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A134799 a(n) = 3^((3^n - 1)/2).

%I A134799 #25 Oct 17 2024 14:46:10
%S A134799 1,3,81,1594323,12157665459056928801,
%T A134799 5391030899743293631239539488528815119194426882613553319203
%N A134799 a(n) = 3^((3^n - 1)/2).
%C A134799 Number of partitions into "bus routes" of the graph G_{n+1} defined below.
%C A134799 These seem to be one-third the reduced denominators of Newton's iteration for 1/sqrt(3), starting with 1/3. - _Steven Finch_, Oct 08 2024
%H A134799 Andrew Howroyd, <a href="/A134799/b134799.txt">Table of n, a(n) for n = 0..7</a>
%H A134799 Author?, <a href="http://image.myany.jp/org/6c511f3698cf6db33d63d6450f331344.jpg">Mitsumata tree</a>
%H A134799 X. Gourdon and P. Sebah, <a href="http://numbers.computation.free.fr/Constants/Sqrt2/sqrt2.html">Pythagoras' Constant</a>.
%F A134799 a(n) is conjectured to be one-third the reduced denominator of b(n) = (3/2)*b(n-1)*(1 - b(n-1)^2); b(0) = 1/3. - _Steven Finch_, Oct 08 2024
%F A134799 Limit_{n -> oo} A376870(n)/(3*a(n)) = 1/sqrt(3) = A020760. - _Steven Finch_, Oct 08 2024
%e A134799 .........|..................G_1
%e A134799 ****
%e A134799 .......__|__................G_2
%e A134799 .........|
%e A134799 ****
%e A134799 .__|_____|_____|__..........G_3
%e A134799 ...|.....|.....|
%e A134799 .........|
%e A134799 .......__|__
%e A134799 .........|
%e A134799 ****.
%e A134799 ..._|_........._|_..........G_4
%e A134799 _|__|_____|_____|__|_
%e A134799 .|._|_....|...._|_.|
%e A134799 ....|.....|.....|
%e A134799 ......_|__|__|_
%e A134799 .......|._|_.|
%e A134799 ..........|
%e A134799 ****
%e A134799 G_1 = o---. = rooted tree with one edge and one leaf node. For n > 0, G_{n+1} is obtained from G_n by splitting each leaf node into three.
%t A134799 3^((3^Range[0, 6] - 1)/2) (* _Paolo Xausa_, Oct 17 2024 *)
%Y A134799 Cf. A131709, A020760, A376870.
%K A134799 nonn,easy
%O A134799 0,2
%A A134799 _Yasutoshi Kohmoto_, Jan 09 2008
%E A134799 Edited by _N. J. A. Sloane_, Jan 29 2008
%E A134799 a(5) from _Andrew Howroyd_, Oct 07 2024