A246642 - OEIS

%I #37 Nov 10 2024 14:45:04

%S 1,41,761,13681,245521,4405721,79057481,1418628961,25456263841,

%T 456794120201,8196837899801,147086288076241,2639356347472561,

%U 47361327966429881,849864547048265321,15250200518902345921,273653744793193961281,4910517205758588957161,88115655958861407267641

%N Sequence appearing in the curvature of a certain four-circle touching problem: (-3 + 5*A007805(n))/2.

%C This sequence is motivated by _Kival Ngaokrajang_'s touching circle problem considered in A240926 and A115032.

%C a(n) appears in a 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.)

%C From Descartes's 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.

%C 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's 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).

%C 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

%H Colin Barker, <a href="/A246642/b246642.txt">Table of n, a(n) for n = 0..796</a>

%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&#39;_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) = (-3 + 5*(S(n, 18) - S(n-1, 18)))/2 = (-3 + 5* A007805(n))/2, n >= 0, with Chebyshev's S-polynomials (see A049310).

%F O.g.f.: (1+22*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)=41, a(-1)=1 and a(0)=1.

%F 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

%e 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).

%t  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 *)

%o (PARI) Vec((1+22*x+x^2)/((1-x)*(1-18*x+x^2)) + O(x^30)) \\ _Colin Barker_, Sep 08 2015

%o (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

%Y Cf. A007805, A049310, A115032, A240926.

%K nonn,easy

%O 0,2

%A _Wolfdieter Lang_, Sep 05 2014