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Replaced by Version 2

This version has been replaced by Beta Function Version 2

Beta Function - Sequence Generating Code

The Beta Function has been defined using program code shown below.

A separate blog will be written to explain how Sequence Generator Code is compiled and executed using a normal programming language ... Work in Progress.

Sequence Generating Code Version 1

The Beta Function is a significant re-design of the code used by the Alpha Function and will create long finite integer strings to define every large Veblen ordinal and FGH function for every finite integer (up to the size of $$f_{SV0}(n)$$). Refer to my other blogs on Unique Ordinal Representation and Program Code Version 4 for more background information.

Beta Function Description

• $$\beta(r,v) = h^m(v)$$ where
• $$h^0(v) = f_{\gamma_0}^{u_0}(v)$$ where $$u_0$$ and $$\gamma_0$$ are in base $$v$$, i.e. only use integers between $$0 ... v-1$$
• $$h^{n + 1}(v) = f_{\gamma_{n + 1}}^{u_{n + 1}}(h^n(v))$$ where
• $$u_{n + 1} < h^n(v)$$ in base $$v$$
• $$\gamma_{n + 1} < \gamma_n$$ in base $$h^n(v)$$ i.e. only use integers between $$0 ... h^n(v) - 1$$
• $$\gamma_m = 0$$, i.e. the function recursively continues until we reach ordinal zero.

and

$$\beta(v^{v+1},v) = f_{\varphi(1,0_{[v]})}(v) = f_{SVO}(v)$$ by definition

The Beta Function can then be uniquely represented by a sequence of finite integers as follows:

$$\beta(r,v) == (v,h_0)$$

The Sequence Generating RuleSet is as follows:

• $$h_x = (d<2,d(0:x<v-u(x),1:(f_0<1,P_h = 1)))$$
• $$f_x = (g_x,g_x((0,0,0):h_u,(h_U,(f_{x+1}<g_x,P_h=d-h_U)))$$
• $$g_x = (q<v+1,g_{[q]},n_0<q,t_x)$$
• $$n_x = (Q,t_A,g_{[Q]},n_{x+1})$$
• $$t_x = (h_T,g_E,g_C,g_x)$$

The convention used for the parameters is:

• Sequence function (e.g. $$h()$$) defines the substitution sequences to replace any instance of the sequence function.
• Instance of a Sequence function (e.g. $$g_0$$) is denoted by a lowercase letter with a subscript.
• Lower case letter (e.g. $$q$$) with no subscript, denotes a finite integer greater or equal to 0
• Upper case letter (e.g. $$Q$$) with no subscript, denotes a finite integer greater or equal to 1
• Upper case letter (e.g. $$P_h$$) with a subscript denotes variables - TO BE EXPLAINED FURTHER - in my blog on Sequence Generator Code ... Work in Progress
• The colon (e.g. $$:$$) denotes decision trees - TO BE EXPLAINED FURTHER - in my blog on Sequence Generator Code ... Work in Progress
• The less than symbol (e.g. $$<$$) denotes constraints - - TO BE EXPLAINED FURTHER - in my blog on Sequence Generator Code ... Work in Progress

These rulesets are intended to be generated in a recursive way beginning with an initial h sequence. For example, a typical start for generating a sequence is:

• $$(v,h_0)$$
• $$(v,(d<2,d(0:x<v-u(x),1:(f_0<1,P_h = 1))))$$
• Let $$d = 1$$
• $$(v,(1,(f_0<1,P_h = 1)))$$
• $$(v,(1,((g_0,g_0((0,0,0):h_u,(h_U,(f_1<g_0,P_h=d-h_U)))<1,P_h = 1)))$$
• $$(v,(1,(((q<v+1,g_{[q]},n_0<q,t_0),g_0((0,0,0):h_u,(h_U,(f_1<g_0,P_h=d-h_U)))<1,P_h = 1)))$$
• $$(v,(1,(((q<v+1,g_{[q]},(Q,t_A,g_{[Q]},n_{x+1})<q,t_0),g_0((0,0,0):h_u,etc.)$$
• $$(v,(1,(((q<v+1,g_{[q]},(Q,(h_T,g_E,g_C,g_A),g_{[Q]},n_{x+1})<q,t_0),g_0((0,0,0):h_u,etc.)$$

The sequences explode quickly as seen in this example:

• $$(v,(1,(((q<v+1,g_{[q]},(Q,(h_T,g_E,g_C,g_A),g_{[Q]},n_{x+1})<q,t_0),g_0((0,0,0):h_u,etc.)$$
• Let $$q = 2$$
• $$(v,(1,(((2,g_{[2]},(Q,(h_T,g_E,g_C,g_A),g_{[Q]},n_{x+1})<q,t_0),g_0((0,0,0):h_u,etc.)$$
• $$(v,(1,(((2,g_1,g_2,(Q,(h_T,g_E,g_C,g_A),g_{[Q]},n_{x+1})<q,t_0),g_0((0,0,0):h_u,etc.)$$
• $$(v,(1,(((2,(q<v+1,g_{[q]},n_0<q,t_1),g_2,(Q,(h_T,g_E,g_C,g_A),g_{[Q]},n_{x+1})<q,t_0),etc.)$$

The sequences are not structured other than the implied sequence order of the elements. Therefore all internal brackets can be dropped:

• $$(v,1,2,q,g_{[q]},n_0,t_1,g_2,Q,h_T,g_E,g_C,g_A,g_{[Q]},n_{x+1},t_0,g_0((0,0,0):h_u,(h_U,f_1))))$$

The sequences are guaranteed to be finite because of sequence limiting rules:

• $$h_x = (d<2,d(0:x<v-u(x),1:(f_0<1,P_h = 1)))$$ - TO BE EXPLAINED FURTHER - in my blog on Sequence Generator Code ... Work in Progress

Example Sequence

Here is an example sequence generated for an arbitrary real number in base $$v = 3$$

$$\beta(14.2246972615,3) = f_{\varphi(2,\varphi(1,0))}(3) == (3,h_0)$$

$$= (3,h_{0}(1,f_{0}(g_{0}(2,g_{Q}(0,h_{Q}(0,1)),g_{2}(2,g_{Q}(0,h_{Q}(0,0)),g_{2}(0,h_{2}(0,0))$$

$$,n_{0}(0),t_{a}(h_{T}(0,0),g_{E}(0,h_{E}(0,0)),g_{C}(0,h_{C}(0,0)),g_{a}(0,h_{a}(0,0)))),n_{0}(0),t_{a}(h_{T}(0,0)$$

$$,g_{E}(0,h_{E}(0,0)),g_{C}(0,h_{C}(0,0)),g_{a}(0,h_{a}(0,0)))),h_{U}(0,0)),f_{1}(g_{1}(0,h_{1}(0,0)),h_{u}(0,0))))$$

$$= (3,1,2,0,0,1,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)$$ with all internal brackets removed

WORK IN PROGRESS

Granularity Examples $$\beta(0,2)$$ to $$\beta(2.255,2)$$

The Beta Function is designed to generate any and every finite integer up to $$f_{SVO}(n)$$ and will use nested FGH functions with descending ordinal values to access individual integers. Here is the first example using base v = 2:

$$\beta(0,2) = 0 = 0$$

$$\beta(0.5,2) = 1 = 1$$

$$\beta(1,2) = 2 = 2$$

$$\beta(1.2,2) = 3 = 3$$

$$\beta(1.5,2) = 4 = 4$$

$$\beta(1.6,2) = 5 = 5$$

$$\beta(1.7,2) = 6 = 6$$

$$\beta(1.9,2) = 7 = 7$$

$$\beta(2.0001,2) = f_{\omega}(2) = 8$$

$$\beta(2.05,2) = f_{\omega}(2) + 1 = 9$$

$$\beta(2.1,2) = f_{\omega}(2) + 2 = 10$$

$$\beta(2.125,2) = f_{\omega}(2) + 3 = 11$$

$$\beta(2.135,2) = f_{\omega}(2) + 4 = 12$$

$$\beta(2.15,2) = f_{\omega}(2) + 5 = 13$$

$$\beta(2.16,2) = f_{\omega}(2) + 6 = 14$$

$$\beta(2.17,2) = f_{\omega}(2) + 7 = 15$$

$$\beta(2.2,2) = f_{\omega}(2).2 = 16$$

$$\beta(2.21,2) = f_{\omega}(2).2 + 1 = 17$$

$$\beta(2.23,2) = f_{\omega}(2).2 + 2 = 18$$

$$\beta(2.24,2) = f_{\omega}(2).2 + 3 = 19$$

$$\beta(2.242,2) = f_{\omega}(2).2 + 4 = 20$$

$$\beta(2.244,2) = f_{\omega}(2).2 + 5 = 21$$

$$\beta(2.247,2) = f_{\omega}(2).2 + 6 = 22$$

$$\beta(2.253,2) = f_{\omega}(2).2 + 7 = 23$$

$$\beta(2.255,2) = f_{\omega}(2).2 + f_{\omega}(2) = 24$$

Granularity Examples $$\beta(2.498,3)$$ to $$\beta(2.58,3)$$

In these examples we can access individual integers around a larger ordinal:

$$\beta(2.4980495,3) = f_{1}^{f_{1}^{2}(3) + 11}(f_{2}(3)) + f_{1}^{f_{1}^{2}(3) + 10}(f_{2}(3)) + f_{1}^{2}(f_{2}(3)) + f_{2}(3).2 + 3$$

$$< f_{2}^2(3) - 1$$ A real number with far greater precision is required to access this finite integer.

$$\beta(2.5,3) = f_{2}^{2}(3)$$

$$\beta(2.54,3) = f_{2}^{2}(3) + 1$$

$$\beta(2.58,3) = f_{2}^{2}(3) + 2$$

Granularity Examples $$\beta(2.999,3)$$ to $$\beta(3.004,3)$$

In these examples we can access individual integers around the ordinal $$\omega$$:

$$\beta(2.9999999999999,3) = f_{1}^{f_{1}^{f_{1}^{2}(3) + 11}(f_{2}(3)) + f_{1}^{f_{1}^{2}(3) + 10}(f_{2}(3)) + f_{1}^{f_{1}^{2}(3) + 9}(f_{2}(3)) + f_{1}^{f_{1}^{2}(3) + 7}(f_{2}(3)) + f_{2}(3) + 2}(f_{2}^{2}(3)) + 1$$

$$< f_{\omega}(3) - 1$$ A real number with far greater precision is required to access this finite integer.

$$\beta(3,3) = f_{\omega}(3)$$

$$\beta(3.002,3) = f_{\omega}(3) + 1$$

$$\beta(3.004,3) = f_{\omega}(3) + 2$$

Granularity Examples $$\beta(9,3)$$ to $$\beta(9.00002,3)$$

This behaviour extends into the Veblen ordinals:

$$\beta(9,3) = f_{\varphi(1,0)}(3)$$

$$\beta(9.00001,3) = f_{\varphi(1,0)}(3) + 1$$

$$\beta(9.00002,3) = f_{\varphi(1,0)}(3) + 2$$

Granularity Examples $$\beta(12.026,3)$$ to $$\beta(12.03,3)$$

Version 1 makes it possible to access Veblen functions of the form:

• $$\varphi(\alpha,(\varphi(\beta,0)\uparrow\uparrow t)^{\gamma_e}.{\gamma_c} + \gamma_a)$$ where
• $$\beta > \alpha$$
• $$0 < \gamma_e < \varphi(\beta,0)\uparrow\uparrow 2$$
• $$0 < \gamma_c < \varphi(\beta,0)\uparrow\uparrow t$$
• $$0 < \gamma_a < (\varphi(\beta,0)\uparrow\uparrow t)^{\gamma_e}.{\gamma_c}$$

Using the following Real Number inputs into the Beta Function generates these ordinals:

$$\beta(12.026783892,3) = f_{\varphi(1,\varphi(2,0) + 1)}(3)$$

$$\beta(12.029332923,3) = f_{\varphi(1,\varphi(2,0).2)}(3)$$

Test Bed for Version 1

Below is the test bed and various results using version 1.

$$\beta(0,2) = 0$$

$$\beta(0.5,2) = 1$$

$$\beta(1,2) = 2$$

$$\beta(1.2,2) = 3$$

$$\beta(1.45,2) = 4$$

$$\beta(1.55,2) = 5$$

$$\beta(1.7,2) = 6$$

$$\beta(1.84,2) = 7$$

$$\beta(2.0001,2) = f_{\omega}(2)$$

Second attempt - Base v = 2 - 20 Apr 2016

$$\beta(0,2) = 0$$

$$\beta(0.5,2) = 1$$

$$\beta(1,2) = 2$$

$$\beta(1.2,2) = 3$$

$$\beta(1.45,2) = 4$$

$$\beta(1.55,2) = 5$$

$$\beta(1.7,2) = 6$$

$$\beta(1.84,2) = 7$$

$$\beta(2.0001,2) = f_{\omega}(2)$$

$$\beta(2.05,2) = f_{\omega}(2) + 1$$

$$\beta(2.1,2) = f_{\omega}(2) + 2$$

$$\beta(2.125,2) = f_{\omega}(2) + 3$$

Next attempt - Base v = 2 - 21 Apr 2016

$$\beta(0,2) = 0 = 0$$

$$\beta(0.5,2) = 1 = 1$$

$$\beta(1,2) = 2 = 2$$

$$\beta(1.2,2) = 3 = 3$$

$$\beta(1.5,2) = 4 = 4$$

$$\beta(1.6,2) = 5 = 5$$

$$\beta(1.7,2) = 6 = 6$$

$$\beta(1.9,2) = 7 = 7$$

$$\beta(2.0001,2) = f_{\omega}(2) = 8$$

$$\beta(2.05,2) = f_{\omega}(2) + 1 = 9$$

$$\beta(2.1,2) = f_{\omega}(2) + 2 = 10$$

$$\beta(2.125,2) = f_{\omega}(2) + 3 = 11$$

$$\beta(2.135,2) = f_{\omega}(2) + 4 = 12$$

$$\beta(2.15,2) = f_{\omega}(2) + 5 = 13$$

$$\beta(2.16,2) = f_{\omega}(2) + 6 = 14$$

$$\beta(2.17,2) = f_{\omega}(2) + 7 = 15$$

$$\beta(2.2,2) = f_{\omega}(2) * 2 = 16$$

$$\beta(2.21,2) = f_{\omega}(2) * 2 + 1 = 17$$

Next attempt - Base v = 2 - 22 Apr 2016

$$\beta(0,2) = 0 = 0$$

$$\beta(0.5,2) = 1 = 1$$

$$\beta(1,2) = 2 = 2$$

$$\beta(1.2,2) = 3 = 3$$

$$\beta(1.5,2) = 4 = 4$$

$$\beta(1.6,2) = 5 = 5$$

$$\beta(1.7,2) = 6 = 6$$

$$\beta(1.9,2) = 7 = 7$$

$$\beta(2.0001,2) = f_{\omega}(2) = 8$$

$$\beta(2.05,2) = f_{\omega}(2) + 1 = 9$$

$$\beta(2.1,2) = f_{\omega}(2) + 2 = 10$$

$$\beta(2.125,2) = f_{\omega}(2) + 3 = 11$$

$$\beta(2.135,2) = f_{\omega}(2) + 4 = 12$$

$$\beta(2.15,2) = f_{\omega}(2) + 5 = 13$$

$$\beta(2.16,2) = f_{\omega}(2) + 6 = 14$$

$$\beta(2.17,2) = f_{\omega}(2) + 7 = 15$$

$$\beta(2.2,2) = f_{\omega}(2) * 2 = 16$$

$$\beta(2.21,2) = f_{\omega}(2) * 2 + 1 = 17$$

$$\beta(2.23,2) = f_{\omega}(2) * 2 + 2 = 18$$

$$\beta(2.24,2) = f_{\omega}(2) * 2 + 3 = 19$$

$$\beta(2.242,2) = f_{\omega}(2) * 2 + 4 = 20$$

$$\beta(2.244,2) = f_{\omega}(2) * 2 + 5 = 21$$

$$\beta(2.247,2) = f_{\omega}(2) * 2 + 6 = 22$$

$$\beta(2.253,2) = f_{\omega}(2) * 2 + 7 = 23$$

$$\beta(2.255,2) = f_{\omega}(2) * 2 + f_{\omega}(2) = 24$$

$$\beta(2.28,2) = f_{1}^{2}(f_{\omega}(2))$$

$$\beta(2.322,2) = f_{1}^{2}(f_{\omega}(2)) + f_{\omega}(2) * 2$$

$$\beta(2.33,2) = f_{1}^{4}(f_{\omega}(2))$$

$$\beta(2.3727,2) = f_{1}^{4}(f_{\omega}(2)) + f_{\omega}(2) * 2$$

$$\beta(2.376,2) = f_{1}^{4}(f_{\omega}(2)) + f_{1}^{2}(f_{\omega}(2))$$

$$\beta(2.3781,2) = f_{1}^{4}(f_{\omega}(2)) + f_{1}^{2}(f_{\omega}(2)) + f_{\omega}(2) * 2$$

Second attempt - Base v = 3 - 22 Apr 2016

$$\beta(0,3) = 0$$

$$\beta(0.5,3) = 1$$

$$\beta(0.7,3) = 2$$

$$\beta(1,3) = 3$$

$$\beta(1.45,3) = 6$$

$$\beta(1.75,3) = f_{1}^{2}(3)$$

$$\beta(2.1,3) = f_{2}(3)$$

$$\beta(2.28,3) = f_{2}(3) * 2$$

$$\beta(2.34,3) = f_{1}^{2}(f_{2}(3))$$

$$\beta(2.5,3) = f_{2}^{2}(3)$$

$$\beta(3,3) = f_{\omega}(3)$$

$$\beta(3.07,3) = f_{\omega}^{2}(3)$$

$$\beta(3.141,3) = f_{\omega + 1}(3)$$

$$\beta(3.214,3) = f_{\omega + 1}^{2}(3)$$

$$\beta(3.224,3) = f_{1}^{f_{\omega}(3) + 2}(f_{\omega + 1}^{2}(3)) + f_{1}^{f_{6}^{f_{2}^{f_{2}^{2}(3) + 1}(f_{\omega}(3)) + f_{1}^{f_{2}(3) + 4}(f_{2}^{f_{1}^{2}(3)}(f_{\omega}(3))) + 3}(f_{\omega}^{2}(3))}(f_{f_{\omega}(3) + 1}^{2}(f_{\omega}^{2}(3)))$$

$$\beta(3.288,3) = f_{\omega + 2}(3)$$

Next attempt - Base v = 3 - 24 Apr 2016

$$\beta(0,3) = 0$$

$$\beta(0.5,3) = 1$$

$$\beta(0.7,3) = 2$$

$$\beta(1,3) = 3$$

$$\beta(1.45,3) = 6$$

$$\beta(1.75,3) = f_{1}^{2}(3)$$

$$\beta(2.1,3) = f_{2}(3)$$

$$\beta(2.28,3) = f_{2}(3).2$$

$$\beta(2.34,3) = f_{1}^{2}(f_{2}(3))$$

$$\beta(2.5,3) = f_{2}^{2}(3)$$

$$\beta(3,3) = f_{\omega}(3)$$

$$\beta(3.07,3) = f_{\omega}^{2}(3)$$

$$\beta(3.141,3) = f_{\omega + 1}(3)$$

$$\beta(3.214,3) = f_{\omega + 1}^{2}(3)$$

$$\beta(3.288,3) = f_{\omega + 2}(3)$$

$$\beta(3.364,3) = f_{\omega + 2}^{2}(3)$$

$$\beta(3.442,3) = f_{\omega.2}(3)$$

$$\beta(3.9485,3) = f_{\omega^2}(3)$$

$$\beta(5.1963,3) = f_{\omega\uparrow\uparrow 2}(3)$$

$$\beta(5.197,3) = f_{\omega\uparrow\uparrow 2}(3) + 2$$

$$\beta(5.2,3) = f_{1}^{6}(f_{\omega\uparrow\uparrow 2}(3)) + 1$$

$$\beta(6,3) = f_{f_{1}^{2}(3) + 3}^{f_{f_{7}(f_{8}^{2}(f_{f_{1}^{2}(3)}(f_{\omega.2 + 1}(3)))) + 1}^{f_{\omega}(f_{\omega.2 + 1}(3))}(f_{\omega.2 + 1}^{2}(3))}(f_{\omega\uparrow\uparrow 2.(\omega) + \omega + 1}^{2}(3))$$

$$\beta(7,3) = f_{\omega\uparrow\uparrow 2^2 + 2}(3) + f_{\omega.2 + 1}(f_{\omega\uparrow\uparrow 2^2}(3)) + f_{\omega^2 + \omega.(f_{\omega}^{f_{\omega}(3)}(f_{\omega^2 + 2}^{2}(3)))}(f_{\omega^2.2 + \omega.2}^{2}(3))$$

$$\beta(8,3) = f_{\omega\uparrow\uparrow 2^2.2}^{f_{f_{1}^{2}(f_{2}^{2}(3)) + 2}(f_{\omega}(f_{\omega + 1}^{f_{\omega.2}(3) + 2}(f_{\omega.2}^{2}(3)))) + f_{\omega.2}^{2}(3)}(f_{\omega\uparrow\uparrow 2^2.(\omega + 1) + \omega^2.2 + 1}^{2}(3))$$

$$\beta(8.5,3) = f_{1}^{3}(f_{\omega\uparrow\uparrow 2^2.(\omega^2 + 2)}(3)) + f_{\omega.(f_{3}^{f_{\omega}(f_{\omega.2 + 2}(3))}(f_{4}^{2}(f_{9}(f_{10}^{2}(f_{\omega^2 + 1}(3))))))}(f_{\omega^2 + \omega}(3))$$

$$\beta(9,3) = f_{\varphi(1,0)}(3)$$

$$\beta(9.00001,3) = f_{\varphi(1,0)}(3) + 1$$

$$\beta(9.001434,3) = f_{\varphi(1,0) + 1}(3)$$

$$\beta(9.004293,3) = f_{\varphi(1,0) + \omega}(3)$$

$$\beta(9.0064395,3) = f_{\varphi(1,0) + \omega\uparrow\uparrow 2}(3)$$

$$\beta(9.00859,3) = f_{\varphi(1,0).2}(3)$$

$$\beta(9.0171825,3) = f_{\varphi(1,0).(\omega)}(3)$$

$$\beta(9.021483,3) = f_{\varphi(1,0).(\omega^2)}(3)$$

$$\beta(9.0257856,3) = f_{\varphi(1,0).(\omega\uparrow\uparrow 2)}(3)$$

$$\beta(9.0344,3) = f_{\varphi(1,0)^2}(3)$$

$$\beta(9.068926,3) = f_{\varphi(1,0)^{\omega}}(3)$$

$$\beta(9.1035865,3) = f_{\varphi(1,0)^{\omega\uparrow\uparrow 2}}(3)$$

Next attempt - Base v = 3 - 25 Apr 2016

$$\beta(0,3) = 0$$

$$\beta(0.5,3) = 1$$

$$\beta(0.7,3) = 2$$

$$\beta(1,3) = 3$$

$$\beta(1.45,3) = 6$$

$$\beta(1.75,3) = f_{1}^{2}(3)$$

$$\beta(2.1,3) = f_{2}(3)$$

$$\beta(2.28,3) = f_{2}(3).2$$

$$\beta(2.34,3) = f_{1}^{2}(f_{2}(3))$$

$$\beta(2.5,3) = f_{2}^{2}(3)$$

$$\beta(3,3) = f_{\omega}(3)$$

$$\beta(3.07,3) = f_{\omega}^{2}(3)$$

$$\beta(3.141,3) = f_{\omega + 1}(3)$$

$$\beta(3.214,3) = f_{\omega + 1}^{2}(3)$$

$$\beta(3.288,3) = f_{\omega + 2}(3)$$

$$\beta(3.364,3) = f_{\omega + 2}^{2}(3)$$

$$\beta(3.442,3) = f_{\omega.2}(3)$$

$$\beta(3.9485,3) = f_{\omega^2}(3)$$

$$\beta(5.1963,3) = f_{\omega\uparrow\uparrow 2}(3)$$

$$\beta(5.197,3) = f_{\omega\uparrow\uparrow 2}(3) + 2$$

$$\beta(5.2,3) = f_{1}^{4}(f_{\omega\uparrow\uparrow 2}(3)) + f_{\omega\uparrow\uparrow 2}(3)$$

$$\beta(6,3) = f_{f_{1}^{2}(3) + 3}^{f_{f_{7}(f_{8}^{2}(f_{f_{1}^{2}(3)}(f_{\omega.2 + 1}(3)))) + 1}^{f_{\omega}(f_{\omega.2 + 1}(3))}(f_{\omega.2 + 1}^{2}(3))}(f_{\omega\uparrow\uparrow 2.(\omega) + \omega + 1}^{2}(3))$$

$$\beta(7,3) = f_{\omega\uparrow\uparrow 2^2 + 2}(3) + f_{\omega.2 + 1}(f_{\omega\uparrow\uparrow 2^2}(3)) + f_{\omega^2 + \omega.(f_{\omega}^{f_{\omega}(3)}(f_{\omega^2 + 2}^{2}(3)))}(f_{\omega^2.2 + \omega.2}^{2}(3))$$

$$\beta(8,3) = f_{\omega\uparrow\uparrow 2^2.2}^{f_{f_{1}^{2}(f_{2}^{2}(3)) + 2}(f_{\omega}(f_{\omega + 1}^{f_{\omega.2}(3) + 2}(f_{\omega.2}^{2}(3)))) + f_{\omega.2}^{2}(3)}(f_{\omega\uparrow\uparrow 2^2.(\omega + 1) + \omega^2.2 + 1}^{2}(3))$$

$$\beta(8.5,3) = f_{1}^{4}(f_{\omega\uparrow\uparrow 2^2.(\omega^2 + 2)}(3)) + 1$$

$$\beta(9,3) = f_{\varphi(1,0)}(3)$$

$$\beta(9.00001,3) = f_{\varphi(1,0)}(3) + 1$$

$$\beta(9.001435,3) = f_{\varphi(1,0) + 1}(3)$$

$$\beta(9.0042933,3) = f_{\varphi(1,0) + \omega}(3)$$

$$\beta(9.0064395,3) = f_{\varphi(1,0) + \omega\uparrow\uparrow 2}(3)$$

$$\beta(9.0085872,3) = f_{\varphi(1,0).2}(3)$$

$$\beta(9.0171825,3) = f_{\varphi(1,0).(\omega)}(3)$$

$$\beta(9.021483,3) = f_{\varphi(1,0).(\omega^2)}(3)$$

$$\beta(9.0257856,3) = f_{\varphi(1,0).(\omega\uparrow\uparrow 2)}(3)$$

$$\beta(9.0343973,3) = f_{\varphi(1,0)^2}(3)$$

$$\beta(9.068926,3) = f_{\varphi(1,0)^{\omega}}(3)$$

$$\beta(9.1035865,3) = f_{\varphi(1,0)^{\omega\uparrow\uparrow 2}}(3)$$

$$\beta(9.13838,3) = f_{\varphi(1,0)\uparrow\uparrow 2}(3)$$

$$\beta(9.278887,3) = f_{\varphi(1,1)}(3)$$

$$\beta(9.421555,3) = f_{\varphi(1,1)\uparrow\uparrow 2}(3)$$

$$\beta(9.566416,3) = f_{\varphi(1,2)}(3)$$

$$\beta(9.8628543,3) = f_{\varphi(1,\omega)}(3)$$

$$\beta(10.32482425,3) = f_{\varphi(1,\omega\uparrow\uparrow 2)}(3)$$

$$\beta(10.8084326,3) = f_{\varphi(1,\varphi(1,0))}(3)$$

$$\beta(10.89120821,3) = f_{\varphi(1,\varphi(1,0)\uparrow\uparrow 2)}(3)$$

$$\beta(11.844667,3) = f_{\varphi(2,0)}(3)$$

$$\beta(11.847177,3) = f_{\varphi(2,0).2}(3)$$

$$\beta(11.8496876,3) = f_{\varphi(2,0).(\omega)}(3)$$

$$\beta(11.859737,3) = f_{\varphi(2,0)^2}(3)$$

$$\beta(11.935378,3) = f_{\varphi(2,0)\uparrow\uparrow 2}(3)$$

$$\beta(12.026783892,3) = f_{\varphi(1,\varphi(2,0) + 1)}(3)$$

$$\beta(12.029332923,3) = f_{\varphi(1,\varphi(2,0).2)}(3)$$

$$\beta(12.211703,3) = f_{\varphi(2,1)}(3)$$

$$\beta(12.9802463,3) = f_{\varphi(2,\omega)}(3)$$

$$\beta(13.58823285,3) = f_{\varphi(2,\omega\uparrow\uparrow 2)}(3)$$

$$\beta(14.2246972615,3) = f_{\varphi(2,\varphi(1,0))}(3)$$

$$\beta(14.8909732862,3) = f_{\varphi(2,\varphi(2,0))}(3)$$

$$\beta(14.9099198273,3) = f_{\varphi(2,\varphi(2,0)\uparrow\uparrow 2)}(3)$$

$$\beta(15.5884574,3) = f_{\varphi(\omega,0)}(3)$$

$$\beta(20,3) = f_{\varphi(\omega^2.2 + \omega + 1,\omega\uparrow\uparrow 2.(\omega + 2) + \omega^2.2 + \omega + 2)^2.(\varphi(1,\varphi(2,\omega^2 + 2)^2.2 + \omega)\uparrow\uparrow 2^{\varphi(2,\omega^2 + 2)^{\omega\uparrow\uparrow 2^2.(\omega.2 + 2)}.(\omega.2)})}(3)$$

$$\beta(27,3) = f_{\varphi(1,0,0)}(3)$$

$$\beta(40,3) = f_{\varphi(\omega + 1,\varphi(1,1,\varphi(1,\varphi(\omega\uparrow\uparrow 2.(\omega^2 + 2) + \omega.2 + 1,\omega\uparrow\uparrow 2^2.2 + \omega\uparrow\uparrow 2.(\omega.2 + 1) + 1)\uparrow\uparrow 2^{\varphi(1,0)^{\omega\uparrow\uparrow 2.2} + \varphi(1,0)^2})),0)}(3)$$

$$\beta(70,3) = f_{\varphi(\varphi(\omega.2 + 1,\varphi(1,\varphi(\omega + 1,2)^{\omega^2 + \omega + 2}.(\omega^2.2 + 2) + \varphi(\omega,1)\uparrow\uparrow 2.(\omega\uparrow\uparrow 2^2 + 1)).2 + \varphi(2,\omega.2 + 1)\uparrow\uparrow 2^{\omega\uparrow\uparrow 2^2.2 + \omega.2}.(\varphi(1,\omega\uparrow\uparrow 2))),0,0)}(3)$$

$$\beta(81,3) = f_{\varphi(1,0,0,0)}(3) = f_{SVO}(3)$$ by definition