A Wallis-like formula for $\frac{\pi}{4}$: $(\frac45)(\frac87)(\frac{10}{11})(\frac{12}{13})(\frac{14}{13})(\frac{16}{17})(\frac{18}{17})\cdots$

Question

I don't quite understand this and would appreciate a clear explanation, if possible.

If we begin with the Wallis product for $\frac{\pi}{2}$,

$$\prod_{n=1}^\infty \left( \frac{2n}{2n-1} \cdot \frac{2n}{2n+1} \right) = \Big(\frac{2}{1} \cdot \frac{2}{3}\Big) \cdot \Big(\frac{4}{3} \cdot \frac{4}{5}\Big) \cdot \Big(\frac{6}{5} \cdot \frac{6}{7}\Big) \cdot \Big(\frac{8}{7} \cdot \frac{8}{9}\Big) \cdot \; \cdots \\ = \Big(\frac{2}{1}\Big) \cdot \Big( \frac{2}{3}\Big) \cdot \Big(\frac{4}{3}\Big) \cdot \Big(\frac{4}{5}\Big) \cdot \Big(\frac{6}{5}\Big) \cdot \Big(\frac{6}{7}\Big) \cdot \Big(\frac{8}{7}\Big) \cdot \Big(\frac{8}{9}\Big) \cdot \; \cdots \\ = \frac{\pi}{2}$$

By removing only factors whose numerator and denominator sum to primes, we obtain

$$for a = 0, b = 1, c = 0, d = 1:\\\prod_{n=1}^\infty \left( \frac{2n}{2n-1}^{ab} \cdot \frac{2n}{2n+1} ^{cd}\right), where \begin{cases}a=1,&(4n-1)\text{ is not prime}\\\hfill b=1,\hfill&(4n-1)\text{ is prime}\\\hfill c=1,\hfill&(4n+1)\text{ is not prime}\\\hfill d=1,\hfill&(4n+1)\text{ is prime}\end{cases}\\=\Big(\frac{2}{1}\Big)^0 \cdot \Big( \frac{2}{3}\Big)^0 \cdot \Big(\frac{4}{3}\Big)^0 \cdot \Big(\frac{4}{5}\Big)^1 \cdot \Big(\frac{6}{5}\Big)^0 \cdot \Big(\frac{6}{7}\Big)^0 \cdot \Big(\frac{8}{7}\Big)^1 \cdot \Big(\frac{8}{9}\Big)^0 \cdot \; \cdots \\ = \Big(\frac{4}{5}\Big) \cdot \Big(\frac{8}{7}\Big) \cdot \Big(\frac{10}{11}\Big) \cdot \Big(\frac{12}{13}\Big) \cdot \Big(\frac{14}{13}\Big) \cdot \Big(\frac{16}{17}\Big) \cdot \Big(\frac{18}{17}\Big) \cdot \Big(\frac{20}{19}\Big) \cdots \\ =\sum_{n=0}^\infty\frac{(-1)^n}{2n+1} \\ = \frac{\pi}{4}$$

After yesterday's question, I understand that this result can be explained by the fact that the infinite product of the factors removed is 2. However, what interests me are the factors remaining. Since we know the infinite product of the remaining factors is precisely $\pi/4$, how can we explain this result in terms of the remaining factors: $(\frac45)(\frac87)(\frac{10}{11})(\frac{12}{13})(\frac{14}{13})\cdots$?

UPDATE:

While I am still unable to explain $\frac{\pi}{4}$ in terms of the remaining factors $(\frac45)(\frac87)(\frac{10}{11})(\frac{12}{13})(\frac{14}{13})\cdots\\$, I do have a general equation for the infinite product in terms of the exponents of the factors, $a, b, c, d,$ described below:

$$\prod_{n=1}^\infty \left( \frac{2n}{2n-1}^{ab} \cdot \frac{2n}{2n+1} ^{cd}\right), where \begin{cases}a=1,&(4n-1)\text{ is not prime}\\\hfill b=1,\hfill&(4n-1)\text{ is prime}\\\hfill c=1,\hfill&(4n+1)\text{ is not prime}\\\hfill d=1,\hfill&(4n+1)\text{ is prime}\end{cases}\\= \frac{(\pi^{2b})(2^{2a-b})(3^{a+b})}{(\pi^d)(2^{c+d})(3^{c+d})}\\=(\pi^{2b-d})(2^{2a-b-c-d})(3^{a+b-c-d})$$

I have included a print command in the code to reflect this expression.

import sympy
product, precision = 1, 10000
a, b, c, d = 0, 1, 0, 1
for k in range(1, precision + 1):
    n =  2 * k
    d1, d2 = n - 1, n + 1
    if sympy.isprime(n + d1):
        #print(product, n, d1)
        product = product * ((n / d1) ** a)
    else:
        print(product, n, d1)
        product = product * ((n / d1) ** b)
    if sympy.isprime(n + d2):
        #print(product, n, d2)
        product = product * ((n / d2) ** c)
    else:
        print(product, n, d2)
        product = product * ((n / d2) ** d)
print(product)

#UPDATE
calculated = ((sympy.pi ** ((2 * b) - d)) * (2 ** ((2 * a) - b - c - d)) * (3 ** (a + b - c - d)))
print(sympy.nsimplify(calculated))                                                                    

Answer 1

The convergence of your product is equivalent to the convergence of $\prod_p\frac{p-\chi(p)}{p+\chi(p)}$ which is almost equivalent to the prime number theorem for $\sum_{n>0}\chi(n)n^{-s}$ (which is a deep theorem with no elementary proof)

Thus you can't expect any other approach than the one given by metamorphy