Problems & Puzzles: Puzzles

Puzzle 763. P(n) = P(n)^P(n+1) mod (P(n)+P(n+1))

Vic Bold sent the following nice pair of puzzle:

(A) p(n)=p(n)^p(n+1) mod (p(n) + p(n+1)) is true for the three consecutive primes 11,13 & 17. Can this be true  for more than three consecutive primes?

(B) p(n+1)=p(n+1)^p(n) mod (p(n)+p(n+1)) is true for the seven consecutive primes 5, 7, 11,13 ,17, 19, 23. Can this be true for more than seven consecutive primes?  

Carlos Rivera made a preliminary search to seize this puzzle-proposal and counted -for each equation- the quantity of solutions for k consecutive primes below 2^32.

Here are his results:

For equation (A), for k = 0,1, 2, & 3 the quantity of solutions are  177793246, 12742978, 327, 9 respectively and none of four or more consecutive primes.

For equation (B), for k = 0,1, 2, 3, 4, 5, 6 & 7 the quantity of solutions are  170223965, 15345649, 721488, 46243, 2909, 153, 7, 1 respectively and none of seven or more consecutive primes.

If the CR's counting are correct here arises perhaps an irregularity. ¿Why are these counting so different if the two equations seems to be so similar?

Q. Can you explain the counting different behavior for the equations (A) and (B)?

Contribution came from Emmanuel Vantieghem


Emmanuel wrote:

No doubt your countings for the cases  k = 0  contain an (non essential) error.
For equation (A), the first term should be  190536562  instead of  177793246
and for equation (B)  it should be  186340417  instead of  170223965.

I continued the countings for p(n) in large intervals and the phenomenom that k <= 4 in case (A) and that k can be bigger than 4 in case (B) seems to persist.
But it is not impossible that for very big and far intervals the number of solutions becomes 'similar' in both cases.
Of course, everything depends on the notion of 'similarity'.  I know of very 'similar' equations that have complete different solutions, like this ones :
   (A')  (p-1)^q = 1 mod (p+q)
   (B')  q^(p-1) = 1 mod (p+q).

Regarding the percentages of the solutions, you could have asked "why are the countings so similar while the equations are so different ?" as well.  


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