Deriving the Z Numbers

So here we are needing to find out how to derive the z numbers so that we can write out the summation formula for any arbitrary power. My first method for finding the equations took a lot of work until I realized that the equations follow a regular pattern much to my relief.

Here are notes on my original method and then next I will detail my current method.

Original Method

To find the formula for the z (the nth column coefficient for the power summation formulas):

  1. Find the lcm of the denominators.
  2. Make a list of results from lcm divided by the denominators multiplied by the numerators.
  3. Make a list of those results divided by n (n being the nth power).
  4. Find formula for those numbers.

Example z4, lcm is 720

4          24          6
5          60          12
6          120        20
7          210        30
8          336        42

The formula for the third column is the summation of the evens with n being substituted with (n-2) to account for the alignment and then that formula is multiplied with the actual n from column one and then everything is divided by 720.

Current Method

Starting from z3 the formulas follow a regular pattern. The formula for the xth z number can be written out by using the formula:

z_x(n)=(n-(x-3)2)(n-(x-3)2-1)...(n-0)C

So now we just need a way of deriving the constant for a given z formula. I have a method for doing this though it relies on knowing the summation formula for a power with enough terms to contain one in the column for the z constant we are looking for.

Finding the Constant

To demonstrate this method we will work through an example for finding the constant for the 10th z number.

We first write out the formula with a place holder value of \frac{1}{1} for C.

z_{10}(n)= \frac{(n-14)(n-13)...(n-1)(n-0)1}{1}

Next we need the formula for a power summation with at least 10 terms, the first of which is:

\sum_{i=1}^{n}i^{16}= \frac{n^{17}}{17}+\frac{n^{16}}{2}+\frac{4n^{15}}{3}-\frac{14n^{13}}{3}+\frac{52n^{11}}{3}-\frac{143n^9}{3}+\frac{260n^7}{3}-\frac{1382n^5}{15}+\frac{140n^3}{3}-\frac{3617n}{510}

With a C of \frac{1}{1} z_{10}(16)=20922789888000

So to get the correct value of C we take the 10th coefficient from our summation formula and divide that by the number we just got.

C= \frac{\frac{3617}{510}}{20922789888000} = \frac{3617}{10670622842880000}

The final formula:

z_{10}(n)= \frac{(n-14)(n-13)...(n-1)(n-0)3617}{10670622842880000}

So now we must ask the question what is the formula for the C values of the z functions? Well I had to know so I googled and there was something about Bernoulli numbers and a whole bunch of mathematical notation that is meaningless to me. Though that is good news because I can go about finding a formula for the C values much as I have with the rest of my math, a fun riddle to be solved.

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