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%I #4 Nov 03 2017 09:54:33
%S 1,2,9,20,41,76,135,232,392,652,1075,1761,2873,4674,7590,12310,19949,
%T 32311,52316,84686,137064,221815,358947,580833,939854,1520764,2460698,
%U 3981545,6442329,10423963,16866384,27290442,44156924,71447467,115604495,187052069
%N Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2) + 2*n - 1, where a(0) = 1, a(1) = 2, b(0) = 3.
%C The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A294532 for a guide to related sequences. Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622)..
%H Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.html">Complementary equations</a>, J. Int. Seq. 19 (2007), 1-13.
%e a(0) = 1, a(1) = 2, b(0) = 3, so that
%e b(1) = 4 (least "new number")
%e a(2) = a(1) + a(0) + b(0) + 3 = 9
%e Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, ...)
%t mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
%t a[0] = 1; a[1] = 3; b[0] = 2;
%t a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 2] + 2n - 1;
%t b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
%t Table[a[n], {n, 0, 40}] (* A294540 *)
%t Table[b[n], {n, 0, 10}]
%Y Cf. A001622, A294532.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Nov 03 2017
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