Revision History for A295141
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| A295141
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Solution of the complementary equation a(n) = 2*a(n-1) + a(n-2) + b(n-2), where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (a(n)) and (b(n)) are increasing complementary sequences.
(history;
published version)
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#4 by Susanna Cuyler at Sun Nov 19 19:05:28 EST 2017
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#3 by Clark Kimberling at Sun Nov 19 18:55:45 EST 2017
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#2 by Clark Kimberling at Sun Nov 19 18:38:35 EST 2017
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| NAME
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allocatedSolution of the complementary equation a(n) = 2*a(n-1) + a(n-2) + b(n-2), where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (a(n)) and (b(n)) are forincreasing Clarkcomplementary Kimberlingsequences.
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| DATA
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1, 2, 8, 22, 57, 142, 348, 847, 2052, 4962, 11988, 28951, 69904, 168774, 407468, 983727, 2374940, 5733626, 13842212, 33418071, 80678377, 194774849, 470228100
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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.
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| LINKS
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Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.html">Complementary equations</a>, J. Int. Seq. 19 (2007), 1-13.
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| FORMULA
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a(n+1)/a(n) -> 1 + sqrt(2).
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| EXAMPLE
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a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4
a(2) =2*a(1) + a(0) + b(0) = 8
Complement: (b(n)) = (3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, ...)
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| MATHEMATICA
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mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4;
a[n_] := a[n] = 2 a[ n - 1] + 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}] (* A295141 *)
Table[b[n], {n, 0, 10}]
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| CROSSREFS
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Cf. A295053, A295142, A295143, A295144.
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| KEYWORD
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allocated
nonn,easy
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| AUTHOR
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Clark Kimberling, Nov 19 2017
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| STATUS
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approved
editing
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#1 by Clark Kimberling at Wed Nov 15 11:20:24 EST 2017
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| NAME
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allocated for Clark Kimberling
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| KEYWORD
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allocated
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| STATUS
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approved
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