Are all odd natural numbers of form $\dfrac{n\cdot 2^{\alpha}-1}{3}$, where $n$ is also an odd natural number? While searching for patterns in odd natural numbers, I realised that if $k$ is an odd natural number then:
$$k=\frac{n \cdot 2^{\alpha}-1}{3},$$
where $n$ is an odd natural number and $\alpha \in \mathbb{N}$.
Even though I have arrived at this conjecture I cannot seem to think of a way to prove it. Can someone furnish the proof for the said conjecture or refer to any earlier stated proofs of this.
 A: Any even number can be expressed as $n.2^\alpha$ where $n$ is odd and $\alpha\geq 1$ - just keep taking out factors of $2$ until what you have left is odd. So the statement is basically saying "if $k$ is odd then $3k+1$ is even", which is immediate.
A: if 3 does not divide n, then there are multiples alphas that make n * 2 ^ alpha -1 divisible by 3.
If n mod 3 is 1, then alpha is even (any even number> = 2), and gives a valid k.
If n mod 3 is 2, then alpha is odd (any odd number> = 1), and gives a valid k.
If 3 divides n, then there is no alpha that gives a valid k.
n=1, alpha=8, k=85
n=1, alpha=6, k=21
n=17, alpha=1, k=11
n=17, alpha=3, k=45
...
The other sense, given k, finding alpha and n is trivial, and is related to how the number (3k + 1) is written in binary code.
The number of the last 0's is the alpha, and odd n is the result of removing those last 0's.
Example:
k=45
3k+1 = 136 (dec) = 10001000 (bin)
alpha = 3 (last 3 zeros)
n = 10001 (bin) = 17 (dec)
And every number, which in binary, ends in 1, is an odd number.
