Problems & Puzzles: Puzzles

Puzzle 39.- The Mirrorable Numbers (By Mike Keith)

We call a number N a Dihedral Calculator Prime (or just a Dihedral Prime) if it has the following property: if we display N on a calculator, then the four numbers

(a) N
(b) N upside down
(c) N in a mirror, and
(d) N upside down and in a mirror

are all primes. Note that (a), (b), (c), (d) need not be all different; the only requirement is that they all be primes.

It's easy to see that the digits of N must be restricted to
0, 1, 2, 5, and 8.

Here are all Dihedral Primes less than 10^6:

101 181
1181 1811
108881 110881 118081 120121 121021 121151 150151
151051 151121 180181 180811 181081 188011 188801

which means the number of such primes with 1,2,3,... digits is:

(**) 1, 1, 2, 2, 1, 14, ...

Now, the puzzles:

(i) How far can you extend the sequence (**) above?

(ii) 120121 is the smallest such prime in which all four
numbers (a, b, c, d) are distinct. How many primes like that are
there for 6,7,8,... digits?

(iii) This idea can be extended so that there are even more than four numbers to look at. For example, place the calculator on a table and stand a mirror on end touching the left edge. Look into this mirror while seated at the bottom of the calculator, and you will see a 2N-digit number consisting of number (c), N-in-a-mirror followed by (a), the digits of N. You can also place a mirror on the right edge of the calculator and do the same (which gives (a)(c)). You can also turn the calculator upside down and look in the same two mirrors, which gives two more numbers.

Thus the final, most challenging puzzle: can you find a number N such that all eight of these numbers (a, b, c, d, plus these four 2N-digit numbers) are primes? It is known that there is no such N less than 10^11, but heuristic arguments suggest that eventually there will be one.

Felice Russo has solved (June 1999) the questions (i) and (ii) of this puzzle:

(i): "The sequence of number of Dihedral primes... is:1, 1, 2, 2, 1, 14, 40, 52, 228, 482"

(ii) "The smallest primes for 7,8,9 and 10 digits in which all four numbers (a,b,c,d) are distinct are ...: 1028011, 10128011, 100252511, 1000202221"

Russo also obtained other results related with this kind of numbers:

"the number of Dihedral primes with four numbers a,b,c,d distinct for 6,7,8,9 and 10 digits is: 4,12,16,132,308"

" Moreover I obtained a more accurate result for the series SUM(1/Pd) for Pd <= 9*10^9. It is: 0.607924 (we could call it the Keith's constant or the Dihedral primes constant, K)

Also for the number (#Pd) of Dihedral primes vs the number of digits d I found a better fit: #P(d)~2/5*K*exp(5/4*K*d), where with "k" I indicated the "Keith's constant

Russo shows that "If the above approximation for #P(d) is correct, then SUM(1/Pd) converges... to <~ 0.661", and adds: "Here below a heuristic to justify the exponential growth observed for #P(d). Since the Diheadral numbers can contain only the digits 0,1,2,5 and 8 we can have 4*5^(d-1) of these numbers with d digits (excluding zero as first digit). According to the Prime Number Theorem the probability that a random number with d digit is prime is given by: 1/(d*ln(10)). Anyway this probability must be increased because the prime numbers are not divisible by 2 and 5. So it became: 5/(2*d(ln(10)).
The number of Dihedral primes with d digits then is given by:
For large d the number of Dihedral primes became roughly : e^(d*ln(5))


Patrick Capelle wrote (Feb. 9, 08):

I would like to mention that the prime number 5 is missing in the list of dihedral primes less than 10^6.
Hence, the Sloane's sequences A038136, A048660, A048662 should be changed.

Reference :


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