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%I #18 Feb 26 2022 12:24:05
%S 1,5,8,4,17,23,10,38,47,19,68,80,31,107,122,46,155,173,64,212,233,85,
%T 278,302,109,353,380,136,437,467,166,530,563,199,632,668,235,743,782,
%U 274,863,905,316,992,1037,361,1130,1178,409,1277,1328
%N Numerators of (n*(n - 3)/6) + 1, arising as the maximum possible number of triple lines for an n-element set.
%C Numerators of (n*(n - 3)/6) + 1, which arises as the maximum possible number of triple lines for an n-element set, according to Green and Tao, cited in Elekes. The fractions for n = 3, 4, 5, 6, ... are 1/1, 5/3, 8/3, 4/1, 17/3, 23/3, 10/1, 38/3, 47/3, 19/1, 68/3, 80/3, 31/1, 107/3, 122/3, 46/1, 155/3, 173/3, 64/1, 212/3, 233/3, 85/1, 278/3, 302/3, 109/1, 353/3, 380/3, 136/1, 437/3, 467/3, 166/1, 530/3, 563/3, 199/1, 632/3, 668/3, 235/1, 743/3, 782/3, 274/1, 863/3, 905/3, 316/1, 992/3, 1037/3, 361/1, 1130/3, 1178/3, 409/1, 1277/3, 1328/3. The corresponding denominators are A169609.
%H György Elekes, Endre Szabó, <a href="http://arxiv.org/abs/1302.5777">On Triple Lines and Cubic Curves --- the Orchard Problem revisited</a>, arXiv:1302.5777 [math.CO], Feb 23, 2013.
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,3,0,0,-3,0,0,1).
%e a(10) = 38 because (10*(10 - 3)/6) + 1 = 38/3.
%t Numerator[Table[(n(n-3))/6+1,{n,3,60}]] (* or *) LinearRecurrence[{0,0,3,0,0,-3,0,0,1},{1,5,8,4,17,23,10,38,47},60] (* _Harvey P. Dale_, Feb 11 2015 *)
%Y Cf. A169609.
%K nonn,easy,frac
%O 3,2
%A _Jonathan Vos Post_, Feb 25 2013