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A016283 a(n) = 6^n/8 - 4^(n-1) + 2^(n-3).

%I A016283 #31 Jul 02 2023 13:50:01
%S A016283 0,0,1,12,100,720,4816,30912,193600,1194240,7296256,44301312,
%T A016283 267904000,1615810560,9728413696,58504691712,351565004800,
%U A016283 2111537479680,12677814747136,76101248090112,456744927232000
%N A016283 a(n) = 6^n/8 - 4^(n-1) + 2^(n-3).
%C A016283 Number of rectangles that can be formed from the vertices of an n-dimensional cube. E.g., a(3)=12 because the three-dimensional cube has six faces plus six rectangles passing through the center of the cube. Cf. A064436: each rectangle on the cube provides an opportunity for a function not to be a linear threshold function, by alternating in value around the rectangle. - _Matthew Cook_, Jan 26 2004
%H A016283 Vincenzo Librandi, <a href="/A016283/b016283.txt">Table of n, a(n) for n = 0..150</a>
%H A016283 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (12, -44, 48).
%F A016283 a(n) = (2^n)*Stirling2(n+3, 3), n >= 0, with Stirling2(n, m) = A008277(n, m).
%F A016283 G.f.: x^2/((1-2*x)*(1-4*x)*(1-6*x)).
%F A016283 E.g.f.: (exp(2*x) - 8*exp(4*x) + 9*exp(6*x))/2!.
%F A016283 a(n) =((6^n - 2^n)/4 - (4^n - 2^n)/2)/2 , n >= 0. - _Zerinvary Lajos_, Jun 05 2009
%p A016283 [seq(9/2*6^n-4*4^n+1/2*2^n,n=0..20)]; # Detlef Pauly (dettodet(AT)yahoo.de), Dec 04 2001
%t A016283 CoefficientList[Series[x^2/((1 - 2 x) (1 - 4 x) (1 - 6 x)), {x, 0, 20}], x] (* _Michael De Vlieger_, Jan 31 2018 *)
%o A016283 (Sage) [((6^n - 2^n)/4-(4^n - 2^n)/2)/2 for n in range(0,21)] # _Zerinvary Lajos_, Jun 05 2009
%o A016283 (Magma) [6^n/8 - 4^(n-1) + 2^(n-3): n in [0..25]]; // _Vincenzo Librandi_, Apr 26 2011
%Y A016283 Third column of triangle A075497.
%Y A016283 Cf. A025966.
%K A016283 nonn
%O A016283 0,4
%A A016283 _N. J. A. Sloane_