Can every odd integer greater than $1$ be written as a product of fractions $\frac{4m+1}{2m+1}$? Can every odd integer greater than $1$ be written as a product of fractions of the form $\frac{4m+1}{2m+1}$, $m$ a positive integer?
Here is a proof if the fractions were instead $\frac{4m-1}{2m+1}$. I tried to mimic it, but it doesn't work.

Since the fractions available have odd numerator and denominator, the same will be true of any product, so even integers aren't reachable. Odd integers can be reached, however.
Let $f(q)$ be the fraction $\frac{4q-1}{2q+1}$. Then $f(1)=1$, $f(4)f(7)=3$ and $3 f(4)=5$. So $1,3,5$ are obtainable.
Suppose now that $m\ge 2$ and we've obtained all odd numbers up to and including $4m-3$. Since $2m+1\le4m-3$ we can use $2m+1$ to obtain
\begin{gather*}
4m-1 = (2m+1) f(m),\\
4m+1=(2m+1) f(3m+1).
\end{gather*}
Hence $4m-1$ and $4m+1$ are obtainable as well, so having obtained $5$ we can then reach all odd positive integers.

Any help appreciated!
 A: $\def\N{\mathbb{N}}$This solution proves by induction on $n \geqslant 0$ that all odd integers $2n + 1$ can be expressed as products of numbers of the form $f(m) = \dfrac{4m + 1}{2m + 1}$.
For $n = 0$, there is $1 = f(0)$. Assume that the proposition holds for all odd integers less than $2n + 1$. Suppose $2n + 2 = 2^k (2n_0 + 1)$ ($k \in \N_+$, $n_0 \in \N$), then $n_0 < n$ and\begin{align*}
&\mathrel{\phantom{=}} (2n_0 + 1) \prod_{j = 0}^{k - 1} f\left(2^j(2^k - 1)n_0 + 2^j(2^{k - 1} - 1) \right)\\
&= (2n_0 + 1) \prod_{j = 0}^{k - 1} \frac{ 2^{j + 2}(2^k - 1)n_0 + 2^{j + 2}(2^{k - 1} - 1) + 1 }{ 2^{j + 1}(2^k - 1)n_0 + 2^{j + 1}(2^{k - 1} - 1) + 1 }\\
&= (2n_0 + 1) · \frac{ 2^{k + 1}(2^k - 1)n_0 + 2^{k + 1}(2^{k - 1} - 1) + 1 }{ 2(2^k - 1)n_0 + 2(2^{k - 1} - 1) + 1 }\\
&= (2n_0 + 1) · \frac{ 2^{k + 1}(2^k - 1)n_0 + (2^k - 1)^2 }{ 2(2^k - 1)n_0 + (2^k - 1) }\\
&= (2n_0 + 1) · \frac{ 2^{k + 1}n_0 + (2^k - 1) }{ 2n_0 + 1 }\\
&= 2^{k + 1}n_0 + (2^k - 1) = 2n + 1.
\end{align*}
End of induction.
A: For integer $a ≥ 0$ define $f(a) = (4a + 1) / (2a + 1)$.
Let $g_k(a) = f(a) \cdot f(2^1a) \cdot f(2^2a) \cdot … \cdot f(2^{k-1}a)$. $g_k(a)$ is a telescopic product, and $g_k(a) = (2^{k+1}a + 1) / (2a+1)$.
We have $g_k((b-1) / 2) = (2^k b - 2^k + 1) / b$.
We have $g_k((cy - 1) / 2) = (2^k cy - (2^k - 1)) / cy = 2^k - (2^k - 1) / cy$.
Pick an odd x > 1. Find the largest k such that x+1 is divisible by $2^k$, so $k ≥ 1$. Let $y = (x + 1) / 2^k$, then y is odd, $y < x$, and $2^k y = x + 1$.
Let $c = 2^k - 1$, then $y \cdot g_k((cy - 1) / 2) = 2^k y - 1 = (x + 1) - 1 = x$, so x is formed by multiplying y by k values of the form $f(a)$.
Together with 1 = f(0) this shows by induction that every odd number is a product of numbers of the form f(a). I think the number of values f(a) used according to this proof is k whenever $2^{k-1}<x<2^k$. Not sure if more or fewer values could be used.
