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A294401
Solution of the complementary equation a(n) = a(n-1) + b(n-2) + 2n, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.
2
1, 3, 9, 19, 32, 48, 67, 89, 115, 144, 176, 211, 249, 290, 334, 381, 431, 485, 542, 602, 665, 731, 800, 872, 947, 1025, 1106, 1190, 1277, 1368, 1462, 1559, 1659, 1762, 1868, 1977, 2089, 2204, 2322, 2443, 2567, 2694, 2824, 2957, 3094, 3234, 3377, 3523, 3672
OFFSET
0,2
COMMENTS
The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A022940 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(1) + a(0) + 4 = 9.
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 10, 11, 12, 13, 15, 16, ...)
MATHEMATICA
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
a[n_] := a[n] = a[n - 1] + b[n - 2] + 2 n;
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}] (* A294401 *)
Table[b[n], {n, 0, 10}]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 31 2017
STATUS
approved