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Bashicu matrix system is very strong, but it is hard to understand how it works. That is why I made Bashicu matrix calculator for showing the calculation process of Bashicu matrix.
After hyp cos found nonteminating sequence of Bashicu matrix, Bashcu updated his system in his blog post. Please note that it is different from the version that User:KurohaKafka posted on the talk page. KurohaKafka thinks (0,0,...,0)(1,1,...,1) terminates for Bashicu's new system, while KurohaKafka's version may not terminate.
Let us call the original version of Bashicu matrix as Bashicu Matrix version 1 (BM1), and the new version that Bashicu made as version 2 (BM2). The problem in BM2 is that it is too difficult to understand. I would suggest that we use BM1 foâ€¦
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Rule
 Post an ordinal which is larger than the previous post
 It should be smaller than \(\omega_1^{CK}\)
Start from \(\omega\)
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I think I saw his face somewhere...http://michaelhalm.tripod.com/andrejoyce.htm
Alfred Jarry (18731907) also helped found the theatre of the absurd
The concept was coined by French writer Alfred Jarry (1873â€“1907)
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A program which is supposed to calculate \(f_{\vartheta(\Omega_\omega)+1}(10)\) was posted at googology thread of Japanese BBS 2ch.net (anonymous author; now GWiki accout was created.).
According to the author of the program, at each loop, the sequence of the pairs changes and the corresponding ordinal decreases as follows. It appears to be reasonable, but I am not yet confident with the calculation. I would like to know if googologists here think this calculation is valid.
\begin{eqnarray*} (0,0)(1,1)(2,2) &=& \psi(\psi_1(0)) \\ (0,0)(1,1)(2,1) &=& \psi(\Omega) \\ (0,0)(1,1)(2,0)(3,1) &=& \psi(\psi(0)) \\ (0,0)(1,1)(2,0)(3,0) &=& \psi(\omega^\omega) \\ (0,0)(1,1)(2,0)(2,0) &=& \psi(\omega^2) \\ (0,0)(1,1)(2,0)(1,1)(2,0) &=& \psi(\omega \cdâ€¦
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A program which calculates \(f_{\epsilon_0+1}(10)\) was posted at googology thread of Japanese BBS 2ch.net (anonymous author; now GWiki accout was created.).
The array is first initialized to \((0,1,2) = \omega^{\omega}\).
At the loop of F, G=1. At the loop of H, E=4 and the array becomes \((0,1,1,1) = \omega^3\).
After that, it goes to the second loop of E and E is decreased to 3 by STEP 1. Therefore, it matches the rightmost position of the sequence (0,1,1,1). In this second loop of E, the array changes to \((0,1,1,0,1,1,0,1,1) = \omega^2 \cdot 3\). After this, the ordinal will decrease as
 \((0,1,1,0,1,1,0,1,0,1,0,1) = \omega^2 \cdot 2 + \omega \cdot 3\)
 \((0,1,1,0,1,1,0,1,0,1,0,0,0) = \omega^2 \cdot 2+\omega \cdot 2+3\)
 \((0,1,1,0,1,1,0,1,0â€¦