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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.
16
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
0,2
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)
LINKS
Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
EXAMPLE
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,...)
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}]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 16 2017
EXTENSIONS
Edited by Clark Kimberling, Dec 02 2017
STATUS
approved