Problems & Puzzles: Puzzles

Puzzle 56.- The Honaker's Constant

At the beginning of this issue, G. L. Honaker, Jr. stated:

"The series 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/101 + .... is convergent."

Soon Mike Keith wrote:

"I computed the value of the sum (dare we call it "Honaker's constant"?) up through all 11-digit palindromic primes. The value I got was:

1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/101 + ... = 1.3239820264 ..."

Then Honaker asked:

"Find a palindromic prime fraction that best approximates this constant."

And made an initial approximation:

"1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/101 + ... ~ 96269/72727 = 1.3237037 [Honaker]"

Questions:

a) Can you demonstrate that the Honaker's series is convergent?
b) Can you improve the Honaker's approximation?

Solution

Jud McCranie (1/6/99) has sent the following argument for the part a):

For odd n, there are 9x10^((n-1)/2) n-digit palindromes.  The reciprocal of each of these is <= 1/10^(n-1). So the sum of the reciprocals of these is <= 9/1 + 90/100 + 900/10000 + ... = 10.

For even n, there are 0.9x10(n/2) n-digit palindromes.  The reciprocal of each of these is <= 1/10^(n-1). So the sum of the reciprocals of these is <= 9/10 + 90/1000 + 900/100000 + ... = 1.  So the sum of the reciprocals of all palindromes is <= 11.  So the sum of reciprocals of all pal-primes converges. 

Actually the sum of the reciprocal of all palindromes (starting with 1) is about 3.37.  The sum for all palindromes > 10 is about 0.541, so a tighter upper bound for pal-primes (including 1-digit ones) is 1/2 + 1/3 + 1/5 + 1/7 + 0.541 = 1.7172, which is a tighter upper bound for the sum of Honaker's series.
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Tiziano Mosconi has improved (17/8/01) the Honaker's approximation to his constant:

9477749 / 7158517  = 1.3239821879...             difference = 1.6153*10^-7    

199171991 / 150434051  = 1.3239821016...          difference = 7.523*10^-8

With "difference" I mean the difference between the fraction and the value of the Honaker's constant computed up through all 11-digits palprimes.

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