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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A296999 Number of nonequivalent (mod D_8) ways to place 4 points on an n X n point grid so that no point is equally distant from two other points on the same row or the same column.

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%I A296999 #14 Jan 22 2018 15:33:53
%S A296999 0,1,17,226,1550,7221,26120,78484,206242,486640,1056377,2137506,
%T A296999 4085167,7430276,12964014,21801632,35520743,56249658,86880957,
%U A296999 131186720,194133425,282024809,402949496,566950056,786640454,1077397347,1458190435,1951789266,2585856152,3393157995
%N A296999 Number of nonequivalent (mod D_8) ways to place 4 points on an n X n point grid so that no point is equally distant from two other points on the same row or the same column.
%C A296999 Rotations and reflections of placements are not counted. If they are to be counted see A296998.
%C A296999 The condition of placements is also known as "no 3-term arithmetic progressions".
%H A296999 Heinrich Ludwig, <a href="/A296999/b296999.txt">Table of n, a(n) for n = 1..1000</a>
%H A296999 <a href="/index/Rec#order_23">Index entries for linear recurrences with constant coefficients</a>, signature (3,-1,-4,4,-4,5,1,-5,6,-10,8,-8,10,-6,5,-1,-5,4,-4,4,1,-3,1).
%F A296999 a(n) = (n^8 - 6*n^6 - 12*n^5 + 64*n^4 + 8*n^3 - 136*n^2 + (n == 1 (mod 2))*(14*n^4 - 96*n^3 + 162*n^2 - 92*n + 93))/192 + (n == 2 (mod 6))*n/6 + (n == 2 (mod 4))*n/4 + (n == 5 (mod 6))*(n + 1)/6.
%F A296999 a(n) = (n^8 - 6*n^6 - 12*n^5)/192 + b(n) + c(n), where
%F A296999   b(n) = (64*n^4 + 8*n^3 - 136*n^2)/192  for n even,
%F A296999   b(n) = (78*n^4 - 88*n^3 + 26*n^2 - 92*n + 93)/192  for n odd,
%F A296999   c(n) = 0          for n == 0, 1, 3, 4, 7, 9  (mod 12),
%F A296999   c(n) = n/4        for n == 6, 10             (mod 12),
%F A296999   c(n) = n/6        for n == 8                 (mod 12),
%F A296999   c(n) = 5/12*n     for n == 2                 (mod 12),
%F A296999   c(n) = (n + 1)/6  for n == 5, 11             (mod 12).
%F A296999 Conjectures from _Colin Barker_, Jan 21 2018: (Start)
%F A296999 G.f.: x^2*(1 + 14*x + 176*x^2 + 893*x^3 + 2861*x^4 + 6847*x^5 + 12704*x^6 + 20412*x^7 + 27052*x^8 + 33142*x^9 + 33910*x^10 + 33289*x^11 + 26586*x^12 + 20709*x^13 + 12212*x^14 + 7178*x^15 + 2639*x^16 + 1094*x^17 + 134*x^18 + 68*x^19 - 3*x^20 + 2*x^21) / ((1 - x)^9*(1 + x)^5*(1 - x + x^2)*(1 + x^2)^2*(1 + x + x^2)^2).
%F A296999 a(n) = 3*a(n-1) - a(n-2) - 4*a(n-3) + 4*a(n-4) - 4*a(n-5) + 5*a(n-6) + a(n-7) - 5*a(n-8) + 6*a(n-9) - 10*a(n-10) + 8*a(n-11) - 8*a(n-13) + 10*a(n-14) - 6*a(n-15) + 5*a(n-16) - a(n-17) - 5*a(n-18) + 4*a(n-19) - 4*a(n-20) + 4*a(n-21) + a(n-22) - 3*a(n-23) + a(n-24) for n>24.
%F A296999 (End)
%t A296999 Array[(#^8 - 6 #^6 - 12 #^5 + 64 #^4 + 8 #^3 - 136 #^2 + Boole[OddQ@ #] (14 #^4 - 96 #^3 + 162 #^2 - 92 # + 93))/192 + Boole[Mod[#, 6] == 2] #/6 + Boole[Mod[#, 4] == 2] #/4 + Boole[Mod[#, 6] == 5] (# + 1)/6 &, 30] (* _Michael De Vlieger_, Jan 21 2018 *)
%Y A296999 Cf. A014409, A296996, A296998.
%K A296999 nonn,easy
%O A296999 1,3
%A A296999 _Heinrich Ludwig_, Jan 21 2018