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nonn,easy,more

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Solution of the complementary equation a(n) = a(n-1)*b(n-2) - 1, where a(0) = 1, a(21) = 3, b(0) = 2, b(1) = 4.

a(2) = a(1) + a*b(0) + b(- 1) = 5

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The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A294381) for a guide to related sequences.

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following See A294381) for a guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:to related sequences.

allocated for Clark KimberlingSolution of the complementary equation a(n) = a(n-1)*b(n-2) - 1, where a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4.

1, 3, 5, 19, 113, 790, 6319, 56870, 568699, 6255688, 75068255, 975887314, 13662422395

0,2

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:

Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.pdf">Complementary equations</a>, J. Int. Seq. 19 (2007), 1-13.

a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that

a(2) = a(1) + a(0) + b(1) = 5

Complement: (b(n)) = (2, 4, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, ...)

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] - 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}] (* A294382 *)

Table[b[n], {n, 0, 10}]

Cf. A293076, A293765, A294381.

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nonn,easy,more

Clark Kimberling, Oct 29 2017

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