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Sum of elements in the simple continued fraction for (1+1/n)^n.
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%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