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3 Generalize from base 10 to arbitrary base $b$, and note equivalence with PNT for AP's mod powers of $b$.

While $\lbrace v_p \rbrace$ is clearly not equidistributed in $(0,1)$, it is equidistributed in $$(\Pi_{10} := [.1,.2) \cup ([.3,.4) \cup ([.7,.8) \cup ([.9,1)$$ by the prime number theorem (PNT) for arithmetic progressions modulo powers of $10$. In particular, the average tends to $0.55$, the average of the midpoints $0.15$, $0.35$, $0.75$, and $0.95$ of these four intervals.

[I see that Frank Thorne wrote much the same thing in a comment as I was editing this...]

[added in reply to further inquiries:] Replacing $10$ by an arbitrary base $b>1$, we likewise use the PNT for arithmetic progressions (APs) modulo powers of $b$ to show that the sequence is equidistributed in $$\Pi_b := \lbrace x \in [0,1) : \gcd(\lfloor bx \rfloor, b) = 1 \rbrace.$$ (I made the intervals half-open in $\Pi_{10}$ for consistency with this definition, though it doesn't change the distribution.) That's a union of $\varphi(b)$ intervals of length $1/b$ whose set of left endpoints is symmetric about $1/2$, so the average is $1/2 + 1/(2b) = (b+1)/2b$ as Timothy Foo surmised.

On further thought, not only does equidistribution in $\Pi_b$ follow from the PNT for APs modulo every power of $b$, but the two statements are readily seen to be equivalent.

2 Typo correction: arithmetic progressions, not analytic

While $\lbrace v_p \rbrace$ is clearly not equidistributed in $(0,1)$, it is equidistributed in $$(.1,.2) \cup (.3,.4) \cup (.7,.8) \cup (.9,1)$$ by the prime number theorem for analytic arithmetic progressions modulo powers of $10$. In particular, the average tends to $0.55$, the average of the midpoints $0.15$, $0.35$, $0.75$, and $0.95$ of these four intervals.

[I see that Frank Thorne wrote much the same thing in a comment as I was editing this...]

1

While $\lbrace v_p \rbrace$ is clearly not equidistributed in $(0,1)$, it is equidistributed in $$(.1,.2) \cup (.3,.4) \cup (.7,.8) \cup (.9,1)$$ by the prime number theorem for analytic progressions modulo powers of $10$. In particular, the average tends to $0.55$, the average of the midpoints $0.15$, $0.35$, $0.75$, and $0.95$ of these four intervals.

[I see that Frank Thorne wrote much the same thing in a comment as I was editing this...]