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A227712
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a(n) = 9*2^n - 3*n - 5.
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1
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4, 10, 25, 58, 127, 268, 553, 1126, 2275, 4576, 9181, 18394, 36823, 73684, 147409, 294862, 589771, 1179592, 2359237, 4718530, 9437119, 18874300, 37748665, 75497398, 150994867, 301989808, 603979693, 1207959466, 2415919015, 4831838116, 9663676321, 19327352734
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OFFSET
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0,1
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COMMENTS
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Denoting by P[n] the path on n vertices, a(n) is the number of vertices of the tree obtained by identifying the roots of 3 identical rooted trees g[n], where g[n] is obtained recursively in the following manner: g[0]=P[2] and g[n] (n>=1) is obtained by identifying the roots of 2 copies of g[n-1] and one of the extremities of P[n+1]; the root of g[n] is defined to be the other extremity of P[n+1]. Most references contain pictures of these trees; however, the small circles have to be viewed as vertices rather than hexagons.
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LINKS
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R. Kopelman, M. Shortreed, Z. Y. Shi, W. Tan, Z. F. Xu, J. S. Moore, A. Bar-Haim, J. Klafter, Spectroscopic evidence for excitonic localization in fractal antenna supermolecules, Phys. Rev. Letters, 78, 1997, 1239-1242.
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FORMULA
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G.f.: (4-6*x+5*x^2)/((1-2*x)*(1-x)^2).
a(0)=4, a(1)=10, a(2)=25, a(n) = 4*a(n-1)-5*a(n-2)+2*a(n-3). - Harvey P. Dale, Apr 15 2015
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EXAMPLE
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a(1) = 10 because g[1] is the rooted tree in the shape of Y (4 vertices) and a "bouquet" of three Y's has 3*4 - 2 = 10 vertices.
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MAPLE
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a := proc (n) options operator, arrow: 9*2^n-3*n-5 end proc: seq(a(n), n = 0 .. 35);
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MATHEMATICA
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Table[9*2^n-3n-5, {n, 0, 40}] (* or *) LinearRecurrence[{4, -5, 2}, {4, 10, 25}, 40] (* Harvey P. Dale, Apr 15 2015 *)
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PROG
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(PARI) Vec((4-6*x+5*x^2)/((1-2*x)*(1-x)^2) + O(x^100)) \\ Altug Alkan, Oct 17 2015
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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