%I #11 May 02 2019 15:00:38
%S 2,8,24,216,4320,19008,103488,1292544,1548288,3264307200,24710676480,
%T 54623600640,16562257920,3345695539200,8216950210560,
%U 33673108783104000,205682009702400000,15655109317676236800,12302792042521559040000
%N Product of elements in the simple continued fraction for (1+1/n)^n.
%C It appears that lim n ->infinity a(n)^(1/A069887(n)) = K (Khinchin constant = 2.68...) - Benoit Cloitre, Jan 29 2006
%H Harvey P. Dale, <a href="/A071599/b071599.txt">Table of n, a(n) for n = 1..449</a>
%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=4320 hence a(5)=4320
%t Table[Times@@ContinuedFraction[(1+1/n)^n],{n,20}] (* _Harvey P. Dale_, May 02 2019 *)
%o (PARI) for(n=1,100,print1(prod(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