# Analysis of Pavel's machine with number of ones 2^(2^31)

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### Pascal Michel

May 27, 2022, 3:59:28 AM5/27/22
to Busy Beaver Discuss
Pavel Kropitz gave in his mail on May 22, 2022, a new (6,2) - busy beaver champion M,
1RB 1RH 0LC 0LD 1LD 1LC 1RE 1LB 1RF 1RD 0LD 0RA
with sigma(M) = 2^(2^31).
I give here more analysis about this machine, using Shawn's configurations
D(k,n) = ...0(D0) (10)^k (01)^n
Notably, I show that s(M) > 8.2 x 10^1,292,913,985, a result that needs an independent confirmation.

We have
(a) ...(A0)... -(11)--> D(3,0)
(b) D(k + 1,0) 00 --('4k + 41)--> D(6,k) 0
(c) D(k + 1,1) 00 --(4k + 45)--> D(6,k) 110
(d) D(k,2) --(16k^2 + 42k + 32)--> D(4k + 6,0)
(e) D(6, 2n) --( t(n) )--> D(8 x (4^n) - 2,0)
with t(n) = 1024 x ((16^n) - 1)/15 - 176 x ((4^n) - 1)/3 + 12n
(f) D(k,1) 11 --(2k + 4)--> ...0 1^(2k + 2) 01 (H1)

So the computation of the machine on a blank tape is
...0(A0)...
--(11)--> D(3,0) 0...
--(49)--> D(6,2) 0...
--(860)--> D(30,0) 0...
--(157)--> D(6,29) 0... = D(6,28) 010...
--( t(14) )--> D((2^31) - 2,1) 0...
--(8,589,934,625)--> D(6,(2^31) - 3) 110... = D(6,(2^31) - 4) 01110...
--( t((2^30) - 2) )--> D(8 x (4^((2^30) - 2) - 2,0) 01110... = D((2^((2^31) - 1)) - 2,1) 110...
--( 2^(2^31) )--> ...0 1^((2^(2^31)) - 2) 01 (H1) 0...

The leading term for s(M) is in t((2^30) - 2). It is
1024 x (16^((2^30) - 2))/15 > 8;275 x 10^1,292,913,985

Pascal