A busy beaver is an 
-state,
2-symbol Turing machine that leaves the maximum
number 
of 1s on an initially all-zero tape when it halts (Rado 1962, Lin and Rado 1965,
Shallit 1998). Alternatively, some authors define a busy beaver as a Turing
machine that performs the maximum number 
of steps before halting from the same initial tape (Wolfram
2002, p. 889). Here 
is a positive integer counting the nonhalting
states, the tape is infinite in both directions, and each transition writes 0 or
1 and moves one cell left or right. The final transition into a separate halting
state counts as a step.
The process leading to the solution of the three-state problem is discussed by Lin and Rado (1965), and the process leading to the solution of the four-state problem is discussed by Brady (1983) and Machlin and Stout (1990).
For 
, 2, ..., 
(also known as Rado's sigma function) is given by 1,
4, 6, 13, 4098, ... (OEIS A028444; Rado 1962;
Wolfram 2002, p. 889).
The equality 
is proved by The bbchallenge Collaboration et al. (2026, p. 37). An earlier
construction gives the lower bound 
(Marxen). The illustration above
shows a 3-state Turing machine with maximal 
(Lin and Rado 1965, Shallit
1998).
The maximum number of steps which an 
-state 2-symbol Turing machine
(not necessarily the same Turing machine
producing a maximum number of 1s) can perform before halting is 
. Its first few terms are 1, 6, 21, 107, 47176870, ... (OEIS
A060843; Wolfram 2002, p. 889).
Turing machines giving maximal 
for 
, 3, and 4 are illustrated above. The equality 
is proved using the Coq proof assistant (The bbchallenge
Collaboration et al. 2026, p. 2).
A set of 5-state rules found by Marxen and Buntrock (1990) that attains both proved maxima 
and 
is illustrated above (Wolfram
2002, p. 889;
The bbchallenge Collaboration et al. 2026, pp. 9, 37).
Pavel Kropitz constructed a 6-state, 2-symbol Turing machine in May 2022 (Ligocki 2022). Goucher (2022) reports that it takes approximately
 |
timesteps to halt, and Ligocki (2022) proves that it takes more than 
steps in Knuth
up-arrow notation. In the preceding display, exponentiation
is right-associative, so the rightmost exponentiation
is applied first. The machine leaves
 |
1s when the machine halts (Goucher 2022, Ligocki 2022).
Ketchersid (2026) reports two 7-state, 2-symbol Turing machines that halt from an initially all-zero tape with more than 
1s, where 
denotes 20 arrows in Knuth
up-arrow notation. This gives the reported lower bounds 
and 
. The submission includes Coq formalizations
of halting and the final number of 1s. As of Oct. 26, 2026, BusyBeaverWiki (2026)
lists the result as not independently verified. The exact values of 
and 
remain unknown.
Brady found a 3-state, 3-symbol Turing machine in December 2004 giving the lower bound 
for 
(Michel 2022). Here 
is the analogous maximum step count for 
nonhalting states and 
tape symbols, one of which is the initial blank symbol.
Historical results and analyses of the machines are collected by Michel (2022).
for the Value 
." IEEE Trans. Elec. Comput. EC-15, 802-803,
1966.Brady, A. H. "The Determination of the Value of Rado's
Noncomputable Function 
for Four-State Turing Machines." Math. Comput. 40, 647-665, 1983.Brady,
A. H. "Busy Beaver Problem of Tibor Rado ('Busy Beaver Game')." 
?" 
."
Jun. 21, 2022.