An Unusual Application of Continued Fractions

Yesterday I turned 27487 days old. This is a prime number of days and this number together with the next three prime numbers form the following sequence:$$27847, 27851, 27883, 27893$$Now let's combine these numbers together to from a continued fraction:

                    1            
27847 + -------------------------
                        1        
         27851 + ----------------
                             1   
                  27883 + -------
                           27893 

This continued fraction is equivalent to the improper fraction:$$ \dfrac{603189725547991331} {21660851250413}$$with both numerator and denominator being prime. The first of the primes, 27847, qualifies it for membership in OEIS A270884:

A270884: smallest of FOUR consecutive prime numbers that when represented as a simple continued fraction, generates prime numbers in the numerator and denominator, when reduced.

The initial members are (permalink):

41, 367, 619, 659, 701, 2267, 2789, 3253, 3463, 6917, 8969, 9221, 11959, 13499, 14431, 17359, 17851, 20143, 22283, 23669, 26107, 27847, 28547, 28879, 29537, 32503, 32717, 32987, 37549

Let's look at the first of these numbers, 41, for another example. The four primes are thus 41, 43, 47 and 53.

            1        
41 + ----------------
               1     
      43 + ----------
                  1  
            47 + ----
                  53 

Here the continued fraction is equal to:$$ \frac{4398061} {107209}$$with both numerator and denominator being prime. Up to one million, the continued fractions of five and six runs of consecutive primes do not produce improper fractions with numerators and denominators that are prime. However, some runs of seven consecutive primes do. Up to 40,000 these are 223, 1579, 5881, 8293, 9013, 12347, 15121, 17783, 25523 and 38903 (permalink). Let's look at the first of these: 223.

                         1                    
223 + ----------------------------------------
                             1                
       227 + ---------------------------------
                                1             
              229 + --------------------------
                                    1         
                     233 + -------------------
                                       1      
                            239 + ------------
                                           1  
                                   241 + -----
                                          251 

This produces the fraction:$$ \frac{39053357680436977}{175123705045789} $$with both numerator and denominator prime. No numbers satisfy for runs of eight and nine consecutive primes but for runs of ten we have 5519, 13037, 18743 and 39857 (permalink). Let's look at 5519. The continued fraction (this time in compact format) is:$$ \begin{align} &5519; 5521, 5527, 5531, 5557, 5563, 5569, 5573, 5581, 5591]\\  \\ &= \frac{27886154132102326260514832393879709899}{5052754705003449540186402260387567} \end{align}$$with the numerator and denominator prime.