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:

 Length of Palprime Quantity of Palprimes 3 15 5 93 7 668 9 5172 11 42042 13 353701 15 3036643 17 27045226

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”:

 Length of Palprime Quantity of Palprimes Number of digits in the total palprimes of each length 3 15 45 5 93 465 7 668 4676 9 5172 46548 11 42042 462462 13 353701 4598113 15 3036643 45549645 17 27045226 459768842

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?

***

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.

*** Records   |  Conjectures  |  Problems  |  Puzzles  Home | Melancholia | Problems & Puzzles | References | News | Personal Page | Puzzlers | Search | Bulletin Board | Chat | Random Link Copyright © 1999-2012 primepuzzles.net. All rights reserved.