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%I #21 May 28 2021 18:38:11
%S 0,0,48,532,3088,11340,33824,83288,183344,364304,681872,1194100,
%T 1992976,3182332,4941360,7420640,10874720,15539952,21812720,30011924,
%U 40650368,54187196,71463440,92990296,119675712,152314920,192393872,240690060
%N Number of sets of four points on an n X n grid (or geoboard), exactly three of which are collinear.
%C The four points build a triangle on an n X n grid, with one of them located on a side of the triangle.
%C The number of sets of four points with the three collinear points in a horizontal or vertical line is 2*n^2*(n-1)*binomial(n,3) = 4*A090448(n). The number of sets of four points with the three collinear points in a diagonal line of slope 1 is 2*n*(n-1)*binomial(n,3) + 4*Sum_{k=3..n-1}(n^2-k)*binomial(k,3). The sum of these two values is a lower bound for this sequence. - _Nathaniel Johnston_, Apr 23 2011
%H Martin Renner, <a href="/A189346/b189346.txt">Table of n, a(n) for n = 1..76</a>
%p A189346 := proc(n)local a,b,j,k,l,m,s,slopes,num,den,tot: tot := 0: slopes := {}: for b from 1 to ceil(n/2)-1 do for a from 0 to b do slopes := slopes union {a/b}: od: od: for s from 1 to nops(slopes) do num := numer(slopes[s]): den := denom(slopes[s]): if(num = 0)then tot := tot + 2*n^2*(n-1)*binomial(n,3): elif(num = den)then tot := tot + 2*(2*add(binomial(k,3)*(n^2-k), k=3..n) - binomial(n,3)*(n^2 - n)): else for j from 1 to n - 2*den do for k from 1 to n - 2*num do tot := tot + 4*(n^2 - 3): for l from 1 to n do for m from 1 to n do if((not l = j or not m = k) and (not l = j + den or not m = k + num) and (not l = j + 2*den or not m = k + 2*num) and (m - k)*den = num*(l - j))then tot := tot - 4: fi: od: od: od: od: fi: od: return tot: end:
%p seq(A189346(n),n=1..15); # _Nathaniel Johnston_, Apr 23 2011
%Y Cf. A000938, A175383, A178256, A189345.
%K nonn
%O 1,3
%A _Martin Renner_, Apr 20 2011
%E a(6)-a(28) from _Nathaniel Johnston_, Apr 23 2011
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