A297830 Solution of the complementary equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n, where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (b(n)) is the increasing sequence of positive integers not in (a(n)). See Comments.

1, 2, 9, 12, 15, 18, 21, 26, 28, 33, 35, 40, 42, 47, 49, 54, 56, 59, 62, 67, 71, 73, 76, 79, 84, 88, 90, 93, 96, 101, 105, 107, 110, 113, 118, 122, 124, 127, 130, 135, 139, 141, 146, 148, 153, 155, 158, 161, 166, 168, 171, 176, 180, 182, 187, 189, 194, 196

Offset: 0

Author

Clark Kimberling, Feb 04 2018

Keywords

• nonn
• easy

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. Conjecture: a(n) - (2 +sqrt(2))*n < 3 for n >= 1.

Guide to related sequences having initial values a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, where (b(n)) is the increasing sequence of positive integers not in (a(n)):

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a(n) = a(1)*b(n-1) - a(0)*b(n-2) + n (a(n)) = A297826; (b(n)) = A297997

a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n (a(n)) = A297830; (b(n)) = A298003

a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 3*n (a(n)) = A297836; (b(n)) = A298004

a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 4*n (a(n)) = A297837; (b(n)) = A298005

a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n - 1 (a(n)) = A297831; (b(n)) = A298006

a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n - 2 (a(n)) = A297832; (b(n)) = A298007

a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n - 3 (a(n)) = A297833; (b(n)) = A298108

a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n - 4 (a(n)) = A297834; (b(n)) = A298109

a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n + 1 (a(n)) = A297835;

a(n) = a(1)*b(n-1) - a(0)*b(n-2)+floor(5*n/2) (a(n)) = A297998;

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For sequences (a(n)) and (b(n)) associated with equations of the form a(n) = a(1)*b(n) - a(0)*b(n-1), see the guide at A297800.

Examples

			a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, so that a(2) = 9.
Complement: (b(n)) = (3,4,5,6,8,10,11,13,14,16,17,19,...)
		

Links

• Clark Kimberling, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A297826, A297836, A297837.

Programs

Mathematica

a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4;
a[n_] := a[n] = a[1]*b[n - 1] - a[0]*b[n - 2] + 2 n;
j = 1; While[j < 100, k = a[j] - j - 1;
While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++]; k
Table[a[n], {n, 0, k}]  (* A297830 *)