How to determine number with same amount of odd and even divisors With given number N, how to determine first number after N with same amount of odd and even divisors?
For example if we have N=1, then next number we are searching for is : 
2
because divisors:
odd : 1
even : 2

I figured out that this special number can't be odd and obviously it can't be prime.
I can't find any formula for this or do i have just compute it one by one and check if it's this special number ? Obviously 1 and number itself is divisors of this number.
Cheers

If $2n$ is a divisor of $2m$1, then
$$
2n \mid 2m \Leftrightarrow n \mid m
$$
In other words, $2n$ is a divisor of $2m$ if and only if $n$ is a divisor of $m$. So for every even divisor of $2m$ there is a divisor of $m$, and for every divisor of $m$ there is an even divisor of $2m$. Now we conclude that the number of even divisors of $2m$ is equal to the number of divisors of $m$ (#).
Now if $2n - 1$ is a divisor of $2m$, then
$$
2n - 1 \mid 2m \Leftrightarrow 2n - 1 \mid m
$$
In other words, $2n - 1$ is a divisor of $2m$ if and only if $2n - 1$ is a divisor of $m$.  So for every odd divisor of $2m$ there is an odd divisor of $m$ and for every odd divisor of $m$ there is an odd divisor of $2m$. Now we conclude that the number of odd divisors of $2m$ is equal to the number of odd divisors of $m$ (##).
If the number of even divisors of $2m$ is equal to the number of odd divisors of it, from ## and # we conclude that the number of divisors of $m$ is equal to the number of odd divisors of it, which means all of the divisors of $m$ are odd, so $m$ is odd.
So if the number of even divisors and odd divisors of $2m$ are equal, then $m$ is odd.
\begin{align}
2m &= 2(2n - 1) \\
&= 4n - 2 \\
&= 2 + 4(n - 1) \\
&= a_1 + d(n - 1), n \in \mathbb{N}
\end{align}
Every term of the sequence of every other even number starting with $2$ has the property.

1.As you said the number has to be even so I wrote $2m$.
