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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.
6
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
1,1
COMMENTS
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
LINKS
Jim Propp and Adam Propp-Gubin, Counting Triangles in Triangles, arXiv:2409.17117 [math.CO], 2024. See p. 9.
FORMULA
a(n) = C(3*(n+1), 3) - 3*C(n+2, 3) where n>0.
a(n) = (n+1)*A002414(n+1) - n*A002414(n). - Bruno Berselli, Dec 11 2012
a(n) = (8n^3 + 15n^2 + 9n + 2)/2 \\ Charles R Greathouse IV, Feb 14 2013
EXAMPLE
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.
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
MATHEMATICA
Table[(8 n^3 + 15 n^2 + 9 n + 2)/2, {n, 100}] (* Wesley Ivan Hurt, Jan 28 2014 *)
PROG
(PARI) a(n)=4*n^3+n*(15*n+9)/2+1 \\ Charles R Greathouse IV, Feb 14 2013
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]
Sequence in context: A214530 A087514 A119625 * A131692 A157864 A229496
KEYWORD
nonn,easy
AUTHOR
Denis Borris, Jul 12 2007
EXTENSIONS
More terms from Wesley Ivan Hurt, Jan 28 2014
STATUS
approved