How can I generalize this?

Question

I'm not sure what to tag this question as or whether its a bit nonsensical, but I'm a bit curious. I asked a question on a pretty (turned out to be) easy question about the Collatz Sequence here:

Collatz $4n+1$ rule?

My question is observe the following sequences (next term is $4n+1$): $$3\ 13\ 53, \cdot\cdot\cdot\\5\ 21\ 85, \cdot\cdot\cdot\\7\ 29\ 117, \cdot\cdot\cdot\\9\ 37\ 149, \cdot\cdot\cdot\\11\ 45\ 181, \cdot\cdot\cdot\\15\ 61\ 245, \cdot\cdot\cdot\\17\ 69\ 277, \cdot\cdot\cdot\\19\ 77\ 309, \cdot\cdot\cdot$$ Basically take some odd number and multiply it by $4$ and add $1$ and you get the next number in the series. I am trying to find all the starting numbers of all the sequences that don't collide with each other. The sequences above do not repeat each other (I think?) and you can see the starting numbers are (if you expand it) $$3, 5, 7, 9, 11, 15, 17, 19, 23, 25, 27, 31, 33, 35, 39, 41, 43, 47, 49, 51, 55, 57, 59, ...$$ With the numbers $13, 37, 53$ colliding. How can I generalize which numbers will not collide or have been covered by others as these are the true "unknown" values of the odd numbers of the Collatz Sequence?

Answer 1

You could notice that the formula describing each of those sequences is of the form $\frac{1}{3}(-1+4^n+3\cdot 4^n\cdot(2j-1))$, where $2j-1$ is the initial odd value.

Then the numbers $2k-1$ which have already been covered in a sequence must be of the form $2k-1=\frac{1}{3}(-1+4^n+3\cdot 4^n\cdot(2j-1))$ for some $n,j\in\mathbb{N}$. Rearranging this yields: $$k=\frac{\frac{1}{3}(-1+4^n+3\cdot 4^n\cdot(2j-1))+1}{2}$$ By varying $j$ and $n$, we then have a formula for all $k$ so that $2k-1$ already appears in a sequence.

EDIT: As requested, to find the formula describing each of the sequences, we note that given an odd number $c=2j-1$, any of the sequences may be described by the recursive relation $a_0=c$; $a_n=4a_{n-1}+1$, for $n\geq 1$. Then, we form the generating function $f(x)=\sum\limits_{n=0}^{\infty}a_nx^n$. From this, we obtain: $$f(x)=\sum\limits_{n=0}^{\infty}a_nx^n=c+\sum\limits_{n=1}^{\infty}a_nx^n=c+\sum\limits_{n=1}^{\infty}(4a_{n-1}+1)x^n$$ Expanding the last term and pulling out an $x$ yields: $$f(x)=c+x(4\sum\limits_{n=1}^{\infty}a_{n-1}x^{n-1}+\sum\limits_{n=1}^{\infty}x^{n-1})$$ Reindexing yields: $$f(x)=c+x(4\sum\limits_{n=0}^{\infty}a_{n}x^{n}+\sum\limits_{n=0}^{\infty}x^{n})$$ Substituting $f(x)=\sum\limits_{n=0}^{\infty}a_nx^n$ and using $\sum\limits_{n=0}^{\infty}x^{n}=\frac{1}{1-x}$ gives us: $$f(x)=c+x(4\cdot f(x)+\frac{1}{1-x})$$ Rearranging and solving for $f(x)$: $$f(x)=\frac{c}{1-4x}+\frac{x}{(1-x)(1-4x)}=c\cdot\frac{1}{1-4x}-\frac{1}{3}\cdot\frac{1}{1-x}+\frac{1}{3}\cdot\frac{1}{1-4x}$$ Then using the power series expansion $\frac{1}{1-rx}=\sum\limits_{n=0}^{\infty}r^nx^n$, we finally obtain: $$f(x)=c\cdot\sum\limits_{n=0}^{\infty}4^nx^n-\frac{1}{3}\cdot\sum\limits_{n=0}^{\infty}x^n+\frac{1}{3}\cdot\sum\limits_{n=0}^{\infty}4^nx^n$$ Combining these: $$f(x)=\sum\limits_{n=0}^{\infty}\frac{1}{3}(3\cdot 4^n\cdot c-1+4^n)x^n$$ Then comparing coefficients on both sides gives us the desired result of: $$a_n=\frac{1}{3}(3\cdot 4^n\cdot c-1+4^n)$$