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This sequence is uncomputable.

No halt state is required. The machine halts whenever the current state/tape symbol combination does not have a corresponding instruction. For example, if the machine is in state B at a tape symbol 1, but there is no instruction for B1, the machine halts. As a specific example, the 2-instruction machine defined by A0:0RB, B0:1LA halts after 3 steps when the machine enters state B at a tape symbol 1. A halt state can be effectively simulated by transitioning to a state not included in the machine's instructions. As another example, the 2-state 2-symbol busy beaver can be given as A0:1RB, A1:1LB, B0:1LA, B1:1RC, where "C" effectively substitutes for "H". So the machine becomes a 3-state 2-symbol machine, but with no instructions for the third state "C".

This sequence effectively combines the various Busy Beaver Sigma functions Sigma(states,symbols) into a single sequence defined only by number of allowable instructions Sigma_i(inst). It is very interesting to observe the number of states and symbols associated with the champion machine for each number of instructions.

This sequence is uncomputable.

No halting instruction is required, as a Turing machine halts whenever the current state/tape symbol combination does not have a corresponding instruction. However, a halting instruction can be implemented by transitioning to a state which has no instructions. For example, the 2-state 2-symbol busy beaver can be given as A0:1RB, A1:1LB, B0:1LA, B1:1RC, where "C" effectively substitutes for "H".

This sequence effectively combines the various Busy Beaver step functions BB(states,symbols) into a single sequence defined only by number of allowable instructions BBi(inst). It is very interesting to observe the number of states and symbols associated with the champion machine for each number of instructions.

a(6)=123 confirmed by Shawn Ligocki on bbchallenge discord on Jul 12 2025.

a(7)=3932963 confirmed and announced on discord on Jul 23 2025. Two different Turing machines - one a 2-state, 4-symbol machine, and the other a 3-state, 3-symbol machine - each take 3932963 steps, writing 2050 non-blank symbols to tape.

A new champion machine for a(8) was discovered by Nick Drozd and confirmed by Shawn Ligocki on Jul 26 2025. This machine takes around 6.889 x 10^1565 steps before halting.

Note that there may be other vertices in the Galebach list of u-uniform tilings with u <= 6 that have this same coordination sequence. See the Galebach link for the complete list of A-numbers for all these tilings.

Note that there may be other vertices in the Galebach list of u-uniform tilings with u <= 6 that have this same coordination sequence. See the Galebach link for the complete list of A-numbers for all these tilings.

Note that there may be other vertices in the Galebach list of u-uniform tilings with u <= 6 that have this same coordination sequence. See the Galebach link for the complete list of A-numbers for all these tilings.

Note that there may be other vertices in the Galebach list of u-uniform tilings with u <= 6 that have this same coordination sequence. See the Galebach link for the complete list of A-numbers for all these tilings.

Note that there may be other vertices in the Galebach list of u-uniform tilings with u <= 6 that have this same coordination sequence. See the Galebach link for the complete list of A-numbers for all these tilings.

Note that there may be other vertices in the Galebach list of u-uniform tilings with u <= 6 that have this same coordination sequence. See the Galebach link for the complete list of A-numbers for all these tilings.

Note that there may be other vertices in the Galebach list of u-uniform tilings with u <= 6 that have this same coordination sequence. See the Galebach link for the complete list of A-numbers for all these tilings.

Note that there may be other vertices in the Galebach list of u-uniform tilings with u <= 6 that have this same coordination sequence. See the Galebach link for the complete list of A-numbers for all these tilings.