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A245961
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Number of 4-cycles in the Lucas cube Lambda(n).
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3
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0, 0, 0, 0, 2, 5, 15, 35, 80, 171, 355, 715, 1410, 2730, 5208, 9810, 18280, 33745, 61785, 112309, 202840, 364245, 650705, 1157015, 2048532, 3612900, 6349200, 11121300, 19421150, 33820061, 58740915, 101777495, 175945280, 303516015, 522541903, 897942115
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OFFSET
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0,5
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COMMENTS
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The vertex set of the Lucas cube Lambda(n) is the set of all binary strings of length n without consecutive 1's and without a 1 in the first and the last bit. Two vertices of the Lucas cube are adjacent if their strings differ in exactly one bit.
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LINKS
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FORMULA
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a(n) = ((n-n^2)*F(n) + (3n^2 - 5n)*F(n-1))/10, where F(n) = A000045(n), the Fibonacci numbers. Formula follows from Eq. (4) of the Klavzar 2005 reference and from the first formula on p. 511 of the Klavzar 2013 reference.
a(n) = Sum(L(i)*b(n-3-i), i=0..n-4), where L(i) = A000032(i) are the Lucas numbers and b(j) = A001629(j+1) is the number of edges in the Fibonacci cube Gamma(j) (see Prop. 9 of the Klavzar 2005 reference).
a(n) = 3*a(n-1)-5*a(n-3)+3*a(n-5)+a(n-6). G.f.: x^4*(x-2) / (x^2+x-1)^3. - Colin Barker, Aug 11 2014
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EXAMPLE
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a(3)=0 because the Lucas cube Lambda(3) is the star-tree on 4 vertices.
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MAPLE
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with(combinat): a := proc (n) options operator, arrow: (1/10)*n*(1-n)*fibonacci(n)+(1/10)*n*(3*n-5)*fibonacci(n-1) end proc: seq(a(n), n = 0 .. 35);
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MATHEMATICA
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Table[((n - n^2) Fibonacci[n] + (3 n^2 - 5 n) Fibonacci[n - 1])/10, {n, 0, 50}] (* Vincenzo Librandi, Aug 11 2014 *)
LinearRecurrence[{3, 0, -5, 0, 3, 1}, {0, 0, 0, 2, 5, 15}, 20] (* Eric W. Weisstein, Jul 29 2023 *)
CoefficientList[Series[(-2 + x) x^3/(-1 + x + x^2)^3, {x, 0, 20}], x] (* Eric W. Weisstein, Jul 29 2023 *)
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PROG
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(Magma) [((n-n^2)*Fibonacci(n) + (3*n^2 - 5*n)*Fibonacci(n-1))/10: n in [0..50]]; // Vincenzo Librandi, Aug 11 2014
(PARI) concat([0, 0, 0, 0], Vec(x^4*(x-2)/(x^2+x-1)^3 + O(x^100))) \\ Colin Barker, Aug 13 2014
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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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