%I #6 Mar 30 2012 18:39:02
%S 2,6,10,16,39,47,48,77,69,228,153,237,142,121,295,300,203,423,281,341,
%T 274,700,703,410,729,642,838,1051,577,619,443,451,605,537,531,1123,
%U 997,2537,805,1192,1312,1255,1037,1074,938,888,859,907,1233,1212,3627,1120
%N Sum of elements in the simple continued fraction for (1+1/n)^n.
%e The continued fraction for (1+1/5)^5 is [2, 2, 20, 1, 9, 2, 3] and 2+2+20+1+9+2+3=39 hence a(5)=39
%o (PARI) for(n=1,100,print1(sum(i=1,length(contfrac((1+1/n)^n)), component(contfrac((1+1/n)^n),i)),","))
%K easy,nonn
%O 1,1
%A _Benoit Cloitre_, Jun 01 2002