Problems & Puzzles:
Puzzle 763. P(n) = P(n)^P(n+1)
Vic Bold sent the following nice pair of
(A) p(n)=p(n)^p(n+1) mod (p(n) +
p(n+1)) is true for the three
primes 11,13 & 17. Can this be true
for more than three consecutive
(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.
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
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
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.