. Where k is the multiple needed to make row number equal the numerator (N). I will calculate k and present it in a table along with the numerator and the row number.
To calculate k for the 2nd row:
The rest will be presented in a table.
Table 2 the values of k, the row number and the numerator, these are necessary in the process of deriving the general statement.
As you can see there is a geometric between the row number and the numerator. With increasing row number increases the value of k increase by
for each individual row, as illustrated in table 2. From this one can identify that if k would be doubled then it would be equal to the row number plus one:
To test this I will use the 2nd and 3rd row numbers and calculate k.
If n=2:
If n=3
Also, when k is doubled then the numerator is doubled (so N is doubled) or
This makes sense as you are multiplying by 2 on the left hand side you have to multiply by 2 on the right hand side, to keep the equation balanced
The rest will be presented in a table.
Table 3 an illustration of the relationship between (n+1), n, k, 2k, N and 2N
As you can see the difference between n and 2k is 1 for each individual row, but in order for this general formula to work it needs to be expressed in terms of n, but since we know (from previous calculations) that
we just need to substitute
with
in the equation
Now I will test the general term by calculating the numerator for the 2nd, 3rd and 6th rows. The rest will be presented in a table.
n=2
n=3
n=6
Table 4 the values of the numerator, for each row, calculated with the general statement I have derived
These values match the values in table 1, hence I can conclude that this general term is valid for the numerators. On top of this I will use Excel’s automated trend line feature to insert a second degree polynomial fit to the graph above, with the Pearson correlation coefficient (R) and the equation.
As you can see in the graph above the fit intersects every point on top of that the Pearson correlation coefficient is exactly one. As you can also see the formula matches the formula I have derived for the numerators (albeit in decimal form), hence I can definitely conclude that the general formula for the numerators is valid.
Denominator:
Just as the numerator, the difference in denominators increases by one every consecutive row. However, the value of the denominator changes within each row depending on its placement in that row or its element number.
Table 5 the values of the denominators for each position within said row, the shaded cells show the difference between the denominators of each row and position.
Now using the values from tables 1 and 5 I can find the 6th and 7th row.
Now onto finding the general statement for the denominators, I will use the same procedure by equating D as the denominator, n as the row number and r as the element number. It is worth noting that the first element in each row is r=0.
Table 6 the values of the denominator (D), numerator (N) and the row number (n) for the first element
It appears to be that the denominator is equal to the difference between the numerator and (n-1) of that row. In other words,
as 1 is the element number
Table 7 the value of the denominator (D) for the first element using the derived equation.
This equation seems to be correct, but just to make sure I will try it with the second and third elements as well.
Table 8 the value of the denominator for r=2 using the derived equation.
The results of the denominator, calculated using the derived equation, do not match the ones in Lacsap’s fractions for the second element. But the formula does seems to work for the 2nd term of the 2nd element but not the rest. I will let the difference be x, hence the equation (for the difference between n and r) should be
Table 9 the value difference between the numerators (N) and the difference between x to see if that would result in D
From the table above I observed that x is too small, hence when subtracted with N it will not yield the same result as the denominator. Hence I will make and equation where h is a number required to be multiplied with x to make it large enough, so that when it is subtracted from N it will equal D.
I will use D=4 and 6. x= 1. N= 6 the previous value seemed to work with the unmodified expression. The rest I will present in a table.
Table 10 showing the values of x, r, h and D for the 2nd element
In the table above you can see in the table above multiplying h by x and then subtracting N by that would give the correct numerator result (at least for the 2nd element). I can also recognize that r=h and x= (n-r). Hence I can substitute h and x to get this general statement.
I also know that:
So I can substitute that as well.
For the first row first element (n=1, r=0)
For the second row first element (n=2, r=0)
For the second row second element (n=2, r=1)
For the second row third element (n=2, r=2)
The remaining values will be put in a table
Table 11 validating the denominator equation by plugging in
I can see that the values for the denominators match that in figure one hence I can say that this form is correct.
I also know that:
Table 12 the first ten terms of Lacsap’s fractions, to validate the general statement.
Although this general statement seems to work there are still some limitations, for instance n must be bigger the zero, if n is 0 then the denominator of an element maybe undefined. As well as the fact that there cannot be zero rows as then the pattern will be non-existent. Furthermore n has to be positive, as if n was to be negative then substituted into the equation then the answer would be positive even though it is supposed to be negative.
n=-3
Table 13 the numerator values for the negative values of n
As you can see a negative n value yielded positive results, not to mention the fact that there can never be any negative rows. r must also be greater than zero as there can never be any negative elements or zero elements as zero elements would imply no pattern. n and r have to also be integers as there can never be any half rows or half elements i.e. the values of n and r cannot be fractions (or decimals).