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 11digit 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^((n1)/2) ndigit
palindromes. The reciprocal of each of these is
<= 1/10^(n1). 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) ndigit
palindromes. The reciprocal of each of these is
<= 1/10^(n1). 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 palprimes
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 palprimes (including 1digit 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.
***
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
11digits palprimes.
***
