%I #29 Oct 31 2021 07:46:22
%S 1,2,6,3,10,20,4,14,30,50,5,18,40,70,105,6,22,50,90,140,196,7,26,60,
%T 110,175,252,336,8,30,70,130,210,308,420,540,9,34,80,150,245,364,504,
%U 660,825,10,38,90,170,280,420,588,780,990,1210,11,42,100,190,315,476,672
%N Triangle read by rows: T(n,k) = number of squares (not necessarily orthogonal) all of whose vertices lie in an (n + 1) X (k + 1) square lattice.
%C Reading down the diagonal gives A002415.
%H Joel B. Lewis, Jun 29 2007, <a href="/A130684/b130684.txt">Table of n, a(n) for n = 1..210</a>
%H Problem solved on the Art of Problem Solving forum, <a href="http://www.artofproblemsolving.com/Forum/viewtopic.php?t=155463">Number of squares in a grid</a>
%F T(n, k) = k*(k+1)*(k+2)*(2*n - k + 1)/12 (k <= n).
%e T(2, 2) = 6 because there are 6 squares all of whose vertices lie in a 3 X 3 lattice: four squares of side length 1, one square of side length 2 and one non-orthogonal square of side length the square root of 2.
%e Triangle begins:
%e 1;
%e 2, 6;
%e 3, 10, 20;
%e 4, 14, 30, 50;
%e 5, 18, 40, 70, 105;
%e 6, 22, 50, 90, 140, 196;
%e 7, 26, 60, 110, 175, 252, 336;
%e ...
%o (PARI) T(n, k) = binomial(k+2,3)*(2*n - k + 1)/2 \\ _Charles R Greathouse IV_, Mar 08 2017
%Y Cf. A002415. For squares whose edges are required to be parallel to the edges of the large square, see A082652.
%K easy,nonn,tabl
%O 1,2
%A _Joel B. Lewis_, Jun 29 2007