Comments on A160910

From: Richard Mathar <mathar(AT)strw.leidenuniv.nl>
Date: Sat, 30 May 2009 19:10:30 +0200

The current entry of  0.237251058
http://research.att.com/~njas/sequences/A160910
seems to be incorrect in at least three decimal digits.
I guess it is defined as the sum over all 1/A001359(n)^2+1/A006512(n)^2.
Gathering twin primes up to prime(1070000) the constant
is at least 0.23725177606 .
Someone with some spare computer cycles---not the cycles that spin the
computer and the earth around the sun in a year but the other, quicker ones---
might test whether the 0.23725177 may turn to 0.23725178 later on.

Richard Mathar

snapshots of results:
prime(900000)  0.23725177594...
prime(1290000) 0.23725177616...

Digits := 40 ;
x := 0.0 ;
for n from 1 do
        p := ithprime(n) ;
        if isprime(p+2) then
                x := evalf(x+1/p^2+1/(p+2)^2) ;
        fi;
        if n mod 10000 = 0 then
        print(n,x) ;
        fi;
od:

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Date: Sun, 31 May 2009 09:29:23 +0430
From: Farideh Firoozbakht <f.firoozbakht(AT)sci.ui.ac.ir>

No, 0.23725177 may not turn to 0.23725178.
Because,

prime(100,000,000)     0.237251776574074...
prime(110,000,000)     0.237251776574342...
prime(120,000,000)     0.237251776574562...
prime(130,000,000)     0.237251776574746...

So the constant c is less than
    0.237251776574747 + lim(sum(1/k^2,{k, prime(130,000,001), n}, n ->  
infinity)
  < 0.237251776574747 + 3.72376*10^(-10) < 0.237251776947124

Hence 0.237251776574746 < c < 0.237251776947124 and we conclude that the
first nine terms of the sequence are:  2, 3, 7, 2, 5, 1, 7, 7, 6

Rigards,
Farideh



_______________________________________________

Date: Tue, 02 Jun 2009 02:54:25 +0430
From: Farideh Firoozbakht <f.firoozbakht(AT)sci.ui.ac.ir>

  We can show that a(9)=6, a(10)=5 and a(11) is in the set {7, 8, 9}.

  Proof:

  s1 = 0.237251776576249072... is the sum up to prime(499,000,000)
  s2 = 0.237251776576250009... is the sum up to prime(500,000,000).
  By using the fact that number of twin primes between the first 10^6*n primes
  and the first 10^6*(n+1) primes is decreasing (up to first 2*10^9 primes), we
  conclude that the sum up to prime(2000,000,000)is less than s2+1500*(s2-s1).
  But since s2-s1 < 10^(-15) so the sum up to prime(2*10^9) is less than
  s2 + 1.5*10^(-12) = 0.237251776576250009... + 1.5*10^(-12) =
  0.237251776577550009... .
  Hence the constant c is less than  0.237251776577550009... +
  lim(sum(1/k^2,{k, prime(2,000,000,001), n}, n -> infinity)
  < 0.237251776577550009... + 2.12514*10^(-11)
  < 0.237251776598801409.
  So we have 0.237251776576250009 < c < 0.237251776598801409, hence a(9)=6,
  a(10)=5 and a(11) is in the set {7, 8, 9}.

  I guess that a(11)=7.

  ---Farideh

_______________________________________________

Date: Wed, 03 Jun 2009 08:25:58 +0200
From: Hagen von EItzen <math(AT)von-eitzen.de>

Your error term sum_{k > N} 1/k^2 < 1/(N-1) can be slightliy 
improved without using additional primes:
Modulo 2*3*5*...*23, only a proportion of 405/5681 of residue classes 
can contain twin primes.
By summing only over these classes, the error term can be redueced to 
405/5681 * 1/(N-2*3*...*23)
If I'm not wrong, this makes c < 0.2725177657771

Hagen