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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
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(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.
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.
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).
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
FORMULA
a(n) = (1 + S(n, 18) - S(n-1, 18))/2 = (1 + A007805(n))/2, n >= 0.
O.g.f.: (1 - 10*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) = 9, a(-1) = 1 and a(0) = 1.
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
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)/((1-x)*(1 - 18*x + x^2)), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)
PROG
(PARI) Vec((1-10*x+x^2)/((1-x)*(1-18*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(n-1) - 19*Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Sep 05 2014
STATUS
approved