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Let $BB(n)$ be the $n$-th Busy Beaver number, let $S(2)=BB(2)$ and for $n>2$ let $S(n)=1+S(n-1)+BB(n)$.

Then is this an uncomputable number?

$$\underset{i=2}{\overset{\infty}{\sum}}\frac{10^{BB(i)}-1}{9\cdot10^{S(i)}}=0.11110111111011111111111110...$$

user3635700
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    In other words, the digits consist of $BB(n)$-length strings of $1$s (for $n\geq2$), separated by single $0$s, correct? – Blue May 14 '21 at 20:57
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    Of course the proof somewhat depends on your definition of computable number, but under any reasonable definition surely you can see that it would not be computable? Suppose we could compute it. Then we could compute $BB(n)$ for each $n$ as per the interpretation given by @Blue. This is a contradiction. – Isky Mathews May 14 '21 at 23:28
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    See also: https://math.stackexchange.com/questions/1266587/example-of-uncomputable-but-definable-number?rq=1 for a similar idea – Isky Mathews May 14 '21 at 23:28
  • @Blue that was my intention yes. – user3635700 May 16 '21 at 10:14

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