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A066258
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a(n) = Fibonacci(n)^2 * Fibonacci(n+1).
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12
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0, 1, 2, 12, 45, 200, 832, 3549, 14994, 63580, 269225, 1140624, 4831488, 20466953, 86698690, 367262700, 1555747893, 6590256856, 27916771136, 118257348165, 500946152850, 2122041977276, 8989114033297, 38078498156832
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OFFSET
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0,3
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COMMENTS
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a(n) is the number of edge covers of a caterpillar graph with spine P_(3n-2), one pendant attached at vertex n counting from the left end of the spine and another pendant at vertex 2n-1. The caterpillar graph for n=3 is as follows:
* *
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*--*--*--*--*--*--*
v
Each pendant edge must be included in an edge cover. Every vertex except v is then incident with at least one edge. Therefore, of the remaining four edges in the spine, only two in the middle have restrictions. At least one of those two has to be in the edge cover to ensure v is incident with one edge in the edge cover, leaving us with 3*2*2 total edge covers. (End)
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LINKS
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FORMULA
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O.g.f.: x*(1-x) / ( (1-4*x-x^2)*(1+x-x^2) ).
a(n) = second term from right in M^(n+1) * [1 0 0 0], where M = the 4 X 4 upper Pascal's triangular matrix [1 3 3 1 / 1 2 1 0 / 1 1 0 0 / 1 0 0 0]. E.g., a(3) = 45 since M^4 * [1 0 0 0] = [125 75 45 27] where 125 = A056570(5), 75 = A066259(4) and 27 = A056570(4). - Gary W. Adamson, Oct 31 2004
a(n) = (1/5)*(Fibonacci(3n+1) - (-1)^n*Fibonacci(n+2)). - Ralf Stephan, Jul 26 2005
a(n) = 3*a(n-1) + 6*a(n-2) - 3*a(n-3) - a(n-4). - Zak Seidov, May 07 2015
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MATHEMATICA
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#[[1]]^2 #[[2]]&/@Partition[Fibonacci[Range[0, 30]], 2, 1] (* or *) LinearRecurrence[ {3, 6, -3, -1}, {0, 1, 2, 12}, 30] (* Harvey P. Dale, Jul 28 2018 *)
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PROG
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(PARI) { for (n=1, 200, a=fibonacci(n)^2 * fibonacci(n+1); write("b066258.txt", n, " ", a) ) } \\ Harry J. Smith, Feb 07 2010
(Magma) [Fibonacci(n)^2*Fibonacci(n+1): n in [0..30]]; // G. C. Greubel, Feb 12 2024
(SageMath) [fibonacci(n)^2*fibonacci(n+1) for n in range(31)] # G. C. Greubel, Feb 12 2024
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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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