

A296367


Number of triangles on a 4 X n grid.


1



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
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OFFSET

1,2


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).


FORMULA

a(n) = 2*n*(n1)*(5*n+2)  4*floor((n1)^2/4)  4*floor((n1)*(n2)/6).
Proof: We will show that a(n) = 6*n^2*(n1) + 2*(n^3  n  2*floor((n1)^2/4)) + 2*(n^3  n  2*floor((n1)*(n2)/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*(n1) 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((n1)^2/4) (since a downwardsloping 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..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 in the "1,2,3" case the number of triangles is n^3  n  2*floor((n1)^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..n2} floor((ni)/2) possibilities, and this sum is equal to floor((n1)*(n2)/6) (see A001840). So in this case there are n^3  n  2*floor((n1)*(n2)/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)/((1x)^2*(1x^2)*(1x^3)).
a(n) = 2*a(n1)  a(n3)  a(n4) + 2*a(n6)  a(n7) for n>7.  Colin Barker, Dec 26 2017


MAPLE

A:=proc(n)
6*n^2*(n1)
+ 2*(n^3  n  2*floor((n1)^2/4))
+ 2*(n^3  n  2*floor((n1)*(n2)/6));
end;
[seq(A(n), n=1..64)];


PROG

(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


CROSSREFS

Cf. A001840, A115514, A262402.
Sequence in context: A231174 A259245 A157923 * A275507 A072254 A290350
Adjacent sequences: A296364 A296365 A296366 * A296368 A296369 A296370


KEYWORD

nonn,easy


AUTHOR

Jared Nash and N. J. A. Sloane, Dec 19 2017, corrected Dec 23 2017


STATUS

approved



