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

%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