login
A295617
Solution of the complementary equation a(n) = 2*a(n-1) - a(n-3) + b(n-2), where a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.
2
1, 3, 5, 13, 29, 60, 115, 210, 370, 636, 1074, 1792, 2963, 4868, 7961, 12977, 21105, 34269, 55582, 90081, 145916, 236274, 382492, 619094, 1001941, 1621418, 2623772, 4245634, 6869882, 11116025, 17986450, 29103053, 47090117, 76193821, 123284627, 199479176
OFFSET
0,2
COMMENTS
The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295613 for a guide to related sequences.
a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622), so that a( ) has the growth rate of the Fibonacci numbers (A000045).
LINKS
Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
EXAMPLE
a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6, so that
b(3) = 7 (least "new number")
a(3) = 2*a(2) - a(0) + b(1) = 13
Complement: (b(n)) = (2, 4, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, ...)
MATHEMATICA
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; a[2] = 5; b[0] = 2; b[1] = 4; b[2] = 6;
a[n_] := a[n] = 2 a[n - 1] - a[n - 3] + 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, 30}] (* A295617 *)
Table[b[n], {n, 0, 20}] (* complement *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 25 2017
STATUS
approved