login
A294481
Solution of the complementary equation a(n) = a(n-2) + b(n-1) + n - 1, where a(0) = 1, a(1) = 3, b(0) = 2.
2
1, 3, 6, 10, 16, 22, 30, 39, 49, 60, 72, 85, 100, 115, 132, 149, 168, 188, 209, 231, 254, 278, 303, 329, 357, 385, 415, 445, 477, 509, 543, 577, 614, 650, 689, 727, 768, 808, 851, 893, 938, 983, 1030, 1077, 1126, 1175, 1226, 1277, 1330, 1383, 1438, 1494
OFFSET
0,2
COMMENTS
The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A294476 for a guide to related sequences.
LINKS
Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
EXAMPLE
a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
a(2) = a(0) + b(1) + 1 = 6
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 10, 12, 13, 15, ...)
MATHEMATICA
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; b[0] = 2;
a[n_] := a[n] = a[n - 2] + b[n - 1] + 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, 40}] (* A294481 *)
Table[b[n], {n, 0, 10}]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 01 2017
STATUS
approved