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A294476
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Solution of the complementary equation a(n) = a(n-2) + b(n-1) + 1, where a(0) = 1, a(1) = 3, b(0) = 2.
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7
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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
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OFFSET
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0,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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,...)
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MATHEMATICA
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mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; b[0] = 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}]
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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