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%I #6 Nov 03 2017 09:53:33
%S 1,2,6,12,23,42,73,124,207,342,562,918,1495,2429,3941,6388,10348,
%T 16756,27125,43903,71052,114980,186058,301065,487151,788245,1275426,
%U 2063702,3339160,5402895,8742089,14145019,22887144,37032200,59919382,96951621,156871043
%N Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-2), 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, which, for the sequences in the following guide, are a(0) = 1, a(1) = 2, b(0) = 3:
%C a(n) = a(n-1) + a(n-2) + b(n-2) A294532
%C a(n) = a(n-1) + a(n-2) + b(n-2) + 1 A294533
%C a(n) = a(n-1) + a(n-2) + b(n-2) + 2 A294534
%C a(n) = a(n-1) + a(n-2) + b(n-2) + 3 A294535
%C a(n) = a(n-1) + a(n-2) + b(n-2) - 1 A294536
%C a(n) = a(n-1) + a(n-2) + b(n-2) + n A294537
%C a(n) = a(n-1) + a(n-2) + b(n-2) + 2n A294538
%C a(n) = a(n-1) + a(n-2) + b(n-2) + n - 1 A294539
%C a(n) = a(n-1) + a(n-2) + b(n-2) + 2n - 1 A294540
%C a(n) = a(n-1) + a(n-2) + b(n-1) A294541
%C a(n) = a(n-1) + a(n-2) + b(n-1) + 1 A294542
%C a(n) = a(n-1) + a(n-2) + b(n-1) + 2 A294543
%C a(n) = a(n-1) + a(n-2) + b(n-1) + 3 A294544
%C a(n) = a(n-1) + a(n-2) + b(n-1) - 1 A294545
%C a(n) = a(n-1) + a(n-2) + b(n-1) + n A294546
%C a(n) = a(n-1) + a(n-2) + b(n-1) + 2n A294547
%C a(n) = a(n-1) + a(n-2) + b(n-1) + n - 1 A294548
%C a(n) = a(n-1) + a(n-2) + b(n-1) + n + 1 A294549
%C a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) A294550
%C a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + 1 A294551
%C a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + n A294552
%C a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - n A294553
%C a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + 2 A294554
%C a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + 3 A294555
%C a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + n + 1 A294556
%C a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + n - 1 A294557
%C a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + 2n A294558
%C a(n) = a(n-1) + a(n-2) + b(n-1) + 2*b(n-2) A294559
%C a(n) = a(n-1) + a(n-2) + 2*b(n-1) + 2*b(n-2) A294560
%C a(n) = a(n-1) + a(n-2) + 2*b(n-1) + b(n-2) A294561
%C a(n) = a(n-1) + a(n-2) + b(n-1) - b(n-2) + 1 A294562
%C a(n) = a(n-1) + a(n-2) + b(n-1) - b(n-2) + n A294563
%C a(n) = a(n-1) + a(n-2) + 2*b(n-1) - b(n-2) - 1 A294564
%C a(n) = a(n-1) + a(n-2) + 2*b(n-1) - b(n-2) - 3 A294565
%C Conjecture: for every sequence listed here, 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(0) + a(1) + b(0) = 6
%e Complement: (b(n)) = (3, 4, 5, 7, 8, 9, 10, 11, 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];
%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}] (* A294532 *)
%t Table[b[n], {n, 0, 10}]
%Y Cf. A001622, A293076, A294413.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Nov 03 2017
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