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

0,1

Comments

Sum of reciprocals of products of successive primes. Differs from A209329 only by the initial term 1/(2*3) = 1/6 = 0.16666...

Links

Table of n, a(n) for n=0..17.

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

Cf. A085548, A209329, A185380.

Sequence in context: A334076 A132884 A319234 * A185951 A188832 A279514

Adjacent sequences: A210470 A210471 A210472 * A210474 A210475 A210476

Keyword

nonn,cons,more

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