A265389 The sums from the following procedure: from the list of positive integers, repeatedly remove the first three numbers and their sum.

6, 16, 27, 36, 46, 57, 66, 75, 87, 96, 106, 117, 126, 136, 147, 156, 165, 177, 186, 196, 207, 216, 227, 237, 246, 255, 267, 276, 286, 297, 306, 316, 327, 336, 345, 357, 366, 376, 387, 396, 406, 417, 426, 435, 447, 456, 466, 477, 486, 497, 507, 516, 525, 537

Offset: 1

Author

Peter Kagey, Dec 08 2015

Keywords

• nonn

Comments

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

Links

• Peter Kagey, Table of n, a(n) for n = 1..10000
• Wieb Bosma, Rene Bruin, Robbert Fokkink, Jonathan Grube, Anniek Reuijl, and Thian Tromp, Using Walnut to solve problems from the OEIS, arXiv:2503.04122 [math.NT], 2025.
• Robbert Fokkink and Gandhar Joshi, Anti-recurrence sequences, arXiv:2506.13337 [math.NT], 2025. See pp. 2, 11, 18.
• William Lowell Putnam Competition, Problem B2, 2015.

Programs

Maple

S:= {$1..1000}: A:= NULL:
while nops(S) >= 3 do
  T:= S[1..3];
  s:= convert(T,`+`);
  S:= S[4..-1] minus {s};
  A:= A, s
od:
A; # Robert Israel, Dec 22 2015

Mathematica

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 *)
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]];
Take[A265389, 100]    (* Peter J. C. Moses, May 03 2018 *)

Ruby

x = (1..10000).to_a
(0...1000).collect do
  y = x.shift(3).reduce(:+); x.delete_at x.index(y); y
end