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Alpha Function and $$f_{\epsilon_0}(m)$$

This blog will compare the growth rate of the Alpha Function to the fast growing hierarchy function $$f_{\epsilon_0}(n)$$. Starting with the following calculations at this link:

$$D(m,m,0,...,0) >> f_{\omega^2}(m)$$ with approximately $$m$$ zeros.

and

$$D(m,m,0,...,0,...,0,...,0) >> f_{\omega^2+1}(m)$$ with approximately $$2^m$$ zeros.

$$f_{\omega^2.2}(m)$$

The following Alpha Functions are approximately equal to:

$$\alpha(2^{m+1}) >> D(m,m,0,...,0) >> f_{\omega^2}(m) = \alpha(2^m)$$

Without proof, the following is proposed:

$$\alpha(2^{2^{m+1}}) >> f_{\omega^2+1}(m)$$

Alternatively the following will be approximately correct:

$$\alpha(D(3,m)) = \alpha(2^{m+1}) >> D(m,m,0,...,0) >> f_{\omega^2}(m) = \alpha(2^m)$$

and

$$\alpha(D(4,m)) = \alpha(2^{2^{m+1}}) >> f_{\omega^2+1}(m)$$

then, without proof, the following is claimed:

$$\alpha(D(3+m,m)) >> f_{\omega^2+m}(m) >> f_{\omega^2+\omega}(m)$$

$$\alpha(D(1,m,0)) >> f_{\omega^2+\omega+1}(m)$$

$$\alpha(D(m,m,0)) >> f_{\omega^2+\omega.2}(m)$$

$$\alpha(D(1,m,0,0)) >> f_{\omega^2+\omega.2+1}(m)$$

$$\alpha(D(m,m,0,0)) >> f_{\omega^2+\omega.3}(m)$$

$$\alpha(\alpha(2^5.m)) = \alpha(D(1,m,0,0,0)) >> f_{\omega^2+\omega.3+1}(m)$$

$$\alpha(\alpha(2^{4+m})) >> f_{\omega^2+\omega.m}(m) = f_{\omega^2.2}(m)$$

$$f_{\omega^{\omega^{\omega}}}(m)$$

Continuing without proof, it is claimed:

$$\alpha(\alpha(D(3,4+m))) = \alpha(\alpha(2^{4+m})) >> f_{\omega^2.2}(m)$$

$$\alpha(\alpha(D(3+m,m))) >> f_{\omega^2.2 + \omega}(m)$$

$$\alpha(\alpha(D(m,m,0))) >> f_{\omega^2.2 + \omega.2}(m)$$

$$\alpha(\alpha(D(m,m,0,0))) >> f_{\omega^2.2 + \omega.3}(m)$$

$$\alpha^3(2^m) = \alpha(\alpha(\alpha(2^m))) >> f_{\omega^2.3}(m)$$

$$\alpha^m(2^m) >> f_{\omega^3}(m)$$

$$\alpha^{m^{m-3}}(2^m) >> f_{\omega^{\omega}}(m)$$

$$\alpha^{m^{(m-3)^m}}(2^m) >> f_{\omega^{\omega^{\omega}}}(m)$$

$$f_{\epsilon_0}(m)$$

Finally without proof, it is proposed:

$$\alpha^{m\uparrow\uparrow m}(2^m) >> f_{\epsilon_0}(m)$$