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A295365 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + b(n-3), where a(0) = 1, a(1) = 2, a[2] = 3, b(0) = 4, b(1) = 5, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences. 2
1, 2, 3, 20, 41, 82, 147, 256, 433, 722, 1191, 1952, 3185, 5182, 8415, 13648, 22117, 35823, 58002, 93891, 151962, 245925, 397962, 643965, 1042008, 1686057, 2728152, 4414299, 7142544, 11556939, 18699582 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295357 for a guide to related sequences.
LINKS
Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
FORMULA
a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).
EXAMPLE
a(0) = 1, a(1) = 2, a[2] = 3, b(0) = 4, b(1) = 5, b(2) = 6, so that
b(2) = 7 (least "new number")
a(3) = a(2) + a(1) + b(2) + b(1) + b(0) = 20
Complement: (b(n)) = (4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, ...)
MATHEMATICA
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 2; a[2] = 3; b[0] = 4; b[1] = 5; b[2]=6;
a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + b[n - 2] + b[n - 3];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
z = 32; u = Table[a[n], {n, 0, z}] (* A295365 *)
v = Table[b[n], {n, 0, 10}] (* complement *)
CROSSREFS
Sequence in context: A136886 A191423 A195686 * A233410 A318765 A055814
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 21 2017
STATUS
approved

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Last modified March 29 02:23 EDT 2024. Contains 371264 sequences. (Running on oeis4.)