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A334694 a(n) = (n/4)*(n^3+2*n^2+5*n+8).

%I A334694 #22 Nov 08 2023 11:17:50
%S A334694 0,4,17,51,124,260,489,847,1376,2124,3145,4499,6252,8476,11249,14655,
%T A334694 18784,23732,29601,36499,44540,53844,64537,76751,90624,106300,123929,
%U A334694 143667,165676,190124,217185,247039,279872,315876,355249,398195,444924,495652,550601,609999,674080,743084,817257,896851,982124,1073340
%N A334694 a(n) = (n/4)*(n^3+2*n^2+5*n+8).
%C A334694 Consider a figure made up of a row of n >= 1 adjacent congruent rectangles in which all possible diagonals of the rectangles have been drawn. The number of regions formed is A306302. If we distort all these diagonals very slightly so that no three lines meet at a point, the number of regions changes to a(n).
%H A334694 Colin Barker, <a href="/A334694/b334694.txt">Table of n, a(n) for n = 0..1000</a>
%H A334694 <a href="/index/St#Stained">Index entries for sequences related to stained glass windows</a>
%H A334694 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A334694 Satisfies the identity a(n) = A306302(n) + Sum_{k=3..(n+1)} binomial(k-1,2)*A333275(n,2*k). E.g. for n=4 we have a(4) = 104 + 8*1 + 2*3 + 1*6 = 124.
%F A334694 From _Colin Barker_, May 27 2020: (Start)
%F A334694 G.f.: x*(4 - 3*x + 6*x^2 - x^3) / (1 - x)^5.
%F A334694 a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>4.
%F A334694 (End)
%t A334694 LinearRecurrence[{5,-10,10,-5,1},{0,4,17,51,124},50] (* or *)
%t A334694 A334694[n_]:=n/4(n^3+2n^2+5n+8);Array[A334694,50,0] (* _Paolo Xausa_, Nov 08 2023 *)
%o A334694 (PARI) concat(0, Vec(x*(4 - 3*x + 6*x^2 - x^3) / (1 - x)^5 + O(x^40))) \\ _Colin Barker_, May 27 2020
%Y A334694 Cf. A306302, A331452, A333275.
%K A334694 nonn,easy
%O A334694 0,2
%A A334694 _Scott R. Shannon_ and _N. J. A. Sloane_, May 19 2020