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Solution of the complementary equation a(n) = a(n-2) + b(n-2), where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n) ) and (b(n) ) are increasing complementary sequences of positive integers such that every positive integer is in one of them and only one term is in both.

A294860: a(n) = a(n-2) + b(n-2); not quite complementary

A294861A022939: a(n) = a(n-2) + b(n-2) + ; offset 1, complementary

A294861: a(n) = a(n-2) + b(n-2) + 1

A022942: a(n) = a(n-2) + b(n-1); with offset 1

A295998: a(n) = 2*a(n-2) + b(n-2)

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

b(1) = 4 (least "new number")

a(2) = a(0) + b(0) = 4

Complement: (b(n)) = (3, 4, 5, 7, 8, 10, 11, 12, 14, 15, 16, ...)

Edited by Clark Kimberling, Dec 02 2017

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allocated for Clark KimberlingSolution of the complementary equation a(n) = a(n-2) + b(n-2), where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

1, 2, 4, 6, 9, 13, 17, 23, 28, 35, 42, 50, 58, 68, 77, 88, 98, 110, 122, 135, 148, 162, 177, 192, 208, 224, 241, 258, 277, 295, 315, 334, 355, 375, 398, 419, 443, 465, 490, 513, 539, 564, 591, 617, 645, 672, 701, 729, 760, 789, 821, 851, 884, 915, 949, 981

0,2

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

A294860: a(n) = a(n-2) + b(n-2)

A294861: a(n) = a(n-2) + b(n-2) + 1

A294862: a(n) = a(n-2) + b(n-2) + 2

A294863: a(n) = a(n-2) + b(n-2) + 3

A294864: a(n) = a(n-2) + b(n-2) + n

A294865: a(n) = a(n-2) + 2*b(n-2)

A294866: a(n) = 2*a(n-1) - a(n-2) + b(n-1)

A294867: a(n) = 2*a(n-1) - a(n-2) + b(n-1) - 1

A294868: a(n) = 2*a(n-1) - a(n-2) + b(n-1) - 2

A294869: a(n) = 2*a(n-1) - a(n-2) + b(n-1) + 1

A294870: a(n) = 2*a(n-1) - a(n-2) + b(n-1) + 2

A294871: a(n) = 2*a(n-1) - a(n-2) + b(n-1) + 3

A294872: a(n) = 2*a(n-1) - a(n-2) + b(n-1) + n

A022942: a(n) = a(n-2) + b(n-1); with offset 1

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

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

b(1) = 4 (least "new number")

a(2) = a(0) + b(0) = 4

Complement: (b(n)) = (3, 4, 5, 7, 8, 10, 11, 12, 14, 15, 16, ...)

mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;

a[0] = 1; a[1] = 2; b[0] = 3;

a[n_] := a[n] = a[n - 2] + b[n - 2];

b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];

Table[a[n], {n, 0, 18}] (* A294860 *)

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

Cf. A294861, A294864, A294865.

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Clark Kimberling, Nov 16 2017

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