A246638 Sequence a(n) = 2 + 3*A001519(n+1) appearing in a certain four circle touching problem together with A246639.

5, 8, 17, 41, 104, 269, 701, 1832, 4793, 12545, 32840, 85973, 225077, 589256, 1542689, 4038809, 10573736, 27682397, 72473453, 189737960, 496740425, 1300483313, 3404709512, 8913645221, 23336226149, 61095033224, 159948873521, 418751587337, 1096305888488, 2870166078125, 7514192345885

Offset: 0

Author

Wolfdieter Lang, Aug 31 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) = A246639(n) appears as the curvature c(n) = a(n) + (4*b(n)/5)*phi (phi = (1+sqrt(5))/2, golden section) of the circle which touches i) a circle of radius 5/4 (in some length units) divided by a chord of length 2 into two unequal parts, and ii) the two touching circles in the smaller part which have curvatures A240926(n) and A240926(n+1), both also touching the circle with radius 5/4. See the illustration of Kival Ngaokrajang's link given in A240926, where the first circles in the smaller (upper) part are shown. The present circles will lie in the region between the large circle and two of these circles in the upper part.

Descartes' theorem on touching circles (see the links) is applied here as c(n) = -4/5 + A(n) + A(n+1) + 2*sqrt((-4/5 )*(A(n) + A(n+1)) + A(n)*A(n+1)), with A(n) = A240926(n), n >= 0.

For the proof for the first formula for a(n) given below use the formula for the curvature A240926(n) = 2 + 2*S(n, 3) - 3* S(n-1, 3) (see the W. Lang link in A240926, part II) in c(n) and compare with a(n) from c(n) = a(n) + (4*b(n)/5)*phi. This is done by using standard S-polynomial identities like the three term recurrence and the Cassini-Simson type identity

S(n, x)*S(n-2, x) = -1 + S(n-1, x)^2 (here for x=3). This implies S(n, 3)*S(n-1, 3) = (-1 + S(n, 3)^2 + S(n-1, 3)^2)/3. See also the mentioned link, part III a).

a(n) appears also in the curvature for the touching circles and chord problem in the smaller part of a circle with radius 5/4 dissected by a chord of length 2, together with A246640, where details are given.

Examples

			a(1) = 8 because c(n) = -4/5 + 5 + 9 + 2*sqrt((-4/5 )*(5 + 9) + 5*9) = 4*(2+(13/5)*phi). This is also 8 + (4*13/5)*phi with A246639(1) = 13.
		

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• 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 (4,-4,1).

Crossrefs

Cf. A246639, A049310, A001519, A115032.

Programs

Magma

I:=[5,8,17]; [n le 3 select I[n] else 4*Self(n-1) - 4*Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017

Mathematica

LinearRecurrence[{4, -4, 1}, {5, 8, 17}, 30] (* or *)  CoefficientList[ Series[(5-12*x+5*x^2)/((1-x)*(1-3*x+x^2)), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)

PARI

Vec((5-12*x+5*x^2)/((1-x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Nov 02 2016

Formula

a(n) = 2 + 3*(S(n,3) - S(n-1,3)) = 2 + 3*A001519(n+1), n >= 0, with Chebyshev's S-polynomials (see A049310).

O.g.f.: (5-12*x+5*x^2)/((1-x)*(1-3*x+x^2)).

a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3), n >=1, a(-2) = 8, a(-1) = 5, a(0) = 5.

a(n) = (2^(-1-n)*(5*2^(2+n)-3*(3-sqrt(5))^n*(-5+sqrt(5))+3*(3+sqrt(5))^n*(5+sqrt(5))))/5. - Colin Barker, Nov 02 2016