This might seem a vague question, but let me give you its background before going on. I have obtained an equation where a=2mn+m+n, where all a,n and m are natural numbers and neither 'm' or 'n' or both of them can be equal to 0 (0 is not a natural number, but a whole number).
Not all values of a i.e. the natural numbers satisfy 2mn+m+n and these are the numbers that I need to find (i.e. the values for 'a' which don't satisfy $2mn+m+m$).
For this, I see that 2mn+m+n can be written as $(2mn+m) + n = (2n+1)m + n$. All the numbers can be represented as (2n+1)m except for $2^x$. This is because, in $(2n+1)m$, $2n+1$ is an odd number, i.e. all the odd numbers can written as such. Similarly, only those even numbers whose prime factorization consists purely of even numbers i.e. 2, CANNOT be written in this form i.e. $2^x$ numbers.
We have some values for 'a' which are not equal to 2mn+m+n and we're looking to find those. Similarly, we see that (2n+1)m also CANNOT be written as $2^x$.
Thus, $a ≠ 2mn+m+n$, $2^x ≠ (2n+1)m$ i.e. $2mn+m$.
Therefore, $2^x+n≠ 2mn+m+n$ (the inequality is maintained here), $2^x+n ≠ a$
Now, a is not alone but, itself is a part of the equation $2a+1$, where 'a' CAN be represented as 2mn+m+n, i.e. '2a+1' is an odd composite number, as in $2a+1$, a can be written as 2mn+m+n, how?? $(2n+1)(2m+1) = 4mn+2m+2n+1 = 2(2mn+m+n)+1$, while the values for a which don't satisfy $2mn+m+n$, $2a+1$ is a prime number.
Therefore, $a ≠ 2^x+n$. Similarly, $2a+1 ≠ 2(2^x+n)+1$ (again, inequality is maintained).
$2^{x+1} + 2n+1≠2a+1$. Thus, $2^{x+1} ≠ 2a-2n$,
$2^{x+1} ≠ 2(a-n)$,
$2^x≠a-n$, where 'a' can be represented as 2mn+m+n.
i.e. $2^x≠2mn+m+n-n$,
giving us, $2^x≠2mn+m$, $2^x≠(2n+1)m$
And, where 'a' CANNOT be represented as $2mn+m+n$, (i.e. where $2a+1$ is prime)
$2^x=a-n$
$2^x$ is given, and there is a pre requisite condition for $a$. So, my question - Is there any way to generalize the values for $n$??
An easier way to do this would be to consider the equivalent equation $$2a+1=(2m+1)(2n+1).$$ It means that the odd number $2a+1$ can't have any non-trivial (i.e. different from $1$ or itself) odd divisors, meaning it must be an odd prime number.