OFFSET
0,1
COMMENTS
Sum of reciprocals of products of successive primes. Differs from A209329 only by the initial term 1/(2*3) = 1/6 = 0.16666...
FORMULA
Equals 1/6 + A209329.
EXAMPLE
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) + ...
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 *)
PROG
(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.
CROSSREFS
KEYWORD
AUTHOR
M. F. Hasler, Jan 23 2013
EXTENSIONS
Corrected and extended by Hans Havermann, Mar 17 2013 using the additional terms of A209329 from R. J. Mathar, Feb 08 2013
STATUS
approved