J. M. Bergot posed
a puzzle about emirps. Here is my reformulation of it:
points out to a pair
of distinct primes such that any of them is the reverse of the
Emirp = (p & p') such that p=reverse(p') and p'=reverse(p)
Here are the first
eight primes generating a couple of emirps:
13, 17, 31, 37, 71, 73, 79, 97, .(sequence A006567 in
For our puzzle we will consider the couple of primes that form an
emirp as a unit.
(13 & 31)
(17 & 71)
The puzzle asks to find emirps such that for both primes of the
emirp, the sum of the prime and its digits produces another emirp.
For Emirp1=(p & p'), p+sod(p)=q and p'+sod(p')=r such that (q &
q')=Emirp2 and (r & r')=Emirp3. This mean that Emirp1 define two
Emirps: Emirp2 and Emirp 3.
Example: Emirp1=(1933, 3391), sod(p)=16 -> Emirp2=(1949, 9491),
Q1. Send your largest example.
Q2. Send and example where both Emirp2 and Emirp3 behave as
Emirp1 (this mean that Emirp1 will define six Emirps)