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.
It’s possible to define an infinite sequence whose nth term must exist before it’s first.
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.