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%I #21 Feb 12 2023 18:59:57
%S 0,5,28,82,180,335,560,868,1272,1785,2420,3190,4108,5187,6440,7880,
%T 9520,11373,13452,15770,18340,21175,24288,27692,31400,35425,39780,
%U 44478,49532,54955,60760,66960,73568,80597,88060,95970,104340,113183,122512,132340
%N a(n) = n*(n+1)*(13*n+2)/6.
%C This sequence gives the number of triangles of all sizes in (5*n^2)-polyiamonds in a tetragonal or hexagonal or heptagonal configuration.
%C It is the sum of (1/2)*Sum{j=0..n-1}(n-j)*(5*n+1-j) triangles oriented in one direction and (1/2)*Sum{j-0..n-1}(n-j)*(5*n-1-3*j) oriented in the opposite direction.
%C Shäfli's notation: 3.3.3.3.3 for a(1).
%C The difference between this sequence and A050409(n) equals A000292(n-1).
%C Also, (1/3)*(A002717(2*n) + A255211(n) - 2*A000330(n) gives A033994(n): a (5*n^2)-polyiamond in pentagonal configuration that does not belong to this sequence because a(1)=6.
%C a(n) is odd only when n mod 4 = 1.
%H Harvey P. Dale, <a href="/A257093/b257093.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) = Sum_{j=0..n-1}(n-j)*(5*n-2*j).
%F G.f.: x*(5+8*x)/(1-x)^4. - _Vincenzo Librandi_, Apr 16 2015
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - _Vincenzo Librandi_, Apr 16 2015
%e Second comment a(0) = 0; a(1) = 3 + 2; a(2) = 16 + 12; a(3) = 46 + 36; a(4) = 100 + 80; a(5) = 185 + 150; a(6) = 308 + 252.
%t Table[n (n + 1) (13 n + 2)/6, {n, 0, 40}] (* _Vincenzo Librandi_, Apr 16 2015 *)
%t CoefficientList[Series[x (5+8x)/(1-x)^4,{x,0,50}],x] (* or *) LinearRecurrence[{4,-6,4,-1},{0,5,28,82},60] (* _Harvey P. Dale_, Feb 12 2023 *)
%o (Magma) [n*(n+1)*(13*n+2)/6: n in [0..40]]; // _Vincenzo Librandi_, Apr 16 2015
%Y Cf. A002411, A011379, A033429, A050409, A000292, A002717, A000330, A033994, A255211.
%K nonn,easy
%O 0,2
%A _Luce ETIENNE_, Apr 16 2015
%E Corrected by _Harvey P. Dale_, Feb 12 2023
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