A296253 - OEIS

A296253

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

1, 4, 14, 43, 93, 185, 342, 608, 1050, 1779, 2973, 4921, 8119, 13296, 21704, 35324, 57389, 93113, 150943, 244540, 396012, 641128, 1037765, 1679569, 2718063, 4398416, 7117320, 11516636, 18634917, 30152577, 48788583, 78942316, 127732124, 206675736, 334409229

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OFFSET

0,2

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). See A296245 for a guide to related sequences.

LINKS

Clark Kimberling, Table of n, a(n) for n = 0..1000

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

FORMULA

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

EXAMPLE

a(0) = 1, a(1) = 4, b(0) = 2, b(1) = 3;

a(2) = a(0) + a(1) + b(1)^2 = 14;

Complement: (b(n)) = (2, 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, ...)

MATHEMATICA

a[0] = 1; a[1] = 4; b[0] = 2; b[1] = 3;

a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n-1]^2;

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}] (* A296253 *)

Table[b[n], {n, 0, 20}] (* complement *)

CROSSREFS