A160910 Decimal expansion of c = sum over twin primes (p, p+2) of (1/p^2 + 1/(p+2)^2).

2, 3, 7, 2, 5, 1, 7, 7, 6, 5, 7

Offset: 0

Author

William Royle (seriesandsequences(AT)yahoo.com), May 29 2009

Keywords

• nonn
• cons
• more

Comments

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).

It appears that c = Sum 1/A001359(n)^2 + 1/A006512(n)^2. - R. J. Mathar, May 30 2009

0.237251776574746 < c < 0.237251776947124. - Farideh Firoozbakht, May 31 2009

c < 0.2725177657771. - Hagen von Eitzen, Jun 03 2009

From Farideh Firoozbakht, Jun 01 2009: (Start)

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) and 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 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).

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... .

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. (End)

From Jon E. Schoenfield, Jan 02 2019: (Start)

Given that the Hardy-Littlewood approximation to the number of twin prime pairs < y is

2 * C_2 * Integral_{x=2..y} dx/log(x)^2

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

t(y) = 2 * C_2 * Integral_{x>y} 2*dx/(x*log(x))^2.

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

.

d s(2^d) t(2^d)*10^10 s(2^d) + t(2^d)

== ==================== ============ ====================

20 0.237251764919808326 115.34589710 0.237251776454398036

21 0.237251771317612979 52.59702970 0.237251776577315949

22 0.237251774173347724 24.08221952 0.237251776581569676

23 0.237251775469086555 11.06766714 0.237251776575853269

24 0.237251776066813995 5.10395459 0.237251776577209454

25 0.237251776340760021 2.36119196 0.237251776576879217

26 0.237251776467109357 1.09553336 0.237251776576662693

27 0.237251776525743797 0.50967952 0.237251776576711749

28 0.237251776552887645 0.23771866 0.237251776576659511

29 0.237251776565549906 0.11113468 0.237251776576663374

30 0.237251776571456873 0.05207020 0.237251776576663892

31 0.237251776574218065 0.02444677 0.237251776576662742

32 0.237251776575513036 0.01149984 0.237251776576663020

33 0.237251776576121140 0.00541938 0.237251776576663078

.

which agrees with all the terms in the Data section and suggests likely values for additional terms.

(End)

Examples

			(1/9 + 1/25) + (1/25 + 1/49) + (1/121 + 1/169) + (1/289 + 1/361) + (1/841 + 1/961) + ... = 0.237251...
		

Links

• Various authors, On the computation of A160910

Crossrefs

Cf. A001359, A005597, A006512, A065421, A152051.

Extensions

R. J. Mathar pointed out that the value of c as originally submitted was incorrect (see link). - N. J. A. Sloane, May 31 2009

More terms from Farideh Firoozbakht and Hagen von Eitzen, Jun 01 2009

Name changed by Michael B. Porter, Jan 04 2019