Goodstein sequence

A Goodstein sequence is a certain class of integer sequences Gk(n).

Suppose we write a nonnegative integer n as a sum of powers of k (possibly duplicated), then we write the k exponents themselves as similar sums of powers. For example, we can write 100 as  +   +   =   +   +. We call this the base-k hereditary representation as n.

In the above representation of 100, we can "bump" the base 2 up by one, forming a different number  +   +. We write this larger number as B[2](100), and in general, B[b](n) means finding the base-b hereditary representation of n and bumping the base.

Now we define the recursive sequence G0(n) = n and Gk(n) = B[k + 1](Gk - 1(n)) - 1. In other words, as k increases, we are repeatedly bumping the base and subtracting one:


 * G0(100) = 100 =  +   +
 * G1(100) = B[2](100) - 1 =  +   +   - 1 = 91507169819870
 * G2(100) = B[3](G1(100)) - 1 =  +   + 2 x   + 2 x 1 - 1 = 3.486030062E156

This rapidly growing sequence is known a Goodstein sequence. Surprisingly, for all values of n, Gk(n) eventually peaks, declines, and returns to zero. This fact is known as Goodstein's theorem. Even more surprisingly, it can be shown that Goodstein's theorem cannot be proved in Peano arithmetic.

Let G(n) be the number of steps it takes for the Goodstein sequence starting on n to terminate (i.e. reach zero). Formally, G(n) as the smallest k for which Gk(n) = 0. G(n) is an extremely fast-growing function. Define the Hardy hierarchy $$H_\alpha (n)$$ for ordinals $$ \alpha \leq \epsilon_0 $$ and nonnegative integers n by

$$ H_0 (n) = n$$

$$ H_{\alpha+1} (n) = H_\alpha (n+1) $$

For limit ordinals $$\alpha, H_\alpha (n) = H_{\alpha[n]} (n)$$

and define $$R^\omega_b (n)$$ to be the ordinal obtained by writing n in base-b hereditary notation, then replacing every instance of b by $$\omega$$, for instance

$$R^\omega_2 (100) = R^\omega_2 (2^{2^2 + 2} + 2^{2^2 + 1} + 2^2) = \omega^{\omega^\omega + \omega} + \omega^{\omega^\omega + 1} + \omega^\omega$$.

Then $$G(n) = H_{R^\omega_2 (n)} (3) - 3$$. Since this function grows faster than any function provably recursive in Peano Arithmetic, Goodstein's Theorem is not provable in Peano Arithmetic.

Some values of G(n):

$$ G(0) = 0$$

$$ G(1) = 1$$

$$ G(2) = 3$$

$$ G(3) = 5$$

$$ G(4) = 3 \times 2^{402653211} - 3$$

$$ G(5) > 10\uparrow\uparrow (10 \uparrow\uparrow (10 \uparrow\uparrow 10^{10^{10^{21}}} ))$$

$$ G(6) > (10\uparrow\uparrow\uparrow\uparrow)^5 (10\uparrow\uparrow\uparrow)^5 (10\uparrow\uparrow)^5 (10^{10^{10^{10^{10^{117}}}}})$$

$$ G(7) > (10 \uparrow\uparrow\uparrow\uparrow\uparrow\uparrow)^7 (10\uparrow\uparrow\uparrow\uparrow\uparrow)^7 (10 \uparrow\uparrow\uparrow\uparrow)^7 (10 \uparrow\uparrow\uparrow)^7 (10 \uparrow\uparrow)^7 (10^{10^{10^{10^{10^{10^{10^{619}}}}}}})$$

$$ G(8) > Ack (Ack (3 \times 2^{402653211}) $$, where Ack is the unary Ackermann function.

$$G(9) > Ack (Ack (Ack (G(5))))$$

$$ G(10)> Ack^5 (G(6))$$

$$ G(11) > Ack^7 (G(7))$$

$$ G(12) > Ack^{3 \times 2^{402653211}}(3 \times 2^{402653211})$$

$$ G(13) > Ack^{G(5)}(G(5))$$

$$ G(14) > Ack^{G(6)}(G(6))$$

$$ G(15) > Ack^{G(7)}(G(7))$$

$$ G(16) > \{3, 3, 3, 3, 3\} $$ using Bowers' array notation.

$$G(17) > \lbrace4, 4, 4, 4, 4, 4\rbrace$$

$$ G(18) > \lbrace 6, 6, 6, 6, 6, 6, 6, 6\rbrace $$

$$ G(19) > \lbrace 8, 8, 8, 8, 8, 8, 8, 8, 8, 8\rbrace$$

$$ G(20) > \lbrace 3 \times 2^{402653211}, 3 \times 2^{402653211} (1) 2\rbrace $$

$$

G(32) > \lbrace 3, 4, 2 (1) 2 \rbrace $$

$$ G(64) > \lbrace3, 4, 4 (1) 2 \rbrace$$

$$ G(128) > \lbrace3, 4, 1, 2 (1) 2 \rbrace $$

$$ G(256) > \lbrace3, 5 (1) 4 \rbrace $$

$$ G(65536) > \lbrace3, 3 (2)(2)(1)(1)1,1,1, 2 \rbrace $$

$$ G(65540) > \lbrace 3, 3 (3 \times 2^{402653211}) 2 \rbrace $$

$$ G(2^{65536}) > \lbrace3, 3 (3, 3, 3, 2) 3 \rbrace $$

$$ G(2^{65536} + 4) > \lbrace3, 3 \times 2^{402653211} (1 (1) 2) 2 \rbrace $$

In general, $$ G(2\uparrow\uparrow (n+1))$$ is comparable to a $$3\uparrow \uparrow n$$ array of 3's.