|
%I #21 Jan 05 2019 05:15:42
%S 2,3,7,2,5,1,7,7,6,5,7
%N Decimal expansion of c = sum over twin primes (p, p+2) of (1/p^2 + 1/(p+2)^2).
%C Compare Viggo Brun's constant (1/3 + 1/5) + (1/5 + 1/7) + (1/11 + 1/13) + (1/17 + 1/19) + (1/29 + 1/31) + ... (see A065421, A005597).
%C It appears that c = Sum 1/A001359(n)^2 + 1/A006512(n)^2. - _R. J. Mathar_, May 30 2009
%C 0.237251776574746 < c < 0.237251776947124. - _Farideh Firoozbakht_, May 31 2009
%C c < 0.2725177657771. - _Hagen von Eitzen_, Jun 03 2009
%C From _Farideh Firoozbakht_, Jun 01 2009: (Start)
%C We can show that a(9)=6, a(10)=5, and a(11) is in the set {7, 8, 9}.
%C Proof: s1 = 0.237251776576249072... is the sum up to prime(499,000,000) and s2 = 0.237251776576250009... is the sum up to prime(500,000,000).
%C 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 the first 2*10^9 primes), we conclude that the sum up to prime(2,000,000,000) is less than s2 + 1500*(s2-s1).
%C But since s2-s1 < 10^(-15), the sum up to prime(2*10^9) is less than s2 + 1.5*10^(-12) = 0.237251776576250009... + 1.5*10^(-12) = 0.237251776577550009... .
%C Hence the constant c is less than
%C 0.237251776577550009... + lim(sum(1/k^2,{k, prime(2,000,000,001), n}, n -> infinity)
%C < 0.237251776577550009... + 2.12514*10^(-11)
%C < 0.237251776598801409.
%C 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}.
%C I guess that a(11)=7. (End)
%C From _Jon E. Schoenfield_, Jan 02 2019: (Start)
%C Given that the Hardy-Littlewood approximation to the number of twin prime pairs < y is
%C 2 * C_2 * Integral_{x=2..y} dx/log(x)^2
%C where C_2 = 0.660161815846869573927812110014555778432623 (see A152051), we can estimate the size of the tail of the summation Sum(1/A001359(j)^2) + 1/A006512(j)^2) for twin primes > y as
%C t(y) = 2 * C_2 * Integral_{x>y} 2*dx/(x*log(x))^2.
%C Let s(y) be the sum of the squares of the reciprocals of all the twin primes <= y, and let s'(y) = s(y) + t(y) be the result of adding to the actual value s(y) the estimated tail size t(y). Evaluating s(y), t(y), and s'(y) at y = 2^d for d = 20..33 gives
%C .
%C d s(2^d) t(2^d)*10^10 s(2^d) + t(2^d)
%C == ==================== ============ ====================
%C 20 0.237251764919808326 115.34589710 0.237251776454398036
%C 21 0.237251771317612979 52.59702970 0.237251776577315949
%C 22 0.237251774173347724 24.08221952 0.237251776581569676
%C 23 0.237251775469086555 11.06766714 0.237251776575853269
%C 24 0.237251776066813995 5.10395459 0.237251776577209454
%C 25 0.237251776340760021 2.36119196 0.237251776576879217
%C 26 0.237251776467109357 1.09553336 0.237251776576662693
%C 27 0.237251776525743797 0.50967952 0.237251776576711749
%C 28 0.237251776552887645 0.23771866 0.237251776576659511
%C 29 0.237251776565549906 0.11113468 0.237251776576663374
%C 30 0.237251776571456873 0.05207020 0.237251776576663892
%C 31 0.237251776574218065 0.02444677 0.237251776576662742
%C 32 0.237251776575513036 0.01149984 0.237251776576663020
%C 33 0.237251776576121140 0.00541938 0.237251776576663078
%C .
%C which agrees with all the terms in the Data section and suggests likely values for additional terms.
%C (End)
%H Various authors, <a href="/A160910/a160910.txt">On the computation of A160910</a>
%e (1/9 + 1/25) + (1/25 + 1/49) + (1/121 + 1/169) + (1/289 + 1/361) + (1/841 + 1/961) + ... = 0.237251...
%Y Cf. A001359, A005597, A006512, A065421, A152051.
%K nonn,cons,more
%O 0,1
%A William Royle (seriesandsequences(AT)yahoo.com), May 29 2009
%E _R. J. Mathar_ pointed out that the value of c as originally submitted was incorrect (see link). - _N. J. A. Sloane_, May 31 2009
%E More terms from _Farideh Firoozbakht_ and _Hagen von Eitzen_, Jun 01 2009
%E Name changed by _Michael B. Porter_, Jan 04 2019
| 