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%I #21 Apr 10 2019 08:43:37
%S 0,10,59,177,394,740,1245,1939,2852,4014,5455,7205,9294,11752,14609,
%T 17895,21640,25874,30627,35929,41810,48300,55429,63227,71724,80950,
%U 90935,101709,113302,125744,139065,153295,168464,184602,201739,219905,239130,259444
%N Triangles in Star of David matchstick arrangement of side n.
%C This is 1/2 the sequence A299965 which counts both the 'standard' and the 'inverted' triangles. - _John King_, Apr 05 2019
%H Colin Barker, <a href="/A045950/b045950.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F a(n) = n*(10*n^2+9*n+1)/2.
%F a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - _Colin Barker_, Dec 02 2014
%F G.f.: x*(x^2+19*x+10) / (x-1)^4. - _Colin Barker_, Dec 02 2014
%F a(n) = A299965(n)/2. - _John King_, Apr 05 2019
%o (PARI) concat(0, Vec(x*(x^2+19*x+10)/(x-1)^4 + O(x^100))) \\ _Colin Barker_, Dec 02 2014
%K nonn,easy
%O 0,2
%A _R. K. Guy_
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