A246641 Sequence a(n) = (1 + A007805(n))/2, appearing in a certain touching problem for three circles and a chord, together with A007805.

1, 9, 153, 2737, 49105, 881145, 15811497, 283725793, 5091252769, 91358824041, 1639367579961, 29417257615249, 527871269494513, 9472265593285977, 169972909409653065, 3050040103780469185, 54730748958638792257, 982103441151717791433, 17623131191772281453529, 316234258010749348372081, 5674593513001715989243921

Offset: 0

Author

Wolfdieter Lang, Sep 05 2014

Keywords

• nonn
• easy

Comments

This sequence is motivated by Kival Ngaokrajang's touching circle problem considered in A240926 and A115032.

a(n), together with b(n) = A007805(n), appears in a sequence of curvatures c(n) = 4*(b(n) + a(n)*phi), with phi = (1+sqrt(5))/2, the golden section, and n >= 0. These are integers in the real quadratic number field Q(sqrt(5)).

The circle with curvature c(n) touches i) the chord of length 2 (in some length units) bisecting a circular disk of radius 5/4, and ii) two touching circles in the larger section with curvatures given by c1(n) and c1(n+1), where c1(n) = A115032(n-1), with c1(0) = 1. See the illustration of Kival Ngaokrajang's link given in A240926, where the first circles in the larger (lower) section are shown.

From Descartes' theorem on touching circles (see the links), one has here: c(n) = c1(n) + c1(n+1) + 2*sqrt(c1(n)*c1(n+1)), with c1(n) = (1 + S(n, 18) - 9*S(n-1, 18))/2, n >= 0, where Chebyshev's S-polynomials (see A049310) appear. See also the W. Lang link in A240926, part I. In this application curvature 0 for the chord is used.

For the proof for the first a(n) formula given below use the above given curvature c1(n) in Descartes' formula and compare it with a(n) from c(n) = 4*(A007805(n) + a(n)* (1+sqrt(5))/2). This can be done by using standard S-polynomial identities like the three term recurrence for S(n+1, 18) and the Cassini-Simson type identity (see a comment on A246638) which implies the formula S(n, 18)*S(n-1, 18) = (-1 + S(n, 18)^2 + S(n-1, 18)^2)/18. See also the W. Lang link in A240926 part IV a).

Also the indices of centered pentagonal numbers which are also centered square numbers. - Colin Barker, Jan 01 2015

Also positive integers y in the solutions to 4*x^2 - 5*y^2 - 4*x + 5*y = 0. - Colin Barker, Jan 01 2015

Examples

			a(1) = 9 because c(1) = 5 + 81 + 2*sqrt(5*81) = 68 + 36*phi, which is indeed 4*(17 + 9*phi), with 17 = A007805(1).
		

Links

• Colin Barker, Table of n, a(n) for n = 0..797
• Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
• Eric Weisstein's World of Mathematics, Descartes' Circle Theorem.
• Wikipedia, Descartes' Theorem.
• Index entries for sequences related to Chebyshev polynomials.
• Index entries for linear recurrences with constant coefficients, signature (19,-19,1).

Crossrefs

Cf. A007805, A049310, A246638, A240926, A115032.

Programs

Magma

I:=[1, 9, 153]; [n le 3 select I[n] else 19*Self(n-1) - 19*Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017

Mathematica

LinearRecurrence[{19,-19,1},{1, 9, 153}, 30] (* or *) CoefficientList[ Series[(1 - 10*x + x^2)/((1-x)*(1 - 18*x + x^2)), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)

PARI

Vec((1-10*x+x^2)/((1-x)*(1-18*x+x^2)) + O(x^100)) \\ Colin Barker, Jan 01 2015

Formula

a(n) = (1 + S(n, 18) - S(n-1, 18))/2 = (1 + A007805(n))/2, n >= 0.

O.g.f.: (1 - 10*x + x^2)/((1-x)*(1 - 18*x + x^2)).

a(n) = 19*a(n-1) - 19*a(n-2) + a(n-3), n >= 1, with a(-2) = 9, a(-1) = 1 and a(0) = 1.

a(n) = (1/2+1/20*(9+4*sqrt(5))^(-n)*(5-2*sqrt(5)+(5+2*sqrt(5))*(9+4*sqrt(5))^(2*n))). - Colin Barker, Mar 04 2016