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%I #36 Mar 08 2024 01:16:28
%S 3,0,1,0,9,3,1,7,6,3,5,8,3,9,9,8,9,4
%N Decimal expansion of Sum_{n>=1} 1/(prime(n)*prime(n+1)).
%C Sum of reciprocals of products of successive primes. Differs from A209329 only by the initial term 1/(2*3) = 1/6 = 0.16666...
%F Equals 1/6 + A209329.
%e 0.3010931763... = Sum_{n>=1} 1/(prime(n)*prime(n+1)).
%e = 1/(2*3) + 1/(3*5) + 1/(5*7)
%e + 0.03731790933454338 (primes 10 < p(n+1) < 100)
%e + 0.0017430141479028 (primes 100 < p(n+1) < 10^3)
%e + 0.00011767024549033 (primes 10^3 < p(n+1) < 10^4)
%e + 9.018426684045269 e-6 (primes 10^4 < p(n+1) < 10^5)
%e + 7.3452282601302 e-7 (primes 10^5 < p(n+1) < 10^6)
%e + 6.19161299373 e-8 (primes 10^6 < p(n+1) < 10^7)
%e + 5.3439026467 e-9 (primes 10^7 < p(n+1) < 10^8)
%e + 4.70035656 e-10 (primes 10^8 < p(n+1) < 10^9) + ...
%t digits = 10;
%t f[n_Integer] := 1/(Prime[n]*Prime[n+1]);
%t s = NSum[f[n], {n, 1, Infinity}, Method -> "WynnEpsilon", NSumTerms -> 2*10^6, WorkingPrecision -> MachinePrecision];
%t RealDigits[s, 10, digits][[1]] (* _Jean-François Alcover_, Sep 05 2017 *)
%o (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.
%o (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.
%Y Cf. A085548, A209329, A185380.
%K nonn,cons,more
%O 0,1
%A _M. F. Hasler_, Jan 23 2013
%E Corrected and extended by _Hans Havermann_, Mar 17 2013 using the additional terms of A209329 from _R. J. Mathar_, Feb 08 2013
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