%I #18 Sep 25 2022 10:38:29
%S 3,5,7,7,7,11,14,19,23,23,23,29,35,42,48,47,48,57,64,74,82,81,82,93,
%T 103,115,125,123,125,139,150,165,177,175,177,193,207,224,238,235,238,
%U 257,272,292,308,305,308,329,347,369,387,383,387,411,430,455,475,471,475
%N Number of polyominoes consisting of 4 regular unit n-gons.
%H Colin Barker, <a href="/A103470/b103470.txt">Table of n, a(n) for n = 3..1000</a>
%H S. Kurz, <a href="http://www.mathe2.uni-bayreuth.de/sascha/oeis/k-polyominoes.html">k-polyominoes</a>.
%H <a href="/index/Rec#order_18">Index entries for linear recurrences with constant coefficients</a>, signature (2,-2,2,-2,2,-1,0,0,0,0,0,1,-2,2,-2,2,-2,1).
%F See the link for a formula.
%F G.f.: -x^3*(x^17 -2*x^16 +3*x^15 -4*x^14 +4*x^13 -4*x^12 +4*x^11 -2*x^10 +2*x^9 -2*x^8 +4*x^7 -x^6 +x^5 +3*x^4 -3*x^3 +3*x^2 -x +3) / ((x -1)^3*(x +1)*(x^2 -x +1)^2*(x^2 +1)*(x^2 +x +1)^2*(x^4 -x^2 +1)). - _Colin Barker_, Jan 19 2015
%e a(3)=3 because there are 3 polyiamonds consisting of 4 triangles and a(4)=5 because there are 5 polyominoes consisting of 4 squares.
%t LinearRecurrence[{2,-2,2,-2,2,-1,0,0,0,0,0,1,-2,2,-2,2,-2,1},{3,5,7,7,7,11,14,19,23,23,23,29,35,42,48,47,48,57},80] (* _Harvey P. Dale_, Feb 11 2020 *)
%o (PARI) Vec(-x^3*(x^17 -2*x^16 +3*x^15 -4*x^14 +4*x^13 -4*x^12 +4*x^11 -2*x^10 +2*x^9 -2*x^8 +4*x^7 -x^6 +x^5 +3*x^4 -3*x^3 +3*x^2 -x +3)/((x -1)^3*(x +1)*(x^2 -x +1)^2*(x^2 +1)*(x^2 +x +1)^2*(x^4 -x^2 +1)) + O(x^100)) \\ _Colin Barker_, Jan 19 2015
%Y Cf. A103469, A103471, A103472, A103473.
%K nonn,easy
%O 3,1
%A _Sascha Kurz_, Feb 07 2005