

A246641


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


2



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

0,2


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(n1), 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(n1, 18))/2, n >= 0, where Chebyshev's Spolynomials (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 Spolynomial identities like the three term recurrence for S(n+1, 18) and the CassiniSimson type identity (see a comment on A246638) which implies the formula S(n, 18)*S(n1, 18) = (1 + S(n, 18)^2 + S(n1, 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


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) 419427.
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).


FORMULA

a(n) = (1 + S(n, 18)  S(n1, 18))/2 = (1 + A007805(n))/2, n >= 0.
O.g.f.: (1  10*x + x^2)/((1x)*(1  18*x + x^2)).
a(n) = 19*a(n1)  19*a(n2) + a(n3), n >= 1, with a(2) = 9, a(1) = 1 and a(0) = 1.
a(n) = (1/2+1/20*(9+4*sqrt(5))^(n)*(52*sqrt(5)+(5+2*sqrt(5))*(9+4*sqrt(5))^(2*n))).  Colin Barker, Mar 04 2016


EXAMPLE

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).


MATHEMATICA

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


PROG

(PARI) Vec((110*x+x^2)/((1x)*(118*x+x^2)) + O(x^100)) \\ Colin Barker, Jan 01 2015
(MAGMA) I:=[1, 9, 153]; [n le 3 select I[n] else 19*Self(n1)  19*Self(n2) + Self(n3): n in [1..30]]; // G. C. Greubel, Dec 20 2017


CROSSREFS

Cf. A007805, A049310, A246638, A240926, A115032.
Sequence in context: A249642 A093849 A165232 * A169958 A012017 A130980
Adjacent sequences: A246638 A246639 A246640 * A246642 A246643 A246644


KEYWORD

nonn,easy


AUTHOR

Wolfdieter Lang, Sep 05 2014


STATUS

approved



