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Sum of the terms of the continued fraction for the sum of the reciprocals of the first n primes.
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%I #6 Apr 09 2014 10:16:38

%S 2,6,31,21,26,38,51,51,330,139,91,119,590,146,198,266,2138,230,247,

%T 331,387,1267,540,320,1302,1021,1291,621,462,954,823,608,447,708,1017,

%U 1505,1107,836,1769,922,1803,791,990,91803,1082,1691,1939,2221,1385,1065

%N Sum of the terms of the continued fraction for the sum of the reciprocals of the first n primes.

%e The sum of the reciprocals of the first 2 primes = 1/2 + 1/3 = 5/6 = 0 + 1/(1 + 1/5). So a(2) = 0 + 1 + 5 = 6.

%K nonn

%O 1,1

%A _Leroy Quet_, Nov 15 2000