Can every odd integer greater than $1$ be written as a product of fractions $\frac{4m+1}{2m+1}$?

Question

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.

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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.

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Any help appreciated!

Answer 1

$\def\N{\mathbb{N}}$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.

Answer 2

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.