A099938 Consider the sequence of circles C0, C1, C2, C3 ..., where C0 is a half-circle of radius 1. C1 is the largest circle that fits into C0 and has radius 1/2. C(n+1) is the largest circle that fits inside C0 but outside C(n), etc. Sequence gives the curvatures (reciprocals of the radii) of the circles.

2, 4, 18, 100, 578, 3364, 19602, 114244, 665858, 3880900, 22619538, 131836324, 768398402, 4478554084, 26102926098, 152139002500, 886731088898, 5168247530884, 30122754096402, 175568277047524, 1023286908188738, 5964153172084900, 34761632124320658

Offset: 1

Author

Hartmut Neubauer (hartmut.f.neubauer(AT)t-online.de), Nov 12 2004

Keywords

• nonn
• easy

Comments

The numbers a(2), a(4), a(6) etc. are squares and a(1), a(3), a(5) ... are twice squares. Furthermore, a(1) - 2, a(3) - 2, a(5) - 2 etc. are squares and a(2) - 2, a(4) - 2, a(6) - 2 etc. are twice square.

C(n) is centered at (x(n), y(n)), where x(n) = sqrt(1 - 2/a(n)) and y(n) = 1/a(n). - David Wasserman, Feb 28 2008

C(n) is tangent to C0 because sqrt(x(n)^2 + y(n)^2) + y(n) = 1 and C(n) is tangent to C(n+1) because sqrt[(x(n+1) - x(n))^2 + (y(n+1) - y(n))^2] = y(n) + y(n+1). - David Wasserman, Feb 28 2008

a(n+1)/a(n) converges to 3 + sqrt(8). - David Wasserman, Feb 28 2008

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Francisco Javier Garcia Capitan Blog, About the A099938 sequence
• David Wasserman, Illustration of this sequence
• Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Crossrefs

Equals 2 * A055997(n-1).

Programs

Mathematica

Table[FullSimplify[2 Cosh[n ArcSinh[1]]^2], {n, 0, 9}] (* Francisco Javier García Capitán, Jun 05 2016 *)

PARI

Vec(-2*x*(2*x^2-5*x+1)/((x-1)*(x^2-6*x+1)) + O(x^30)) \\ Colin Barker, Jun 05 2016

Formula

a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3). G.f.: -2*x*(2*x^2-5*x+1) / ((x-1)*(x^2-6*x+1)). - Colin Barker, Jan 07 2013

a(n) = 1/2*(2 + (3 - 2*sqrt(2))^n*(3 + 2*sqrt(2)) + (3 - 2*sqrt(2))*(3 + 2*sqrt(2))^n). - Colin Barker, Jun 05 2016

Extensions

More terms from David Wasserman, Feb 28 2008