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%I #14 Mar 21 2018 17:16:45
%S 3,12,40,120,336,896,2304,5760,14080,33792,79872,186368,430080,983040,
%T 2228224,5013504,11206656,24903680,55050240,121110528,265289728,
%U 578813952,1258291200,2726297600,5888802816,12683575296,27246198784,58384711680,124822487040
%N a(n) = 2^(n-2)*binomial(n+1,2).
%C Also the number of 4-cycles in the (n+1)-folded cube graph for n > 3. - _Eric W. Weisstein_, Mar 21 2018
%H Colin Barker, <a href="/A052482/b052482.txt">Table of n, a(n) for n = 2..1000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FoldedCubeGraph.html">Folded Cube Graph</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/GraphCycle.html">Graph Cycle</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,-12,8).
%F a(n) = (1/2) * Sum_{k=0..n-1} Sum_{i=0..n-1} (k+1) * C(n-1,i). - _Wesley Ivan Hurt_, Sep 20 2017
%F From _Colin Barker_, Sep 22 2017: (Start)
%F G.f.: x^2*(3 - 6*x + 4*x^2) / (1 - 2*x)^3.
%F a(n) = 2^(n-3)*n*(1 + n).
%F a(n) = 6*a(n-1) - 12*a(n-2) + 8*a(n-3) for n>4.
%F (End)
%t Table[2^(n - 2) Binomial[n + 1, 2], {n, 2, 28}] (* _Michael De Vlieger_, Sep 21 2017 *)
%t LinearRecurrence[{6, -12, 8}, {3, 12, 40}, 20] (* _Eric W. Weisstein_, Mar 21 2018 *)
%t CoefficientList[Series[(-3 + 6 x - 4 x^2)/(-1 + 2 x)^3, {x, 0, 20}], x] (* _Eric W. Weisstein_, Mar 21 2018 *)
%o (PARI) Vec(x^2*(3 - 6*x + 4*x^2) / (1 - 2*x)^3 + O(x^40)) \\ _Colin Barker_, Sep 22 2017
%Y Essentially the same as A080929.
%Y Cf. A301459 (6-cycles in the n-folded cube graph).
%K nonn,easy
%O 2,1
%A _N. J. A. Sloane_, Mar 16 2000
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