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 A246642 Sequence appearing in the curvature of a certain four circle touching problem:  (-3 + 5*A007805(n))/2. 4
 1, 41, 761, 13681, 245521, 4405721, 79057481, 1418628961, 25456263841, 456794120201, 8196837899801, 147086288076241, 2639356347472561, 47361327966429881, 849864547048265321, 15250200518902345921, 273653744793193961281, 4910517205758588957161, 88115655958861407267641 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS This sequence is motivated by Kival Ngaokrajang's touching circle problem considered in A240926 and A115032. a(n) appears in a circle curvature c(n) = (4/5)*(2*(a(n) + 2) + a(n)*phi), with phi = (1+sqrt(5))/2, the golden section. c(n) is the curvature of the circle which touches i) the larger part of a circle of radius 5/4 (in some length units), obtained from the bisection of the circle with a chord of length 2 and ii) two touching circles in the larger part of this bisected disk of radius 5/4 having curvatures c1(n) and c1(n+1) with c1(n) = A115032(n-1) and c1(0) = 1, n >= 0.   See the illustration of Kival Ngaokrajang's link given in A115032, where the first circles in the larger (lower) part are shown. From Descartes' theorem on touching circles (see the links) one has here: c(n) = -4/5 + c1(n) + c1(n+1) + 2*sqrt((-4/5)*( c1(n) + c1(n+1)) + 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. For the proof for the first a(n) formula given below use the curvature c1(n) = (1 + S(n, 18) - 9*S(n-1, 18))/2  (see the W. Lang link in A240926, part I) in c(n) from Descartes' formula and compare it with a(n) in c(n) = (4/5)*(2*(a(n) + 2) + 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 mentioned W. Lang link part IV b). Also the first of four consecutive positive integers the sum of the squares of which is equal to the sum of the squares of five consecutive positive integers. For example 41^2 + ... + 44^2 = 7230 = 36^2 + ... + 40^2. - Colin Barker, Sep 08 2015 LINKS Colin Barker, Table of n, a(n) for n = 0..796 Eric Weisstein's World of Mathematics, Descartes' Circle Theorem. Wikipedia, Descartes' Theorem. Index entries for linear recurrences with constant coefficients, signature (19, -19, 1). FORMULA a(n) = (-3 + 5*(S(n, 18) - S(n-1, 18)))/2 = (-3 + 5* A007805(n))/2, n >= 0, with Chebyshev's S-polynomials (see A049310). O.g.f.: (1+22*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)=41, a(-1)=1 and a(0)=1. a(n) = (-6+(5-2*sqrt(5))*(9+4*sqrt(5))^(-n)+(5+2*sqrt(5))*(9+4*sqrt(5))^n)/4. - Colin Barker, Mar 03 2016 EXAMPLE a(1) = 41 because the two curvatures of the circles in the larger part are c1(1) = 5 and c1(2) = 81 (from A115032), and c(1) = -4/5 + 5 + 81 + 2*sqrt((-4/5)*( 5 + 81) + 5*81) = (4/5)*(213 + 41*sqrt(5))/2 = (4/5)*(86 + 41*phi)) (by Descartes). This is indeed (4/5)*(2*(a(1) + 2) + a(1)*phi). MATHEMATICA CoefficientList[Series[(1+22*x+x^2)/((1-x)*(1-18*x+x^2)), {x, 0, 50}], x] (* or *) LinearRecurrence[{19, -19, 1}, {1, 41, 761}, 30] (* G. C. Greubel, Dec 20 2017 *) PROG (PARI) Vec((1+22*x+x^2)/((1-x)*(1-18*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 08 2015 (MAGMA) I:=[1, 41, 761]; [n le 5 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 CROSSREFS Cf. A007805, A049310, A115032, A240926. Sequence in context: A200914 A221790 A299552 * A167737 A268993 A125551 Adjacent sequences:  A246639 A246640 A246641 * A246643 A246644 A246645 KEYWORD nonn,easy AUTHOR Wolfdieter Lang, Sep 05 2014 STATUS approved

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Last modified June 19 00:37 EDT 2021. Contains 345125 sequences. (Running on oeis4.)