets. This means that the formula for trios can, once again, be adapted to fit in with quartets. The formula for quartets is:

   (n-1)(n-2)(n-3)

I know how many set of brackets to use for each device by obeying this rule: The number of brackets to be used is 1 less than the device selected. Therefore, if I wanted to know how many ‘tenios’ there are for the number 17, I would apply the following formula: (n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)(n-9)

This would give me an answer of 11440!!

Conclusion:

By looking at all of my findings, and the sequences and formulas, I can say that I believe I have completed the task successfully.

        My aim was to discover the formula and the theory behind trios, and the relationship they have with other devices. I have experimented with quartets, quintets, and sextets etc. quite thoroughly and have found links between them all. I have discovered that every device tracks back to triangular numbers because their formulas have been derived from the formula for trios.

        I was very glad when I discovered the formula for trios and quartets, but I was euphoric when I discovered the ‘word formula’ which enables me to work out the formula for any device, whether it is ‘tenios’ or ‘hundredios’!

        This enables me to work out, without having to physically count out how many trios, quintets, octets etc. a given number has.

Overall, I believe that I have completed a significant and thorough investigation, and I have achieved from it, much more than I expected as far as formulas and sequences go. I have discovered, for myself, the link between triangular numbers and trios, which I think, essentially, was the aim of the investigation.