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A294476 Solution of the complementary equation a(n) = a(n-2) + b(n-1) + 1, where a(0) = 1, a(1) = 3, b(0) = 2. 7
1, 3, 6, 9, 14, 18, 25, 30, 38, 44, 54, 61, 72, 81, 93, 103, 116, 127, 141, 154, 169, 183, 199, 215, 232, 249, 267, 285, 304, 323, 344, 364, 386, 407, 430, 453, 477, 501, 526, 551, 577, 603, 630, 657, 686, 714, 744, 773, 804, 834, 867, 898, 932, 964, 999 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

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:

A294476:  a(n) = a(n-2) + b(n-1) + 1

A294477:  a(n) = a(n-2) + b(n-1) + 2

A294478:  a(n) = a(n-2) + b(n-1) + 3

A294479:  a(n) = a(n-2) + b(n-1) + n

A294480:  a(n) = a(n-2) + b(n-1) + 2n

A294481:  a(n) = a(n-2) + b(n-1) + n - 1

A294482:  a(n) = a(n-2) + b(n-1) + n + 1

For a(n-2) + b(n-1), with offset 1 instead of 0, see A022942.

LINKS

Clark Kimberling, Table of n, a(n) for n = 0..1000

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, 7, 8, 10, 11, 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] + 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}]  (* A294476 *)

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

CROSSREFS

Cf. A293076, A293765, A293358, A294414.

Sequence in context: A230876 A000791 A027424 * A258087 A191130 A319738

Adjacent sequences:  A294473 A294474 A294475 * A294477 A294478 A294479

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, Nov 01 2017

STATUS

approved

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Last modified September 26 02:39 EDT 2020. Contains 337346 sequences. (Running on oeis4.)