A295862 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.

1, 3, 9, 18, 34, 60, 104, 175, 291, 479, 784, 1278, 2078, 3373, 5470, 8863, 14354, 23239, 37616, 60879, 98520, 159425, 257972, 417425, 675426, 1092881, 1768338, 2861251, 4629622, 7490908, 12120566, 19611511, 31732115, 51343665, 83075820, 134419526, 217495388

Offset: 0

Author

Clark Kimberling, Dec 08 2017

Keywords

• nonn
• easy

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622). Following is a guide to related sequences:

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Complementary equation: a(n) = a(n-1) + a(n-2) + b(n); initial values (a(0), a(1); b(0), b(1), b(2)):

A295862: (1,3; 2,4,5)

A295947: (2,4; 1,3,5)

A295948: (3,4; 1,2,5)

A295949: (1,2; 3,4,5)

A295950: (1,4; 2,3,5)

A295951: (2,3; 1,4,5)

A295952: (1,5; 2,3,4)

Complementary equation: a(n) = a(n-1) + a(n-2) + b(n) + 1; initial values (a(0), a(1); b(0), b(1), b(2)):

A295953: (1,3; 2,4,5)

A295954: (2,4; 1,3,5)

A295955: (3,4; 1,2,5)

A295956: (1,2; 3,4,5)

A295957: (1,4; 2,3,5)

A295958: (2,3; 1,4,5)

A295959: (1,5; 2,3,4)

Complementary equation: a(n) = a(n-1) + a(n-2) + b(n) - 1; initial values (a(0), a(1); b(0), b(1), b(2)):

A295860: (1,3; 2,4,5)

A295961: (2,4; 1,3,5)

A295962: (3,4; 1,2,5)

A295963: (1,2; 3,4,5)

A295964: (1,4; 2,3,5)

A295965: (2,3; 1,4,5)

A295966: (1,5; 2,3,4)

Examples

			a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, b(2) = 5, so that
b(3) = 6 (least "new number");
a(2) = a(1) + a(0) + b(2) = 9;
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 19, 20, ...)
		

Links

• Clark Kimberling, Table of n, a(n) for n = 0..3000
• Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Crossrefs

Cf. A001622, A000045, A295947.

Programs

Mathematica

a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4; b[2] = 5;
a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n];
j = 1; While[j < 6, k = a[j] - j - 1;
 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
Table[a[n], {n, 0, k}]  (*A295862*)
Table[b[n], {n, 0, 20}] (*complement*)

Formula

a(n) = H + R, where H = f(n-1)*a(0) + f(n)*a(1) and R = f(n-1)*b(2) + f(n-2)*b(3) + ... + f(2)*b(n-1) + f(1)*b(n), where f(n) = A000045(n), the n-th Fibonacci number.