## FANDOM

6 Pages

$\Sigma(5,3)>2.35\cdot 10^{15887327066011}$ (Wythagoras 2017)

## Historical table of boundsEdit

Bound Discoverer Year and month
$2.35\cdot 10^{15887327066011}$ Wythagoras June 2017

## Analysis of the current record holderEdit

The following machine yields this bound:

0 1 _ l 0
0 _ 1 r 0
0 2 1 r 1
1 1 2 l 2
1 2 1 l 3
1 _ 2 r 4
2 1 1 l 2
2 _ 1 l 0
3 1 1 l 3
3 _ 2 l 0
3 2 _ l 4
4 _ 2 l 3
4 1 2 l 3
4 2 2 r halt


This machine simulates a function C(k,l) which is represented by 1k+221l with the head on the first and there can possibly be some extra symbols after the last one. The function C(k,l) satisfies:

• C(k,l) = C(2k+4,l-1) if l>0.
• C(k,0) = 2k+4, followed by a blank where the head is.

From the recursion follows that $C(2,l) = 3 \cdot 2^{l+2} - 4$

After 1268 steps, the configuration C(2,42)2_2 is reached.

This therefore yields 52776558133244 ones, followed by 2_2.

This then yields C(2,52776558133244)_2.

This finally gives $3 \cdot 2^{52776558133246} - 4$ ones followed by 22. The machine then halts. So this machine leaves a total of $3 \cdot 2^{52776558133246} - 2 > 2.35 \cdot 10^{15887327066011}$ ones.