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## Calculating Alpha Numbers

The Alpha Function is reasonably complicated so it is usually the case that the ordinal and FGH function cannot be predicted for an arbitrary real number input to the function.

This blog will illustrate how to approach calculating Alpha numbers and will partially explain the pseudo code algorithm for how it works.

It will be useful to refer to the blog on Sequence Generating Code for more background information.

## Alpha Number for $$\zeta_0$$

The following Alpha Function is a good place to start. The sequence for the $$\zeta_0$$ ordinal is shown here. The table calculates the Alpha Number required to generate this sequence.

$$\alpha(100.78626719) = J_8(<2,<0,0,<0,1>,<0,0>,0>,0,<0,0>,<0,0>,<0,0>>,2,3)$$

$$= f_{\varphi(2,0)}^2(3) = f_{\zeta_0}^2(3)$$

Integer Range Divisor Cummulative Value
2 n/a n/a n/a 2
0 1 2 2 0
0 6 7 14 0
0 2 3 42 0
1 6 7 294 .0034014
0 1 2 588 0
0 6 7 4116 0
0 1 2 8232 0
0 6 7 57624 0
0 2 3 172872 0
0 6 7 1210104 0
0 2 3 3630312 0
0 6 7 25412184 0
0 2 3 76236552 0
0 6 7 533655864 0
2 (*) adjusted to 0 6 7 3.73 x $$10^{9}$$ 0
3 (*) adjusted to 0 6 7 2.615 x $$10^{10}$$ 0
2.003401361

A final calculation is required with this sum total as follows: $$10^{2.003401361} = 100.7862672$$ to arrive at the Alpha Number. Some sequence numbers are adjusted for fine tuning purposes. The rationale is that the numbers 2 and 3 are minimum values, therefore sequence numbers of 0 are sufficient to represent the minimums in each case.

## Alpha Number for $$\zeta_1$$

Here is the calculation for another ordinal that uses similar calculations.

$$\alpha(100.8426651) = J_8(<2,<0,0,<0,1>,<0,1>,0>,0,<0,0>,<0,0>,<0,0>>,2,3)$$

$$= f_{\varphi(2,1)}^2(3) = f_{\zeta_1}^2(3)$$

Integer Range Divisor Cummulative Value
2 n/a n/a n/a 2
0 1 2 2 0
0 6 7 14 0
0 2 3 42 0
1 6 7 294 .0034014
0 1 2 588 0
1 6 7 4116 0.000242954
0 1 2 8232 0
0 6 7 57624 0
0 2 3 172872 0
0 6 7 1210104 0
0 2 3 3630312 0
0 6 7 25412184 0
0 2 3 76236552 0
0 6 7 533655864 0
2 (*) adjusted to 0 6 7 3.73 x $$10^{9}$$ 0
3 (*) adjusted to 0 6 7 2.615 x $$10^{10}$$ 0
2.003644315

The final calculation in this case is: $$10^{2.003644315} = 100.8426651$$ to arrive at the Alpha Number.

## Alpha Number for $$\zeta_0.2$$

A final calculation illustrates the range possible .

$$\alpha(100.7862763) = J_8(<2,<0,0,<0,1>,<0,0>,0>,0,<0,0>,<0,1>,<0,0>>,2,3)$$

$$= f_{\varphi(2,0).2}^2(3) = f_{\zeta_0.2}^2(3)$$

Integer Range Divisor Cummulative Value
2 n/a n/a n/a 2
0 1 2 2 0
0 6 7 14 0
0 2 3 42 0
1 6 7 294 .0034014
0 1 2 588 0
0 6 7 4116
0 1 2 8232 0
0 6 7 57624 0
0 2 3 172872 0
0 6 7 1210104 0
0 2 3 3630312 0
1 6 7 25412184 3.93512E-08
0 2 3 76236552 0
0 6 7 533655864 0
2 (*) adjusted to 0 6 7 3.73 x $$10^{9}$$ 0
3 (*) adjusted to 0 6 7 2.615 x $$10^{10}$$ 0
2.0034014

Final calculation of this Alpha Number is: $$10^{2.0034014} = 100.7862763$$