There is interesting association between $2\cos(\frac{\pi}{60})$ which is $\frac{1}{2}\sqrt{8+\sqrt{15}+\sqrt3+\sqrt{10-2\sqrt5}}$ and first few Prime numbers with infinite expansion of balloon nested radical as a ratio to $\pi$
By applying formula $$\pi = \lim_{n \to \infty} n\sin(\frac{180^\circ}{n})$$ we can get increasingly accurate values of $\pi$ as follows
$\pi = \lim_{n\to\infty} 60\cdot2^n\cdot\sqrt{2\color{red}-\sqrt{2+\sqrt{2+\sqrt{2+...\text{(n times)}+\frac{1}{2}\sqrt{8+\sqrt{15}+\sqrt3+\sqrt{10-2\sqrt5}}}}}}$
Representing first few Prime numbers with ratio as infinite expansion of balloon nested radical, 60 and $2^n$ over $\pi$ and will be as follows. (With some arithmetic manipulation and rearrangement)
$7 = \frac {\lim_{n\to\infty} 60\cdot2^n\cdot\sqrt{2\color{red}-\sqrt{2+\sqrt{2+\sqrt{2+...\text{(n times)}+\frac{1}{2}\sqrt{8+\sqrt{15}-\sqrt3+\sqrt{10+2\sqrt5}}}}}}}{\pi}$
$11 = \frac {\lim_{n\to\infty} 60\cdot2^n\cdot\sqrt{2\color{red}-\sqrt{2+\sqrt{2+\sqrt{2+...\text{(n times)}+\frac{1}{2}\sqrt{8+\sqrt{15}+\sqrt3-\sqrt{10-2\sqrt5}}}}}}}{\pi}$
$13 = \frac {\lim_{n\to\infty} 60\cdot2^n\cdot\sqrt{2\color{red}-\sqrt{2+\sqrt{2+\sqrt{2+...\text{(n times)}+\frac{1}{2}\sqrt{8-\sqrt{15}+\sqrt3+\sqrt{10+2\sqrt5}}}}}}}{\pi}$
$17 = \frac {\lim_{n\to\infty} 60\cdot2^n\cdot\sqrt{2\color{red}-\sqrt{2+\sqrt{2+\sqrt{2+...\text{(n times)}+\frac{1}{2}\sqrt{8+\sqrt{15}-\sqrt3-\sqrt{10+2\sqrt5}}}}}}}{\pi}$
$19 = \frac {\lim_{n\to\infty} 60\cdot2^n\cdot\sqrt{2\color{red}-\sqrt{2+\sqrt{2+\sqrt{2+...\text{(n times)}+\frac{1}{2}\sqrt{8-\sqrt{15}-\sqrt3+\sqrt{10-2\sqrt5}}}}}}}{\pi}$
$23 = \frac {\lim_{n\to\infty} 60\cdot2^n\cdot\sqrt{2\color{red}-\sqrt{2+\sqrt{2+\sqrt{2+...\text{(n times)}+\frac{1}{2}\sqrt{8-\sqrt{15}+\sqrt3-\sqrt{10+2\sqrt5}}}}}}}{\pi}$
$29 = \frac {\lim_{n\to\infty} 60\cdot2^n\cdot\sqrt{2\color{red}-\sqrt{2+\sqrt{2+\sqrt{2+...\text{(n times)}+\frac{1}{2}\sqrt{8-\sqrt{15}-\sqrt3-\sqrt{10-2\sqrt5}}}}}}}{\pi}$
Note:
Infinite expansion happens between left and right extreme of nested radical where right extreme contains $2\cos\frac{7\pi}{60}, 2\cos\frac{11\pi}{60}, 2\cos\frac{13\pi}{60}, 2\cos\frac{17\pi}{60}, 2\cos\frac{19\pi}{60}, 2\cos\frac{23\pi}{60}, 2\cos\frac{29\pi}{60}$ respectively for $7,11,13,17,19,23,29$
Interesting observations
- 60 as a composite of first 3 prime numbers $2^2\cdot3\cdot5$
- $\pi$ i.e $180^\circ$ within the cosine angle able to eliminate the composite numbers from 6 to 30
- Value of transcendental number $\pi$ as denominator and irrational number as numerator ($2\cos\frac{x\pi}{60}$ where $x$ is a Prime number along with infinite expansion of nested Radical and $60$ and $2^n$) and I got enthralled by this part
On further exploration we may get the angles as follows
$\sqrt{2-\frac{1}{2}\sqrt{8+\sqrt{15}-\sqrt3+\sqrt{10+2\sqrt5}}} = 2\sin\frac{2\pi}{60}$ which is equal to $2\cos(\frac{\pi}{2}-\frac{2\pi}{60}) = 2\cos\frac{58\pi}{60} \text{ or } 2\cos\frac{29\pi}{30}$
As we are dealing with double angle of cosine values in nested radical, further simplification yields same set of prime numbers in reverse order.
$(30 - 1 = 29)$ ; $(30 - 7 = 23)$ ; $(30 - 11 = 19)$ ; $(30 - 13 = 17)$ ; $(30 - 17 = 13)$ ; $(30 - 19 = 11)$ ; $(30 - 23 = 7)$ ; $(30 - 29 = 1)$
But what happens next? That is quite more interesting; it is set of numbers which contain prime numbers.
$60-1 = 59$ ; $60-7 = 53$ ; $60-11 = 49$ ; $60-13 = 47$ ; $60-17 = 43$ ; $60-19 = 41$ ; $60-23 = 37$ ; $60-29 = 31$ ; $60-31 = 29$ ; $60-37 = 59$ ; $60-1 = 59$ ; $60-1 = 59$ ; $60-1 = 59$ ; $60-1 = 59$ ; $60-1 = 59$ ; $60-1 = 59$ ;
It is understandable that first 3 prime numbers $2\cdot3\cdot5 = 30$ along with 1 helps to generate further prime numbers by interaction of angle $\pi$ with half angle cosine formula along with $2^n\cdot30$. Here we start to get nonprime numbers but the set of nonprime numbers help to generate prime numbers further
Simpler expression as a set will be as follows
$2^1\cdot3\cdot5 - \{1,7,11,13,17,19,23,29\}= \{29,23,19,17,13,11,7,1\} ---->set(1) $ it has count of $2^3$ out of which $7$ are prime numbers
$2^2\cdot3\cdot5 \{1,7,11,13,17,19,23,29\}= \{59,53,49,47,43,41 ,37,31\}----> set(2)$ here 49 is the only composite number ----> set(2) also has 7 prime numbers out of 8
$2^3\cdot3\cdot5 (set(1)+set(2) \text {after sorting in ascending order} = \{119,113,109,107,103,101,97,91,89,83,79,77,73,71,67,61\}$ ---->set (3) has 13 out of 2^4 as prime numbers
Above set has 3 composite numbers from prime numbers $119 =17\cdot7, 91= 13\cdot7$ and $77 = 11\cdot7$
But previous composite number 49 generated a prime number $120-49 = 71$
Observation : Interaction of cosine angle with $180^\circ$ or $\pi$ in terms of multiples of $2^n\cdot2\cdot3\cdot5$ forms set of numbers in 2^n which contain prime numbers and composite numbers which generate prime numbers in further interaction with higher multiples of $30$s
Question:
I was able to get the nested radical for $2\cos(\frac{\pi}{60})$ where we can get permutation of signs to get other angles associated with prime numbers
Is it possible to get nested radical expression of $2\cos(\frac{\pi}{2^n\cdot60})$ and so on, in such way to get more permutations and thereby prime numbers? ( For me it was not possible). Therefore please help me in this regard to predict appearance of prime numbers in terms of changing signs in nested radical in permutation with $+$ and $-$ signs in half angles developing further in nested radical.