A112323 - OEIS

%I #12 Oct 11 2019 16:48:11

%S 3,5,2,10,17,19,13,31,38,11,41,49,31,73,82,93,89,19,115,107,5,136,148,

%T 107,31,38,178,161,139,67,220,236,103,256,56,181,47,283,302,313,269,

%U 52,287,61,74,65,69,388,341,289,423,359,137,456,467,478,101,107,349,211

%N a(n) = numerator of sum of reciprocals of the terms of the continued fraction for prime(n+1)/prime(n).

%H Harvey P. Dale, <a href="/A112323/b112323.txt">Table of n, a(n) for n = 1..1000</a>

%e prime(6)/prime(5) = 13/11 = 1 + 1/(5 + 1/2).

%e So a(5) is 17, the numerator of 17/10 = 1 + 1/5 + 1/2.

%t f[n_] := Plus @@ (1/# &) /@ ContinuedFraction[Prime[n + 1]/Prime[n]]; Table[Numerator[f[n]], {n, 60}] (* _Ray Chandler_, Sep 07 2005 *)

%Y Cf. A071866, A110021, A109374, A112324, A112768.

%K nonn,frac

%O 1,1

%A _Leroy Quet_, Sep 03 2005

%E Extended by _Ray Chandler_, Sep 07 2005