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A295618
Solution of the complementary equation a(n) = 2*a(n-1) - a(n-3) + b(n-2), where a(0) = 2, a(1) = 4, a(2) = 6, b(0) = 1, b(1) = 3, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.
2
2, 4, 6, 13, 27, 55, 105, 192, 339, 584, 988, 1651, 2733, 4494, 7354, 11993, 19511, 31688, 51404, 83319, 134973, 218566, 353838, 572729, 926920, 1500031, 2427363, 3927837, 6355675, 10284020, 16640237, 26924834, 43565684, 70491168, 114057540
OFFSET
0,1
COMMENTS
The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295613 for a guide to related sequences.
a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622), so that a( ) has the growth rate of the Fibonacci numbers (A000045).
LINKS
Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
EXAMPLE
a(0) = 2, a(1) = 4, a(2) = 6, b(0) = 1, b(1) = 3, b(2) = 5, so that
b(3) = 7 (least "new number")
a(3) = 2*a(2) - a(0) + b(1) = 13
Complement: (b(n)) = (1, 3, 5, 7, 8, 9, 10, 11, 12, 14, 15, 16, ...)
MATHEMATICA
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 2; a[1] = 4; a[2] = 6; b[0] = 1; b[1] = 3; b[2] = 5;
a[n_] := a[n] = 2 a[n - 1] - a[n - 3] + 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, 30}] (* A295618 *)
Table[b[n], {n, 0, 20}] (* complement *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 25 2017
STATUS
approved