

A262402


a(n) = number of triangles that can be formed from the points of a 3 X n grid.


3



0, 18, 76, 200, 412, 738, 1200, 1824, 2632, 3650, 4900, 6408, 8196, 10290, 12712, 15488, 18640, 22194, 26172, 30600, 35500, 40898, 46816, 53280, 60312, 67938, 76180, 85064, 94612, 104850, 115800, 127488, 139936, 153170, 167212, 182088, 197820, 214434, 231952, 250400, 269800, 290178, 311556
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OFFSET

1,2


LINKS

Andrew Howroyd, Table of n, a(n) for n = 1..1000
Jared Nash, Formula for number of triangles in a 3 X n grid
Ran Pan, Exercise T, Project P, 2015.
Index entries for linear recurrences with constant coefficients, signature (3,2,2,3,1).


FORMULA

a(n) = n^2*(8*n7)/2  (n mod 2)/2.  Jared Nash, Dec 14 2017
Proof of Jared Nash's formula, from N. J. A. Sloane, Dec 16 2017 (Start).
On a 3Xn grid of points, a triangle can either have 2 points on one line and 1 point on another line (for a total of 6*n*binomial(n,2) ways) or one point on each line (in n^3  Q ways, where Q is the number of degenerate triangles formed by collinear triples with one point on each line).
Q is equal to n (for vertical triples) plus 2*floor((n1)^2/4) (since a downwardsloping diagonal passing through the points (1,i), (2,i+d), (3,i+2d), say, where i >= 1, i+2d <= n, can be drawn in Sum_{i=1..n2} floor((ni)/2) ways, and this sum is equal to floor((n1)^2/4), as can be seen by considering the row sums of the triangle A115514).
So a(n) = 6*n*binomial(n,2) + n^3  (n + 2*floor((n1)^2/4)), which simplifies to give the above formula. (End)
For another proof, see the link.
G.f.: 2*x^2*(4*x^2 + 11*x + 9) / ((1  x)^3*(1  x^2)).  N. J. A. Sloane, Dec 16 2017
From Colin Barker, Dec 20 2017: (Start)
a(n) = n^2*(8*n  7) / 2 for n even.
a(n) = (8*n^3  7*n^2  1) / 2 for n odd.
a(n) = 3*a(n1)  2*a(n2)  2*a(n3) + 3*a(n4)  a(n5) for n>5.
(End)


MAPLE

A:=n> if n mod 2 = 0 then n^2*(8*n7)/2 else n^2*(8*n7)/21/2; fi; [seq(A(n), n=1..30)]; # N. J. A. Sloane, Dec 16 2017


PROG

(PARI) concat(0, Vec(2*x^2*(9 + 11*x + 4*x^2) / ((1  x)^4*(1 + x)) + O(x^40))) \\ Colin Barker, Dec 20 2017


CROSSREFS

Cf. A115514.
The old, incorrect, formula proposed for this problem is now in A296363.
Sequence in context: A143666 A139757 A285918 * A296363 A164603 A229714
Adjacent sequences: A262399 A262400 A262401 * A262403 A262404 A262405


KEYWORD

nonn,easy


AUTHOR

Ran Pan, Sep 21 2015


EXTENSIONS

Terms corrected and entry revised by N. J. A. Sloane, Dec 16 2017 following an email from Jared Nash, Dec 14 2017.


STATUS

approved



