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.
17, 72, 190, 395, 711, 1162, 1772, 2565, 3565, 4796, 6282, 8047, 10115, 12510, 15256, 18377, 21897, 25840, 30230, 35091, 40447, 46322, 52740, 59725, 67301, 75492, 84322, 93815, 103995, 114886, 126512, 138897, 152065, 166040, 180846, 196507, 213047, 230490
Offset: 1
Examples
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.
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
- Jim Propp and Adam Propp-Gubin, Counting Triangles in Triangles, arXiv:2409.17117 [math.CO], 2024. See p. 9.
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Crossrefs
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).
Cf. A260260 (comment). [Bruno Berselli, Jul 22 2015]
Programs
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Maple
A130748:=n->(8*n^3 + 15*n^2 + 9*n + 2)/2; seq(A130748(n), n=1..100); # Wesley Ivan Hurt, Jan 28 2014
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Mathematica
Table[(8 n^3 + 15 n^2 + 9 n + 2)/2, {n, 100}] (* Wesley Ivan Hurt, Jan 28 2014 *) -
PARI
a(n)=4*n^3+n*(15*n+9)/2+1 \\ Charles R Greathouse IV, Feb 14 2013
Formula
a(n) = C(3*(n+1),3) - 3*C(n+2,3) where n>0.
a(n) = (8*n^3 + 15*n^2 + 9*n + 2)/2. - Charles R Greathouse IV, Feb 14 2013
From Elmo R. Oliveira, Aug 25 2025: (Start)
G.f.: x*(17 + 4*x + 4*x^2 - x^3)/(x-1)^4.
E.g.f.: -1 + exp(x)*(2 + 32*x + 39*x^2 + 8*x^3)/2.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 4. (End)
Extensions
More terms from Wesley Ivan Hurt, Jan 28 2014
Comments