Let $\mu(x)=0$ if $\exists y<x\mu(y)=0$ or $x$ is the 20 millionth prime. $\mu(x)=\mu(x+1)+1$ otherwise. The elements of the sequence only become defined once the 20 millionth prime is found; till then they’ve yet to be constructed.
Bonus: The sequence $\nu$ is defined as follows.
Let $\nu(x)=0$ if $\exists y\leq x\mu(y)$ is perfect. $\nu(x)=\nu(x+1)+1$ otherwise.
We’re not sure if this is a sequence, even though it is algorithmically definable. Its existence relies on the existence of an odd perfect number.
do it now…
Let $\mu(x)=0$ if $\exists y<x\mu(y)=0$ or $x$ is the 20 millionth prime. $\mu(x)=\mu(x+1)+1$ otherwise. The elements of the sequence only become defined once the 20 millionth prime is found; till then they’ve yet to be constructed.
Bonus: The sequence $\nu$ is defined as follows.
Let $\nu(x)=0$ if $\exists y\leq x\mu(y)$ is perfect. $\nu(x)=\nu(x+1)+1$ otherwise.
We’re not sure if this is a sequence, even though it is algorithmically definable. Its existence relies on the existence of an odd perfect number.