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A295614
Solution of the complementary equation a(n) = 2*a(n-1) - a(n-3) + b(n-1), where a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.
2
1, 3, 5, 15, 34, 71, 136, 248, 436, 748, 1261, 2100, 3468, 5692, 9302, 15155, 24638, 39995, 64857, 105099, 170227, 275622, 446171, 722142, 1168690, 1891238, 3060364, 4952069, 8012932, 12965533, 20979032, 33945168, 54924840, 88870686, 143796243, 232667686
OFFSET
0,2
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) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, so that
b(3) = 7 (least "new number")
a(3) = 2*a(2) - a(0) + b(2) = 15
Complement: (b(n)) = (2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, ...)
MATHEMATICA
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; a[2] = 5; b[0] = 2; b[1] = 4; b[2] = 6;
a[n_] := a[n] = 2 a[n - 1] - a[n - 3] + b[n - 1];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 30}] (* A295614 *)
Table[b[n], {n, 0, 20}] (* complement *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 25 2017
STATUS
approved