A093568 Let M = the 2 X 2 matrix [0 1 / -1 2+sqrt(8)]. Perform the operation M^n * [1 1] = [x y]; then a(n) = floor(x), a(n+1) = floor(y).

1, 3, 17, 80, 371, 1714, 7904, 36451, 168098, 775200, 3574898, 16485939, 76026256, 350601298, 1616826563, 7456127936, 34384543809, 158567136067, 731245317072, 3372197587729, 15551164985523, 71715469249777, 330721752004000

Offset: 1

Author

Gary W. Adamson, Mar 31 2004, suggested by Herb Conn

Keywords

• nonn

Comments

Lim_{n->infinity} a(n)/a(n-1) = 4.61158178931... = e^(arccosh(1+sqrt(2))). This constant is found in the Descartes circle formula given ("curvature" = 1/r), which states that given 4 mutually tangent circles with curvatures a,b,c,d, (a^2 + b^2 + c^2 + d^2) = (1/2)*(a + b + c + d)^2. Let a=b=1. Find the radius of circle c such that the radius of the innermost circle d = 1/(r of c). The answer is r of c = 4.611581789... and r of d = 1/4.611581... = 0.216845335437. Let C = 4.611581... C + 1/C = 2 + sqrt(8), where C may be found through an iterative method:

1. Begin with the formula for the radius r of the innermost tangential circle surrounded by 3 mutually tangent circles with radii a, b, c. Given a = b = 1, with c variable, then r = f(c) = (2*sqrt(1 + 2/c) - 2 - 1/c) / (4/c - 1/c^2).

2. Using an iterative operation, take any real number 1 or > 1 (we can use 1). Find f(c), then repeatedly use 1/f(c) as the new c. After about a dozen of such operations, the fixed constant becomes 4.61158178931... This constant C has the property that C + 1/C = 4.828427124... = 2 + sqrt(8).

Examples

			M^5 * [1 1] = [371.676... 1714.013...]. a(5) = 371, a(6) = 1714. a(10)/a(9) = 775200/168098 = 4.611595...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..200

Programs

Mathematica

Table[ Floor[ MatrixPower[{{0, 1}, {-1, 2 + Sqrt[8]}}, n].{1, 1}][[1]], {n, 24}] (* Robert G. Wilson v, Apr 09 2004 *)