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A137368 a(n) = least m such that sum of m reciprocal primes starting with n-th prime is >1. 1

%I #4 Mar 30 2012 17:34:57

%S 3,9,27,66,144,253,424,651,977,1392,1866,2479,3169,3981,4978,6137,

%T 7420,8829,10477,12279,14295,16613,19124,21906,24904,28056,31494,

%U 35320,39486,44102,49116,54103,59468,65143,71315,77649,84504,91720,99303,107365

%N a(n) = least m such that sum of m reciprocal primes starting with n-th prime is >1.

%F a(n)=m: sum(1/prime(i), i=n,n+m-1))>1, while sum(1/prime(i), i=n,n+m-2))<1.

%e a(1)=3 because 1/2+1/3+1/5=31/30 (3 terms), while 1/2+1/3<1,

%e a(2)=9 because 1/3+1/5+1/7+1/11+1/13+1/17+1/19+1/23+1/29 =3343015913/3234846615 (9 terms), while 1/3+...+1/23<1,

%e a(3)=27 because 1/p(3)+...1/p(29)>1 (27 terms) while 1/p(3)+...1/p(28)<1.

%t ss={};Do[s=1/Prime[n];k=1;While[s<1,k++;s+=1/Prime[n+k-1]];AppendTo[ss,k],{n,1,30}]

%K nonn

%O 1,1

%A _Zak Seidov_, Apr 09 2008

%E a(31)-a(40) from _Donovan Johnson_, Sep 05 2008

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