User:Sligocki/Goodstein sequences


 * This article is now available on my blog: https://www.sligocki.com/2009/10/14/goodstein-sequences.html

Goodstein sequences are very long sequences which eventually reach 0, but run for so long that Peano arithmetic cannot be used to prove that they reach 0.

Introduction
Let $$G(n)$$ be the Goodstein sequence starting with n and ending at 0.
 * $$ G(3) = (2 + 1, 3, 3, 2, 1, 0) $$
 * $$ G(4) = (2^2, 2 \cdot 3^2 + 2 \cdot 3 + 2, \dots, 2\cdot 23^2, \dots, (24 \cdot 2^{24} - 1)^2, \dots, 3, 2, 1, 0) $$

Let $$g_n$$ be the base of the hereditary notation of the last term of $$G(n)$$ (Alternatively, it is the length of the Goodstein sequence + 1). We shall call these the Goodstein numbers

Now, in fact, if $$ f(n) = (n+1) 2^{n+1} - 1 $$, then $$g_4 = f^3(2)$$. I show that this function has a meaning.

Growth of Goodstein Numbers
Let us define a family of functions:
 * $$ f_0(B) = B + 1 $$
 * $$ f_k(B) = f_{k-1}^{B+1}(B) $$

Ah ha,
 * $$ f_1(B) = 2B + 1 $$
 * $$ f_2(B) = (B+1) \cdot 2^{B+1} - 1 $$ and
 * $$ f_3(2) = f_2^3(2) = g_4 $$

In fact:

Notice the pattern? If $$ B^k $$ appears in $$ G(n) $$ then $$ g_n = f_k(B) $$ (where B is the base and k<B). Likewise, if $$ B^B $$ appears, then $$ g_n = f_{B+1}(B) $$.

In fact, let's rename our functions (here $$\omega$$ is a label, not a variable — in fact, it is actually an ordinal number, or generalized index, who's properties I hope to take advantage of):
 * $$ f_{\omega^0}(B) = B + 1 $$
 * $$ f_{\omega^k}(B) = f_{\omega^{k-1}}^{B+1}(B) $$

And add a new one:
 * $$ f_{\omega^\omega}(B) = f_{\omega^{B+1}}(B) $$

Thus, if we have a value of the form $$\alpha$$ at base B in $$ G(n) $$, then $$ g_n = f_\alpha(B) $$.

Derivation of Growth Function

 * Note: It turns out that the function I define here is a variant of the fast-growing hierarchy much like the Hardy hierarchy.

Goodstein talks about the hereditary form of a number and the unique ordinal number associated with each hereditary form. For example:
 * $$ 266 = 2^5 + 2^3 + 2 = 2^{2^2 + 1} + 2^{2 + 1} + 2 $$ is in form $$ \omega^{\omega^\omega + 1} + \omega^{\omega + 1} + \omega $$

Let us identify the hereditary form with that ordinal number.

If N has hereditary form $$\alpha$$ with base B, then let $$ f_\alpha(B) $$ be the base at which the the Goodstein sequence starting at N in base B will end.

For some values of $$\alpha$$:
 * $$ f_0(B) = B $$
 * $$ f_1(B) = B + 1 $$
 * $$ f_k(B) = f_1^k(B) = B + k $$


 * $$ f_\omega(B) = f_B(B+1) = f_1^{B+1}(B) = 2B + 1 $$
 * $$ f_{\omega \cdot k}(B) = f_\omega^k(B)) = (B + 1) 2^k - 1 $$


 * $$ f_{\omega^2}(B) = f_{\omega \cdot B + B}(B+1) = f_{\omega \cdot B}(f_B(B+1)) = f_{\omega \cdot B}(f_\omega(B)) = f_{\omega \cdot (B+1)}(B) = f_\omega^{B+1}(B) = (B + 1) 2^{B + 1} - 1 $$
 * $$ f_{\omega^2 \cdot k}(B) = f_{\omega^2}^k(B)) > 2 \uparrow \uparrow k $$


 * $$ f_{\omega^3}(B) = f_{\omega^2 \cdot B + \omega \cdot B + B}(B+1) = f_{\omega^2 \cdot B}(f_{\omega^2 \cdot B + B}(B+1)) = f_{\omega^2 \cdot B}(f_{\omega^2}(B)) = f_{\omega^2 \cdot (B+1)}(B) = f_{\omega^2}^{B+1}(B) > 2 \uparrow \uparrow (B+1) $$


 * $$ f_{\omega^k}(B) = f_{\omega^{k-1} \cdot (B+1)}(B) = f_{\omega^{k-1}}^{B+1}(B) > 2 \uparrow^{k-1} (B+1) $$


 * $$ f_{\omega^\omega}(B) = f_{\omega^{B+1}}(B) > 2 \uparrow^B (B+1) = 2 \to B + 1 \to B $$
 * $$ f_{\omega^\omega \cdot k}(B) = f_{\omega^\omega}^k(B) > 2 \to B + 1 \to k \to 2 $$

Note: $$ h(n) = 3 \uparrow^n 3 < 2 \uparrow^n (n-1) = f_{\omega^\omega}(n-1) $$. So Graham's number $$ \mathcal{G} = h^{64}(4) < f_{\omega^\omega}^{64}(4) = f_{\omega^\omega \cdot 64}(4) $$.

Now $$ G(12) = (2^{2+1} + 2^2, \dots, g_4^{g_4 + 1}, g_4 (g_4 + 1)^{g_4 + 1}, \dots) $$ where we remember that $$ g_4 = f_{\omega^\omega}(2) = 3 \cdot 2^{402653211} - 1 > 64 $$. So, $$g_{12} = f_{\omega^\omega \cdot g_4}(g_4 + 1) > f_{\omega^\omega \cdot 64}(4) > \mathcal{G} $$


 * $$ f_{\omega^{\omega + 1}}(B) = f_{\omega^\omega \cdot B + \omega^B \cdot B + \cdots + B}(B+1) = f_{\omega^\omega \cdot B}(f_{\omega^{B+1}}(B)) = f_{\omega^\omega \cdot B}(f_{\omega^\omega}(B)) = f_{\omega^\omega \cdot (B + 1)}(B) = f_{\omega^\omega}^{B+1}(B) > 2 \to B + 1 \to B + 1 \to 2 $$
 * $$ f_{\omega^{\omega + 1} \cdot k}(B) = f_{\omega^{\omega + 1}}^k(B) > 2 \to B + 1 \to k \to 3 $$


 * $$ f_{\omega^{\omega + 2}}(B) = f_{\omega^{\omega + 1} \cdot B + \omega^\omega \cdot B + \omega^B \cdot B + \cdots + B}(B+1) = f_{\omega^{\omega + 1} \cdot B}(f_{\omega^{\omega+1}}(B)) = f_{\omega^{\omega + 1} \cdot (B + 1)}(B) = f_{\omega^{\omega + 1}}^{B+1}(B) > 2 \to B + 1 \to B + 1 \to 3 $$


 * $$ f_{\omega^{\omega + k}}(B) = f_{\omega^{\omega + (k-1)} \cdot (B + 1)}(B) = f_{\omega^{\omega + (k-1)}}^{B+1}(B) > 2 \to B + 1 \to B + 1 \to k + 1 $$


 * $$ f_{\omega^{\omega \cdot 2}}(B) = f_{\omega^{\omega + (B+1)}}(B) > 2 \to B + 1 \to B + 1 \to B + 2 $$

...


 * $$ f_{\omega^{\omega^\omega}}(B) = f_{\omega^{\omega^{B+1}}}(B) >> 2 \to \overbrace{B + 1 \to \cdots \to B + 1}^{B^2 + 2} $$

Now let's look back at the table:

= f_{\omega^{\omega^{g_0+1}}}(g_0) = f_{\omega^{\omega^{g_0} \cdot (g_0 + 1)}}(g_0) = f_{\omega^{\omega^{g_0} \cdot g_0 + g_0 + 1}}(g_0) = f_{\omega^{\omega^{g_0} \cdot g_0 + g_0}}^{g_0 + 1}(g_0) = f_{\omega^{\omega^{g_0} \cdot g_0 + g_0}}^{g_0}(f_{\omega^{\omega^2 \cdot 2 + 2}}(g_0)) = ? $$
 * $$ g_{16} = f_{\omega^{\omega^\omega}}(g_0)


 * $$ \begin{array}{rcl}

g_6 &=& f_{\omega^\omega + \omega}(2) = f_{\omega^\omega}(f_\omega(2)) \\ &=& f_{\omega^\omega}(5) = f_{\omega^6}(5) = f_{\omega^5}^6(5) \\ &=& f_{\omega^5}^5(f_{\omega^4}^5(f_{\omega^3}^5(f_{\omega^2}^5(f_{\omega^2}(5))))) \\ &=& f_{\omega^5}^5(f_{\omega^4}^5(f_{\omega^3}^5(f_{\omega^2}^5(2^{B+1} - 1)))) \\ &=& f_{\omega^5}^5(f_{\omega^4}^5(f_{\omega^3}^5(f_{\omega^2}^4(2^{2^{B+1} - 1} - 1)))) \\ &=& f_{\omega^5}^5(f_{\omega^4}^5(f_{\omega^3}^5(f_{\omega^2}^3(2^{2^{2^{B+1} - 1} - 1} - 1)))) \\ \end{array} $$