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 A294476 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + 1, where a(0) = 1, a(1) = 3, b(0) = 2. 7
 1, 3, 6, 9, 14, 18, 25, 30, 38, 44, 54, 61, 72, 81, 93, 103, 116, 127, 141, 154, 169, 183, 199, 215, 232, 249, 267, 285, 304, 323, 344, 364, 386, 407, 430, 453, 477, 501, 526, 551, 577, 603, 630, 657, 686, 714, 744, 773, 804, 834, 867, 898, 932, 964, 999 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS The complementary sequences a() and b() are uniquely determined by the titular equation and initial values.  The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2: A294476:  a(n) = a(n-2) + b(n-1) + 1 A294477:  a(n) = a(n-2) + b(n-1) + 2 A294478:  a(n) = a(n-2) + b(n-1) + 3 A294479:  a(n) = a(n-2) + b(n-1) + n A294480:  a(n) = a(n-2) + b(n-1) + 2n A294481:  a(n) = a(n-2) + b(n-1) + n - 1 A294482:  a(n) = a(n-2) + b(n-1) + n + 1 For a(n-2) + b(n-1), with offset 1 instead of 0, see A022942. LINKS Clark Kimberling, Table of n, a(n) for n = 0..1000 Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13. EXAMPLE a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that a(2)  = a(0) + b(1) + 1 = 6 Complement: (b(n)) = (2, 4, 5, 7, 8, 10, 11, 12, 13, 15,...) MATHEMATICA mex := First[Complement[Range[1, Max[#1] + 1], #1]] &; a = 1; a = 3; b = 2; a[n_] := a[n] = a[n - 2] + b[n - 1] + 1; b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]]; Table[a[n], {n, 0, 40}]  (* A294476 *) Table[b[n], {n, 0, 10}] CROSSREFS Cf. A293076, A293765, A293358, A294414. Sequence in context: A230876 A000791 A027424 * A258087 A191130 A319738 Adjacent sequences:  A294473 A294474 A294475 * A294477 A294478 A294479 KEYWORD nonn,easy AUTHOR Clark Kimberling, Nov 01 2017 STATUS approved

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Last modified July 26 15:49 EDT 2021. Contains 346294 sequences. (Running on oeis4.)