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A295620
Solution of the complementary equation a(n) = a(n-1) + 3*a(n-2) -2*a(n-3) - 2*a(n-4) + b(n-4), where a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, and (a(n)) and (b(n)) are increasing complementary sequences.
4
1, 2, 3, 4, 12, 20, 49, 85, 177, 304, 578, 979, 1765, 2953, 5150, 8538, 14570, 23997, 40352, 66149, 110094, 179867, 297172, 484313, 795934, 1294823, 2119684, 3443689, 5621258, 9123343, 14860404, 24100573, 39192618, 63526879, 103182816, 167177109, 271286602
OFFSET
0,2
COMMENTS
The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.
LINKS
Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
EXAMPLE
a(0) = 1, a(1) = 2, a(2) = 3, a(3) = 4, b(0) = 5, b(1) = 6, b(2) = 7, b(3) = 8, so that
b(4) = 9 (least "new number")
a(4) = a(3) + 3*a(2) -2*a(1) - 2*a(0) + b(0) = 12
Complement: (b(n)) = (5, 6, 7, 8, 9, 10, 11, 13, 14, 15, ...)
MATHEMATICA
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 2; a[2] = 3; a[3] = 4;
b[0] = 5; b[1] = 6; b[2] = 7; b[3] = 8;
a[n_] := a[n] = a[n - 1] + 3*a[n - 2] - 2*a[n - 3] - 2 a[n - 4] + b[n - 4];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
z = 36; Table[a[n], {n, 0, z}] (* A295620 *)
Table[b[n], {n, 0, 20}] (*complement *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 25 2017
STATUS
approved