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%I #32 May 05 2018 04:19:24
%S 6,16,27,36,46,57,66,75,87,96,106,117,126,136,147,156,165,177,186,196,
%T 207,216,227,237,246,255,267,276,286,297,306,316,327,336,345,357,366,
%U 376,387,396,406,417,426,435,447,456,466,477,486,497,507,516,525,537
%N The sums from the following procedure: from the list of positive integers, repeatedly remove the first three numbers and their sum.
%C This sequence is a solution, along with three other sequences, of a system of four complementary equations; see A297464. It is the "anti-tribonacci" sequence, in analogy with the anti-Fibonacci sequence, A075326. - _Clark Kimberling_, Apr 22 2018
%H Peter Kagey, <a href="/A265389/b265389.txt">Table of n, a(n) for n = 1..10000</a>
%H William Lowell Putnam Competition, <a href="http://kskedlaya.org/putnam-archive/2015.pdf">Problem B2</a>, 2015.
%p S:= {$1..1000}: A:= NULL:
%p while nops(S) >= 3 do
%p T:= S[1..3];
%p s:= convert(T,`+`);
%p S:= S[4..-1] minus {s};
%p A:= A, s
%p od:
%p A; # _Robert Israel_, Dec 22 2015
%t f[n_] := Block[{a = {}, r = Range@ n, s}, Do[If[Length@ r > 4, s = Total@ Take[r, 3 ]; AppendTo[a, s]; r = Drop[#, 3] &@ DeleteCases[r, x_ /; x == s], Break[]], {k, n}]; a]; f@ 184 (* _Michael De Vlieger_, Dec 22 2015 *)
%t morph = Nest[Flatten[# /. {0 -> {1, 2, 0}, 1 -> {1, 1, 0}, 2 -> {1, 0, 0}}] &, {0}, 9]; A265389 = Accumulate[Prepend[Drop[Flatten[morph /. Thread[{0, 1, 2} -> {{1, 1, 4}, {1, 2, 3}, {1, 3, 2}}]], 1] + 8, 6]];
%t Take[A265389, 100] (* _Peter J. C. Moses_, May 03 2018 *)
%o (Ruby)
%o x = (1..10000).to_a
%o (0...1000).collect do
%o y = x.shift(3).reduce(:+); x.delete_at x.index(y); y
%o end
%K nonn
%O 1,1
%A _Peter Kagey_, Dec 08 2015