Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.
%I #6 Apr 17 2018 23:08:00
%S 2,5,7,10,12,16,20,22,25,28,31,36,38,40,43,47,50,51,56,60,63,66,68,71,
%T 76,78,81,85,86,89,91,95,99,103,106,109,110,114,117,121,124,128,133,
%U 135,137,139,142,146,148,151,154,156,159,164,167,170,174,176,178
%N Solution (b(n)) of the system of 3 complementary equations in Comments.
%C Define sequences a(n), b(n), c(n) recursively, starting with a(0) = 1, b(0) = 2:
%C a(n) = least new;
%C b(n) = least new k >= a(n) + n;
%C c(n) = a(n) + b(n);
%C where "least new k" means the least positive integer not yet placed.
%C ***
%C The sequences a,b,c partition the positive integers. Let x = be the greatest solution of 1/x + 1/(x+1) + 1/(2x+1) = 1. Then
%C x = 1/3 + (2/3)*sqrt(7)*cos((1/3)*arctan((3*sqrt(111))/67));
%C x = 2.07816258732933084676..., and a(n)/n - > x, b(n)/n -> x+1, and c(n)/n -> 2x+1.
%H Clark Kimberling, <a href="/A298869/b298869.txt">Table of n, a(n) for n = 0..1000</a>
%e n: 0 1 2 3 4 5 6 7 8 9
%e a: 1 4 6 8 11 14 15 17 19 21
%e b: 2 5 7 10 12 16 20 22 25 28
%e c: 3 9 13 18 23 30 35 39 44 49
%t z = 400;
%t mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
%t a = {1}; b = {2}; c = {}; AppendTo[c, Last[a] + Last[b]]; n = 0;
%t Do[{n++, AppendTo[a, mex[Flatten[{a, b, c}], 1]],
%t AppendTo[b, mex[Flatten[{a, b, c}], a[[n]] + n]],
%t AppendTo[c, Last[a] + Last[b]]}, {z}];
%t Take[a, 100] (* A298868 *)
%t Take[b, 100] (* A298869 *)
%t Take[c, 100] (* A298870 *)
%t (* _Peter J. C. Moses_, Apr 08 2018 *)
%Y Cf. A299634, A298868, A298870.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Apr 17 2018