A294860 Solution of the equation a(n) = a(n-2) + b(n-2), where a( ) and b( ) are increasing sequences of positive integers such that every positive integer is in one of them and only one term is in both.

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

Offset: 0

Author

Clark Kimberling, Nov 16 2017

Keywords

• nonn
• easy

Comments

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); not quite complementary

A022939: a(n) = a(n-2) + b(n-2); offset 1, complementary

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); offset 1

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

Examples

			a(0) = 1, a(1) = 2, b(0) = 3, so that a(2) = 4
(b(n)) = (3,4,5,7,8,10,11,12,14,15,...)
		

Links

• Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Crossrefs

Cf. A294861, A294864, A294865.

Programs

Mathematica

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}]

Extensions

Edited by Clark Kimberling, Dec 02 2017