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%I #29 Sep 08 2022 08:46:09
%S 1,9,153,2737,49105,881145,15811497,283725793,5091252769,91358824041,
%T 1639367579961,29417257615249,527871269494513,9472265593285977,
%U 169972909409653065,3050040103780469185,54730748958638792257,982103441151717791433,17623131191772281453529,316234258010749348372081,5674593513001715989243921
%N Sequence a(n) = (1 + A007805(n))/2, appearing in a certain touching problem for three circles and a chord, together with A007805.
%C This sequence is motivated by Kival Ngaokrajang's touching circle problem considered in A240926 and A115032.
%C 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)).
%C 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.
%C 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.
%C 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).
%C Also the indices of centered pentagonal numbers which are also centered square numbers. - _Colin Barker_, Jan 01 2015
%C Also positive integers y in the solutions to 4*x^2 - 5*y^2 - 4*x + 5*y = 0. - _Colin Barker_, Jan 01 2015
%H Colin Barker, <a href="/A246641/b246641.txt">Table of n, a(n) for n = 0..797</a>
%H Giovanni Lucca, <a href="http://forumgeom.fau.edu/FG2016volume16/FG201654.pdf">Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences</a>, Forum Geometricorum, Volume 16 (2016) 419-427.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/DescartesCircleTheorem.html">Descartes' Circle Theorem</a>.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Descartes'_theorem">Descartes' Theorem</a>.
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (19,-19,1).
%F a(n) = (1 + S(n, 18) - S(n-1, 18))/2 = (1 + A007805(n))/2, n >= 0.
%F O.g.f.: (1 - 10*x + x^2)/((1-x)*(1 - 18*x + x^2)).
%F 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.
%F 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
%e 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).
%t 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 *)
%o (PARI) Vec((1-10*x+x^2)/((1-x)*(1-18*x+x^2)) + O(x^100)) \\ _Colin Barker_, Jan 01 2015
%o (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
%Y Cf. A007805, A049310, A246638, A240926, A115032.
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, Sep 05 2014
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