Is the following inductive argument valid?
Let:
- $x$ be a positive real
- $n,t \ge 1$ be an integers
- $p_n$ be the $n$th prime
- $p\#$ be the primorial of $p$
- gcd$(a,b)$ is the greatest common divisor of $a$ and $b$
- $H_n(x)$ be the count of integer $i$ such that $0 < i \le x$ and gcd$(i,p_n\#)=1$
- $x \% t$ be the modulo operation such that if $c = x \% t$, then $\lfloor x\rfloor \equiv c \pmod t$ and $0 \le c < t$
- $c_n(x)$ be recurrence relations defined as follows:
- $c_0(x) = \dfrac{x \% t}{t}$
- $c_n(x) = c_{n-1}(x) - c_{n-1}\left(\dfrac{x}{p_{n}}\right)$
Observation:
- $H_n(x)$ can be defined as a recurrence relation:
- $H_0(x) = \lfloor x \rfloor$
- $H_n(x) = H_{n-1}(x) - H_{n-1}\left(\frac{x}{p_{n}}\right)$
Inductive Argment to be evaluated:
For any given positive integer $t$: $$\frac{H_n(x)}{t} = H_n\left(\frac{x}{t}\right) + c_n(x)$$
(1) Base Case $n=0$
- $\dfrac{\lfloor x\rfloor}{t} - \dfrac{\lfloor x \rfloor \% t}{t} = \left\lfloor\dfrac{\lfloor x\rfloor}{t}\right\rfloor = \left\lfloor\dfrac{x}{t}\right\rfloor$
- $\dfrac{H_0(x)}{t} = \dfrac{\lfloor x\rfloor}{t}= \left\lfloor\dfrac{x}{t}\right\rfloor + \dfrac{\lfloor x \rfloor \% t}{t} = H_0\left(\dfrac{x}{t}\right) + c_0(x)$
(2) Assume up to $n\ge0$:
- $\dfrac{H_n(x)}{t} = H_n\left(\dfrac{x}{t}\right) + c_n(x)$
(3) Inductive Step:
- $\dfrac{H_{n+1}(x)}{t} = \dfrac{H_{n}(x) - H_{n}\left(\frac{x}{p_{n+1}}\right)}{t} = H_n\left(\dfrac{x}{t}\right) + c_n(x) - H_n\left(\dfrac{\frac{x}{p_{n+1}}}{t}\right) - c_n\left(\dfrac{x}{p_{n+1}}\right) = H_{n+1}\left(\dfrac{x}{t}\right) + c_{n+1}(x)$