A210473 Decimal expansion of Sum_{n>=1} 1/(prime(n)*prime(n+1)).
3, 0, 1, 0, 9, 3, 1, 7, 6, 3, 5, 8, 3, 9, 9, 8, 9, 4
Offset: 0
Examples
0.3010931763... = Sum_{n>=1} 1/(prime(n)*prime(n+1)).
= 1/(2*3) + 1/(3*5) + 1/(5*7)
+ 0.03731790933454338 (primes 10 < p(n+1) < 100)
+ 0.0017430141479028 (primes 100 < p(n+1) < 10^3)
+ 0.00011767024549033 (primes 10^3 < p(n+1) < 10^4)
+ 9.018426684045269 e-6 (primes 10^4 < p(n+1) < 10^5)
+ 7.3452282601302 e-7 (primes 10^5 < p(n+1) < 10^6)
+ 6.19161299373 e-8 (primes 10^6 < p(n+1) < 10^7)
+ 5.3439026467 e-9 (primes 10^7 < p(n+1) < 10^8)
+ 4.70035656 e-10 (primes 10^8 < p(n+1) < 10^9) + ...
Programs
-
Mathematica
digits = 10; f[n_Integer] := 1/(Prime[n]*Prime[n+1]); s = NSum[f[n], {n, 1, Infinity}, Method -> "WynnEpsilon", NSumTerms -> 2*10^6, WorkingPrecision -> MachinePrecision]; RealDigits[s, 10, digits][[1]] (* Jean-François Alcover, Sep 05 2017 *) -
PARI
S(L=10^9,start=3)={my(s=0,q=1/precprime(start));forprime(p=1/q+1,L,s+=q*q=1./p);s} \\ Using 1./p is maybe a little less precise, but using s=0. and 1/p takes about 50% more time. -
PARI
{my( tee(x)=printf("%g,",x);x ); t=vector(8,n,tee(S(10^(n+1),10^n))); s=1/2/3+1/3/5+1/5/7; vector(#t,n,s+=t[n])} \\ Shows contribution of sums over (n+1)-digit primes (vector t) and the vector of partial sums; the final value is in s.
Formula
Equals 1/6 + A209329.
Extensions
Corrected and extended by Hans Havermann, Mar 17 2013 using the additional terms of A209329 from R. J. Mathar, Feb 08 2013
Comments