login
This site is supported by donations to The OEIS Foundation.

 

Logo

Annual Appeal: Please make a donation to keep the OEIS running. In 2018 we replaced the server with a faster one, added 20000 new sequences, and reached 7000 citations (often saying "discovered thanks to the OEIS").
Other ways to donate

Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
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 (list; graph; refs; listen; history; text; internal format)
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*n-7)/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((n-1)^2/4) (since a downward-sloping 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..n-2} floor((n-i)/2) ways, and this sum is equal to floor((n-1)^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((n-1)^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(n-1) - 2*a(n-2) - 2*a(n-3) + 3*a(n-4) - a(n-5) for n>5.

(End)

MAPLE

A:=n-> if n mod 2 = 0 then n^2*(8*n-7)/2 else n^2*(8*n-7)/2-1/2; 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

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified December 18 06:13 EST 2018. Contains 318215 sequences. (Running on oeis4.)