(This article was original published on in Oct 2009 on my Wikipedia User page. I’m re-posting on my blog to bring my writing together in one place.)

Milton Green discovered a class of Turing machines which produce extremely large results in the busy beaver game.1

The machines defined recursively and the number of 1s they leave on the tape is likewise defined recursively. We examine those numbers here:

## Definition

Let us define the numbers $$B_n$$ for $$n$$ odd.

• $B_n(0) = 1$
• $B_1(m) = m+1$
• $B_n(m) = B_{n-2}[B_n(m-1) + 1] + 1$

Then, Green’s numbers $$BB_n$$ are defined as:

• $$BB_n = B_{n-2}[B_{n-2}(1)]$$ for odd $$n$$
• $$BB_n = B_{n-3}[B_{n-3}(3) + 1] + 1$$ for even $$n$$

## Examples

• $B_1(m) = m+1$
• $B_3(m) = 3m + 1$
• $B_5(m) = \frac{7}{2} \cdot 3^m - \frac{5}{2}$
• $$B_7(m) > 3 \uparrow\uparrow m$$ and $$B_7(1) = B_5(2) + 1 = 29$$
• $$B_9(m) > 3 \uparrow\uparrow\uparrow m$$ and $$B_9(1) = B_7(2) + 1 = B_5(30) + 2 = 720618962331271$$

• $BB_3 = 3$
• $BB_4 = 7$
• $BB_5 = 13$
• $BB_6 = 35$
• $BB_7 = B_5(8) = 22961$
• $BB_8 = B_5(93) + 1 = 3(7 \cdot 3^{92} - 1) / 2 = 824792557184288824246737061810550733633916929$
• $BB_9 = B_7(B_7(1)) = B_7(29) > 3 \uparrow\uparrow 29 > 10 \uparrow\uparrow 28$
• $BB_{10} = B_7(B_7(3) + 1) + 1 > 3 \uparrow\uparrow 3 \uparrow \uparrow 3 = 3 \uparrow\uparrow\uparrow 3$
• $BB_{11} = B_9(B_9(1)) = B_9(720618962331271) > 3 \uparrow\uparrow\uparrow 720618962331271$

## Bounds

We will show that $$B_n(m)$$ grows like $$3 \uparrow^{n/2} m$$ and furthermore that $$BB_n$$ grows like $$3 \uparrow^{n/2} 3$$ (See Knuth’s up-arrow notation and Up-arrow Properties).

Claim: $$B_{2k+3}(m) \ge 3 \uparrow^k m$$ for any $$k \ge -2$$ and $$m \ge 0$$

Proof by induction:

Base Case:

$B_{2k+3}(0) = 1 \ge 1 = 3 \uparrow^k 0$ $B_1(m) = m+1 \ge m+1 = 3 \uparrow^{-2} m$

Inductive Step:

Assume that $$B_{2k^\prime+3}(m^\prime) \ge 3 \uparrow^{k^\prime} m^\prime$$ (for all $$k^\prime = k, m^\prime < m$$ or $$k^\prime < k, m^\prime < 3 \uparrow^k (m-1)$$)

$B_{2k+3}(m) = B_{2k+1}[B_{2k+3}(m-1) + 1] + 1 > B_{2k+1}[3 \uparrow^k (m-1)] > 3 \uparrow^{k-1} [3 \uparrow^k (m-1)] = 3 \uparrow^k m \,$

QED

Claim: $$B_{2k+3}(m) < \frac{1}{2} (3 \uparrow^k (m+2)) < (3 \uparrow^k (m+2))$$ for any $$k \ge 1$$ and $$m \ge 0$$ (In fact the bound can be tightened to $$m$$+1 for $$k$$ ≥ 2)

Proof by induction:

Base Case:

$B_{2k+3}(0) = 1 < \frac{1}{2} (3 \uparrow^k 2)$ $B_5(m) = \frac{9}{2} 3^m < \frac{1}{2} (3 \uparrow (m+2))$ $B_7(m) < \frac{1}{2} (3 \uparrow^2 (m+1))$

(left as an exercise to the reader)

Inductive Step:

Assume that $$B_{2k^\prime+3}(m^\prime) < \frac{1}{2} (3 \uparrow^{k^\prime} (m^\prime+2))$$ (for $$k^\prime = k, m^\prime < m$$ or $$k^\prime < k, m^\prime < 3 \uparrow^k m + 1$$)

$\begin{array}{rll} B_{2k+3}(m) = B_{2k+1}[B_{2k+3}(m-1) + 1] + 1 & < & B_{2k+1}[\frac{1}{2} (3 \uparrow^k (m+1)) + 1] + 1 \\ & < & \frac{1}{2} (3 \uparrow^{k-1} [\frac{1}{2} (3 \uparrow^k (m+1)) + 3]) + 1 \\ & \le? & \frac{1}{2} (3 \uparrow^{k-1} [\frac{1}{2} (3 \uparrow^k (m+1)) + 4]) \\ & \le? & \frac{1}{2} (3 \uparrow^{k-1} (3 \uparrow^k (m+1))) \\ & = & \frac{1}{2} (3 \uparrow^k (m+2)) \end{array}$

≤? statements seem obvious, but grungy to prove (left as an exercise to the reader :) )

Claim: $$3 \uparrow^k 3 < BB_{2k+4}, BB_{2k+5} < 3 \uparrow^{k+1} 3$$ for $$k \ge 1$$

Proof:

$BB_{2k+5} = B_{2k+3}(B_{2k+3}(1)) > 3 \uparrow^k (3 \uparrow^k 1) = 3 \uparrow^k 3$ $BB_{2k+4} > B_{2k+1}(B_{2k+1}(3)) > 3 \uparrow^{k-1} (3 \uparrow^{k-1} 3) = 3 \uparrow^k 3$ $BB_{2k+5} = B_{2k+3}(B_{2k+3}(1)) < \frac{1}{2} (3 \uparrow^k (\frac{1}{2} (3 \uparrow^k (1+2))+2)) < 3 \uparrow^k (3 \uparrow^k 3) = 3 \uparrow^{k+1} 3$ $BB_{2k+4} = B_{2k+1}[B_{2k+1}(3) + 1] + 1 < \frac{1}{2} (3 \uparrow^{k-1} (\frac{1}{2} (3 \uparrow^{k-1} (3+2))+3)) + 1 < 3 \uparrow^{k-1} (3 \uparrow^{k-1} 5) < 3 \uparrow^k 4 < 3 \uparrow^{k+1} 3$

QED

## References

1. Milton W. Green. “A Lower Bound on Rado’s Sigma Function for Binary Turing Machines”, Preceedings of the IEEE Fifth Annual Symposium on Switching Circuits Theory and Logical Design, 1964, pages 91–94, doi: 10.1109/SWCT.1964.3