Number of Primes

$\text{ Let } f(n) = \sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}} \\ f^{2}(n) = n+f(n) \\ f^{2}(n)-f(n)-n=0 \\ f(n) = \dfrac{1}{2}+\dfrac{\sqrt{1+4n}}{2} \quad \small{\color{#3D99F6}{\text{-ve is neglected since f(n) is +ve }}} \\ \text{for f(n) to be a prime number , Let } 1+4n = m^2 \\ f(n) = \dfrac{1}{2} \left( 1+m \right) \\ 1+m \text{ must be an even no. , Let 2k } \\ f(n) = k \\ \text{ now , } n=\dfrac{1}{4}\left[ (2k-1)^2-1\right] \\ n = k^2-k \le 2016 \quad \small{\color{#3D99F6}{\text{where k is prime}}} \\ \text{Possible values of k } = \left( 2,3,5,7,11,13,17,19,23,29,31,37,41,43 \right) \\ \text{So our answer is } 14$