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A Turing Machine is a computational model concept that runs on the unrestricted grammar of Type-0. It accepts recursive enumerable language and comprises of an infinite tape length, where reading and writing operations can be performed accordingly.

$1^n$Assume that BB is a computable function. Now, if that so, then there must exist a Turing Machine F that will compute BB.

So, let F be TM that halts with $1^{BB\left(n\right)}$on the tape on input: for all value of 鈥渘鈥�

Now, construct a Turing Machine M which will halt if started with a blank tape as shown in the following steps:

1. M writes n times 1s on the tape.
2. M increases 1s to double on tape.
3. M runs F on the input $1^{2n}$. Therefore, M will always halt with $BB\left(2n\right)$when it starts with a blank tape.

We will run the 1st step up to n times, which correspond to 鈥渘鈥� states. And we have to run step 2 and step 3 for some constant state 鈥渃鈥�.

So according to this, $BB\left(n+c\right)$is the maximum number of 1s that a$\left(n+c\right)$ stage Turing Machine will halt. This implies that$BB\left(n+c\right)猢�BB\left(2n\right)鈭赌n鈥�鈥�鈥�鈥�鈥�\left(1\right)$

is monotonically increasing because$BB\left(n+1\right)猢�BB\left(n\right)鈭赌n鈭�N.$

But if that so then . This contradicts the above equation given in (1).

Hence, we can conclude that $BB\left(k\right)$is not a computable function.