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A295062
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Solution of the complementary equation a(n) = 4*a(n-2) + b(n-2), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.
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2
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1, 3, 6, 16, 29, 71, 124, 293, 506, 1183, 2036, 4745, 8158, 18995, 32649, 75998, 130615, 304012, 522481, 1216070, 2089947, 4864304, 8359813, 19457242, 33439279, 77828996, 133757146, 311316015
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OFFSET
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0,2
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COMMENTS
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The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295053 for a guide to related sequences.
The sequence a(n+1)/a(n) appears to have two convergent subsequences, with limits 1.71... and 2.32... .
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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
a(2) = 4*a(0) + b(0) = 6
Complement: (b(n)) = (2, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17, ...)
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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; b[1]=4;
a[n_] := a[n] = 4 a[n - 2] + b[n - 2];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 18}] (* A295062 *)
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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