Number of positive integer solution, n=2x+y n is arbitnary positive integer.
then Cosider this equation   n=2x+y.
Where x and y are positive integer.
Finally, if gcd (x,y)=1, what's numer of pair of (x,y)?
Without the condition $\gcd(x,y)=1$ then the number of solutions is $\lfloor\frac{n-1}2\rfloor$, because $x$ can only range from $1 \text{ to }\lfloor\frac{n-1}2\rfloor$. One is the low value because $x$ must be positive. The high value of $x$ comes from $y$'s minimum values: $y=1=n-2\lfloor\frac{n-1}2\rfloor$ for odd $n$ and $y=2=n-2\lfloor\frac{n-1}2\rfloor$ for even $n$.

Remembering that $\gcd(x,y)=1$ is a condition: consider the case where $n$ is prime. Let $g=gcd(x,y), u=\frac{x}g, v=\frac{y}g$, then $n=2gu+gv$. Because $n$ is prime $g=1$. Therefore $\lfloor\frac{n-1}2\rfloor$ is the number of solutions when $n$ is prime.

If $n$ is composite the number of solutions to discard because $g&gt;1$ must be accounted for. 

Consider $n$ even, then: $1 \leq x \leq \frac{n}2-1$. Each even $x$ in this range must be discarded. Similarly if $3|n$ then each $x$ divisible by 3 must be discarded. This is true for all prime factors of $n$ with the exception of $\frac{n}2$ being prime, because $n\over 2$ isn't in the range. 

Let $P_n = \{\text{distinct prime factors of n not including }\frac{n}2\}$ and $m=\frac{n}2-1$ then the solution for even n is: $$m - \sum_{i \in P_n} \lfloor\frac{m}i\rfloor$$

For odd $n$, with $P_n = \{\text{distinct prime factors of n}\}$ and $m=\frac{n-1}2$:
$$m - \sum_{i \in P_n} \lfloor\frac{m}i\rfloor$$.
For an arbitrary natural number x, there are infinitely many values for y which satisfy gcd(x,y)=1. This can be shown by the fact that there are infinitely many prime numbers, each of which, by definition, can only have a gcd of 1 or itself with another number, and the latter case only occurs when the number is a multiple of that prime; therefore, every prime which x is not a multiple of will have a gcd of 1 with x, and since the prime factorization for a given natural number must be finite, there will be an infinite set of primes for which this is true. It follows that there are infinitely many pairs (x,y)  that satisfy these conditions.