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%I A228012 #33 Dec 24 2015 18:30:55
%S A228012 1,4,5,8,9,14,16,23,25,33,36,46,49,60,64,77,81,95,100,116,121,138,144,
%T A228012 163,169,189,196,218,225,248,256,281,289,315,324,352,361,390,400,431,
%U A228012 441,473,484,518,529,564,576,613,625,663,676,716,729,770,784,827,841
%N A228012 The 2-color Rado number for the equation x_1 + x_2 + ... + x_n = 2*x_0
%C A228012 For n=1, the Rado number is infinity (since the positive integers can be colored using two colors in such a way that no monochromatic solution to the equation x_1 = 2*x_0 exists).
%D A228012 D. Schaal and D. Vestal, Rado numbers for x_1 + x_2 + ... + x_(m-1) = 2*x_m, Congressus Numerantium, 191(2008), 105-116.
%F A228012 For n >= 5, a(n) = ceiling(ceiling(n/2)*n/2).
%F A228012 Conjecture: For n >= 5, a(n) = (1-(-1)^n+i*(-i)^n-i*i^n+n-(-1)^n*n+2*n^2)/8, where i=sqrt(-1). G.f.: x^2*(x^9-2*x^7-x^6+x^5+x^4-3*x-1) / ((x-1)^3*(x+1)^2*(x^2+1)). - _Colin Barker_, Aug 12 2013
%e A228012 For n=4, we have a(4) = 5, meaning that the 2-color Rado number for the equation E:x_1 + x_2 + x_3 + x_4 = 2*x_0 is 5.  The coloring (or partition) Red = {1,4} and Blue = {2,3} avoids a monochromatic solution to E; however, any 2-coloring of the integers {1,2,3,4,5} will have a monochromatic solution to E.
%K A228012 easy,nonn
%O A228012 2,2
%A A228012 _Donald Vestal_, Aug 08 2013

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