OFFSET
0,2
COMMENTS
From Husnain Raza, Aug 30 2023: (Start)
Note that since p_n > n*log(n), we can place a bound on the tail of the sum:
Sum_{n >= N} (prime(2n))^(-2) <= Sum_{n >= N} (2*n*log(2n))^(-2) <= Integral_{x=N..oo} (2*x*log(2x))^(-2) dx.
Taking the sum over all primes < 10^12, we see that the constant lies between 0.14471558668870 and 0.14471558668873. (End)
LINKS
FORMULA
Equals Sum_{n>=1} 1/A031215(n)^2 = Sum_{n>=1} 1/prime(2n)^2.
EXAMPLE
1/3^2 + 1/7^2 + 1/13^2 + 1/19^2 + 1/29^2 + ... = 0.14471558...
PROG
(PARI) sum(n=1, 2500000, 1./prime(2*n)^2)
(PARI) \\ see Raza link
CROSSREFS
KEYWORD
AUTHOR
Terry D. Grant, Apr 14 2017
EXTENSIONS
a(8)-a(12) from Husnain Raza, Aug 31 2023
STATUS
approved