# turning $2x$ into a perfect even

So I am trying to generate a sequence with an equation (that I don't think exists) and it involves all the even numbers, and one way to find the sequence is to get rid of all odd prime numbers so...

$$2\frac{x}{n_k}$$ where $$n_m$$ is $$x$$ without any $$2$$'s in its prime factorization.

then I tried to make an equation for $$n_k$$.

$$n_m=n_{m-1}(\frac{(n_{m-1}\mod{2})+1}2)$$
where $$n_1=x$$
or
$$n_m=n_{m-1}(\frac{3-\cos(\pi n_{m-1})}{4})$$
where $$n_1=x$$

now that I have a recursive formula I was wondering if any one could make a standard formula from this (and show how you did it if you want)

• Are you trying to list all numbers without $2$ in their prime factorization? If so, just list all the odd integers – gt6989b May 21 at 16:39
• @gt6989b no, I'm not I am trying to get a sequence from this, 1,2,1,3,1,2,1,4,... you can get this by seeing how many $2$'s are in each even number 2,4,6,8,10,12,14,16 – spydragon May 21 at 16:42
• So what are you trying to do? – gt6989b May 21 at 16:43
• accidentally replied too early – spydragon May 21 at 16:44
• So, if we were to include the odd numbers in this list, this would be $a_n$ is the maximum integer value of $k$ such that $2^k$ divides evenly into $n$? That would be oeis.org/A007814. If you were to skip the odd numbers, that would be the ruler function oeis.org/A001511 which is just the previous sequence after adding $1$ to every entry. – JMoravitz May 21 at 16:52