A296367 Number of triangles on a 4 X n grid.
0, 48, 200, 516, 1056, 1884, 3052, 4628, 6668, 9232, 12380, 16176, 20672, 25936, 32024, 38996, 46912, 55836, 65820, 76932, 89228, 102768, 117612, 133824, 151456, 170576, 191240, 213508, 237440, 263100, 290540, 319828, 351020, 384176, 419356, 456624, 496032, 537648, 581528, 627732, 676320
Offset: 1
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (2,0,-1,-1,0,2,-1).
Programs
-
Maple
A:=proc(n) 6*n^2*(n-1) + 2*(n^3 - n - 2*floor((n-1)^2/4)) + 2*(n^3 - n - 2*floor((n-1)*(n-2)/6)); end; [seq(A(n),n=1..64)];
-
PARI
concat(0, Vec(4*x^2*(12 + 26*x + 29*x^2 + 18*x^3 + 5*x^4) / ((1 - x)^4*(1 + x)*(1 + x + x^2)) + O(x^40))) \\ Colin Barker, Dec 26 2017
Formula
a(n) = 2*n*(n-1)*(5*n+2) - 4*floor((n-1)^2/4) - 4*floor((n-1)*(n-2)/6).
Proof: We will show that a(n) = 6*n^2*(n-1) + 2*(n^3 - n - 2*floor((n-1)^2/4)) + 2*(n^3 - n - 2*floor((n-1)*(n-2)/6)), which is equivalent to the above formula. The argument is similar to that used to prove the formula for A262402.
There are three cases. On a 4Xn grid of points, a triangle can either have (I) 2 points on one line and 1 point on another line, (II) one point on each of lines 1,2,3 or 2,3,4, or (III) one point on each of lines 1,2,4 or 1,3,4.
(I) The triangle can be drawn in 4*3*binomial(n,2)*n = 6*n^2*(n-1) ways.
(II) Suppose the points are on lines 1,2,3. The number of triangles is n^3 - Q, where Q is the number of degenerate triangles formed by collinear triples with one point on each of lines 1,2,3. 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, d >= 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 in the "1,2,3" case the number of triangles is n^3 - n - 2*floor((n-1)^2/4). The same number arises in the "2,3,4" case.
(III) Suppose the points are on lines 1,2,4. The number of triangles is n^3 - Q, where Q is the number of degenerate triangles formed by collinear triples with one point on each of lines 1,2,4. There are n vertical triples. If the three points are on a downward sloping line, through points (1,i), (2,i+d, i+3d), say, with i >= 1, d >= 1, i+3d <= n, there are Sum_{i=1..n-2} floor((n-i)/2) possibilities, and this sum is equal to floor((n-1)*(n-2)/6) (see A001840). So in this case there are n^3 - n - 2*floor((n-1)*(n-2)/6) triangles. The same number arises in the "1,3,4" case. QED.
G.f.: 4*x^2*(5*x^4+18*x^3+29*x^2+26*x+12)/((1-x)^2*(1-x^2)*(1-x^3)).
a(n) = 2*a(n-1) - a(n-3) - a(n-4) + 2*a(n-6) - a(n-7) for n>7. - Colin Barker, Dec 26 2017