Problems & Puzzles: Conjectures

Conjecture 14.- Enoch Haga Observation about Palprimes

Patrick De Geest has been collecting the quantity of Palprimes of every (odd) length. Please see here.

The results of this collection up today are:

Enoch Haga had the inspired idea of multiplying this two columns to get a third one: “the number of digits in the total palprimes of each length”:

Can you see it?… can you see the rate of growth of the numbers of the last column?… it’s around 10 (ten) !!!…

Enoch Haga wrote [Feb 28, 1999] to Patrick:

“I'm surprised to see that the successive number of digits in the total of palprimes from 1-digit to 17-digits seems to increase by approximately a constant multiplier of 10. E.g. 45 is ~ 10*4 = 40 (45 actual), 465 is ~ 10*45 = 450 (465 actual), and so on [Sloane A039657]…”

Enoch, immediately realized the predictive capability of his discovery and added:

“… Therefore, it is easy to guess at the approximate number of 19-digit and 21-digit palprimes (I will not hazard a guess beyond that!). Simply divide the estimated total number of digits by the number of digits; thus 4597688420/19 = ~ 241.983.600 19-digit palprimes.  The same procedure yields an estimate of ~ 2.189.370.676 21-digit palprimes.”

Questions:

a)  Can you give some explanation about the Enoch's observation?

b) Can you predict if the rate of growth of the third column is going to change, and how?

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Jud McCranie has obtained an explanation of the Enoch's observation. Here is his explanation:

1.- P(D), the quantity of primes of D digits is given by:

P(D) = pi(10^D) - pi(10^(D-1)

But as pi(x) = x/ln(x) then, after some algebraic steps:

P(D) = (10 -D/(D-1)) * (10^(D-1)) / (D*ln(10))

When D>>>>1, then, approximately:

P(D) = 9 * (10^(D-1)) / (D*ln(10))

2.- FPal(D), the fraction of Palindromic numbers of D digits is given by:

FPal(D) = (10^((3-D)/2))/9

3.- Then, Ppal(D) the quantity of Palindromic primes of D digits is given by a combination of the two previous results:

Ppal(D) = P(D) * FPal(D) = (10^(D-1))/(D*ln(10)) * (10^((3-D)/2)) =
= 10^((D+1)/2)/(D*ln(10))

Then approximately: D* Ppal(D) = 10^((D+1)/2)/(ln(10))

As you can see, D* Ppal(D) is the "Total Digits of all the Palprimes of D digits", that is to say, D* Ppal(D) is the third column calculated by Enoch Haga.

But 10^((D+1)/2) increases by a factor of 10 every time D increases two units, so this explains the Haga's observation.

Regarding the quantity of Palprimes of D digits the answer is given - without any simplification - by:

Ppal(D) = P(D) * FPal(D) =
Ppal(D) = (10 -D/(D-1)) * (10^(D-1))/(D*ln(10)) * (10^((3-D)/2))/9 =

then:

D*Ppal(D) = (10 -D/(D-1)) * 10^((D+1)/2) /(9*ln(10))

This answer the two questions posed.

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