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A130748 Place n points on each of the three sides of a triangle, 3n points in all; a(n) = number of nondegenerate triangles that can be constructed using these points (plus the 3 original vertices) as vertices.

%I A130748 #34 Aug 26 2025 15:53:49
%S A130748 17,72,190,395,711,1162,1772,2565,3565,4796,6282,8047,10115,12510,
%T A130748 15256,18377,21897,25840,30230,35091,40447,46322,52740,59725,67301,
%U A130748 75492,84322,93815,103995,114886,126512,138897,152065,166040,180846,196507,213047,230490
%N A130748 Place n points on each of the three sides of a triangle, 3n points in all; a(n) = number of nondegenerate triangles that can be constructed using these points (plus the 3 original vertices) as vertices.
%C A130748 Define b(1)=1 and b(n) = a(n-1) for n>1. Then (b(n)) is the principal diagonal of the convolution array A213833. - _Clark Kimberling_, Jul 04 2012
%H A130748 Michael De Vlieger, <a href="/A130748/b130748.txt">Table of n, a(n) for n = 1..10000</a>
%H A130748 Jim Propp and Adam Propp-Gubin, <a href="https://arxiv.org/abs/2409.17117">Counting Triangles in Triangles</a>, arXiv:2409.17117 [math.CO], 2024. See p. 9.
%H A130748 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F A130748 a(n) = C(3*(n+1),3) - 3*C(n+2,3) where n>0.
%F A130748 a(n) = (n+1)*A002414(n+1) - n*A002414(n). - _Bruno Berselli_, Dec 11 2012
%F A130748 a(n) = (8*n^3 + 15*n^2 + 9*n + 2)/2. - _Charles R Greathouse IV_, Feb 14 2013
%F A130748 From _Elmo R. Oliveira_, Aug 25 2025: (Start)
%F A130748 G.f.: x*(17 + 4*x + 4*x^2 - x^3)/(x-1)^4.
%F A130748 E.g.f.: -1 + exp(x)*(2 + 32*x + 39*x^2 + 8*x^3)/2.
%F A130748 a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 4. (End)
%e A130748 5 points are put on each side of a triangle (n = 5); we then have 18 vertices to construct with: 5 * 3 + 3 originals. The number of total arrangements = combi(18,3) : combi[3(n+1),3]. But these include degenerates along the 3 sides: 7 points on each side, so combi(7,3) on each side : 3 * combi[n+2, 3] combi[18,3] - 3 * combi[7,3] = 816 - 105 = 711.
%p A130748 A130748:=n->(8*n^3 + 15*n^2 + 9*n + 2)/2; seq(A130748(n), n=1..100); # _Wesley Ivan Hurt_, Jan 28 2014
%t A130748 Table[(8 n^3 + 15 n^2 + 9 n + 2)/2, {n, 100}] (* _Wesley Ivan Hurt_, Jan 28 2014 *)
%o A130748 (PARI) a(n)=4*n^3+n*(15*n+9)/2+1 \\ _Charles R Greathouse IV_, Feb 14 2013
%Y A130748 Cf. A002414, A213833, A220084 (for a list of numbers of the form n*P(k,n)-(n-1)*P(k,n-1), where P(k,n) is the n-th k-gonal pyramidal number).
%Y A130748 Cf. A260260 (comment). [_Bruno Berselli_, Jul 22 2015]
%K A130748 nonn,easy,changed
%O A130748 1,1
%A A130748 _Denis Borris_, Jul 12 2007
%E A130748 More terms from _Wesley Ivan Hurt_, Jan 28 2014