Help me figure this out How do I find the number of times needed for these patterns of numbers to be added to a result. It's kinda like the process of figuring out binary.
1, 2, 4, 8, 16, 32 ,64, 128, 256... etc.
For example, the number 17 would be 2 because 16 + 1 = 17, because 2 numbers are being added.
The number 25 would be 3 because 16 + 8 + 1 = 25 because 3 numbers are being added.

Question

dustin_whitelock · Answer

http://openstudy.com/updates/584d6affe4b071fda882243d

holsteremission · Answer

Sounds like you want a formula that gives the number of $1$s in the binary expansion for any given decimal integer. Let's say $a_n$ denotes this number for any (non-negative) integer $n$.

Starting with $n=0$, you have $0_{10}=0_2$, so $a_0=0$.

Now assume $n>0$, and consider the cases when $n$ is even and when $n$ is odd.

If $n$ has $a_n$ $1$s in its binary expansion, then $2n$ will also have $a_n$ $1$s. Why? Consider for example $n=3$ and $n=6$.
$$3_{10}=11_2=2^1+2^0\implies 6_{10}=2\times3_{10}=2^2+2^1=110_2$$Any time you double $n$, you add a new digit to $n$'s binary expansion, but that new digit is $0$. So $a_{2n}=a_n$.

The odd case can be treated similarly. Odd integers can be written in the form $2n+1$. If $2n$ has $a_{2n}=a_n$ $1$s, then the expansion for $2n+1$ is the same but adds $1$ as the last digit. This means $a_{2n+1}=a_{2n}+1=a_n+1$.

So you have the recursive formula for the number of $1$s in a binary expansion:
$$\begin{cases}a_0=0\[1ex]a_{2n}=a_n\[1ex]a_{2n+1}=a_n+1\end{cases}$$Let's verify that this works for the examples you provide.
$$\begin{align*}
n=17\implies a_{17}&=a_{2\times8+1}\[1ex]&=a_8+1\[1ex]&=1+1\[1ex]&\stackrel{\color{red}\checkmark}{=}2\end{align*}$$$$\begin{align*}
n=25\implies a_{25}&=a_{2\times12+1}\[1ex]
&=a_{12}+1\[1ex]
&=a_6+1\[1ex]
&=a_3+1\[1ex]&=a_{2\times1+1}+1\[1ex]&=a_1+2\[1ex]
&=1+2\[1ex]&\stackrel{\color{red}\checkmark}=3\end{align*}$$

steve816 · Answer

Thank you so much!