Further more, I clear pattern that the answers increase by the same number, as the number of consecutives in the sum is also visible.
The formula for this is
4n+2
(A pattern of + 4)
My initial theory was incorrect, however this has not come as a surprise, as, as I stated, it is common practice to wait for three terms before a fourth or fifth can be predicted. Now that I have a third term I am going to submit a new theory. I now believe that since n has increased by 1 still, but the variable has increased by 1 then 2, my new formula for five consecutives is 5n+5.
My second theory, however, has been proven correct. The increase in the answers is equal to the number of consecutives in the sum
8+9+10+11=38
4n+2
4x9+2=38
43+44+45+46=178
4n+2
4x44+2=178
The formula for this is
5n+5
7+8+9+10+11=45
5n+5
5x8+5=45
42+43+44+45+46=220
5n+5
5x43+5=220
This time my theory was correct. I can now surmise that for six consecutive numbers the formula shall be 6n+9. From this data a suitable generic formula can be created.
xn-x+n1
where n is the term, x is the number of consecutives, and n1 represents the first term
This formula can be used to create the formula for any number of consecutives.
I shall prove this by using the formulae previously worked out as examples.
xn-x+n1
2n-2+1
2n-1
3n-3+3
3n
4n-4+6
4n+2
5n-5+10
5n+5
I shall also attempt to back up my theory formula
6n-6+15
6n+9
From this it would appear that my formula is accurate and correct, however it is limited by the fact that it requires prior knowledge about the first term. I shall first go through the first five terms for six consecutive numbers, and shall then proceed to find a way of making a formula which requires no prior knowledge.
Using the formula I devised earlier, I shall now attempt to prove my generic formula correct.
3+4+5+6+7+8=33
6n+9
64+9=33
33+34+35+36+37+38=213
6n+9
634+9=213
I now have a formula which can be used to create a formula for any number of consecutives. I am now going to attempt to refine it.
NB The second number in the sequence represents the term.
Results for Formula
I shall now move to prove the four additional formulae in that table.
0+1+2+3+4+5+6=21
71+14=21
0+1+2+3+4+5+6+7=28
81+20=28
0+1+2+3+4+5+6+7+8=36
91+27=36
0+1+2+3+4+5+6+7+8+9=45
101+35=45
It has been proven correct
Exemplar Formula
As explained before I wish to find a generic formula which requires no use of prior knowledge of the first term. This could prove essential in working with extremely large numbers.
Instead of re-writing all of the sums, I am simply going to use the first term collected from prior research.
I believe the formula x2-x can be used to determine the first term of each set of
2
consecutives. I shall now move to prove this.
22-2 = 1
2
32-3 = 3
2
42-4 = 6
2
This proves the formula to be correct.
I shall amalgamate this with the initial formula, so as to replace n1. The formula shall now only require the mathematician to know the term and the number of consecutives.
xn-x+(x2-x)
2
This can be used to work out the formula for any term of any number of consecutives.
I shall now prove that this formula works.
Term 4 of 5 consecutives is 3+4+5+6+7=25
xn-x+(x2-x)
2
54-5+(52-5) = 25
2
This formula is without a doubt 100% accurate.
Numbers Which can be Made Using Consecutives
I am now going to refer back to one of the initial tasks put forward, which was to see which numbers could and could not be made using consecutive sums, and to see which could be made more than once. By using earlier work, a list can be compiled stating which numbers can or cannot be made using variable consecutives. This however would be limited and inefficient, so instead I am going to look at patterns within the numbers, after which I may decide to go back and compile some lists.
Two Consecutive Numbers
No list is required for this, as there is a very simple rule as to which numbers can and cannot be made. All odd numbers can be made. No even numbers can be made.
Three Consecutive Numbers
Again the pattern here is very simple. All answers follow the three times tables. If a number is not part of the three times tables i.e. if it is not divisible by three, then it cannot be made using three consecutive numbers.
Four Consecutive Numbers
There are several patterns which can be defined here. No odd numbers can be made using four consecutive numbers. No number from the four times table i.e. which is divisible by four, can be made using consecutive numbers. In accordance with this, no numbers divisible by eight, twelve, or divisible by any number which is divisible by four, can be made using four consecutive numbers.
Five Consecutive Numbers
There is a clearly definable pattern here. Only numbers divisible by five occur. In addition, all numbers, excluding five, which are divisible by five are included. It would be interesting to see if only numbers divisible by ten appear in a stream of ten consecutive numbers. However, the first term for this which I worked out earlier would immediately disprove this, as 45 is not divisible by 10. Perhaps only numbers divisible by 45 would appear? An investigation into this must be planned and set for a later date.