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%I #32 May 23 2024 04:31:37
%S 1,17,569,19193,647441,21840257,736741769,24852657833,838359690401,
%T 28280555553137,953993651593049,32181259154185433,1085576868377967281,
%U 36619982192463218657,1235309202728347728809,41670933053031653811593,1405693940978521100034881
%N A special solution of X(n)^2 - 280*Y(n)^2 = 3^(2*n), n >= 0; here the X member.
%C The member Y(n) = A248163(n-1) with A248163(-1) = 0.
%C This pair of sequences (X(n), Y(n)) appears in the solution of the touching circles and chord problem proposed by _Kival Ngaokrajang_ in A249457. The curvatures (inverse radii) b(n) (for bend) of the circles in the lower section (on the left hand side) are here considered.
%C The derivation of the solution follows the lines given in the Wolfdieter Lang link in A240926, part I. Now the original radius of the large circle is R = 1 l.u. (length units) and the larger sagitta is 7/5 l.u. The circle radii are R(n), n >= 0, starting with R(0) = 7/10. Then a rescaling is done by a factor of 10/7 in the lengths: r = (10/7)*R = 10/7 l.u., such that the larger sagitta has 2 l.u. and r(n) = (10/7)*R(n).
%C The (nonlinear) recurrence for the curvature b(n) = 1/r(n) is written for bhat(n) := 3^n*b(n) and found to be: bhat(n) = 17*bhat(n-1) - 7*3^(n-1) + 140*sqrt((bhat(n-1) - 3^(n-1))*bhat(n-1)/(7*10)), n >= 1 with input bhat(0) = 1. This looks like A249457(n)/10. We search therefore for a positive integer solution which will then be the unique solution.
%C Define Y(n) := sqrt((bhat(n)-3^(n))*bhat(n)/(7*10)). This means (for positive bhat(n)) bhat(n) = (3^n + sqrt(9^n + 280*Y(n)^2))/2. Isolating the root and squaring yields X(n)^2 - 289*Y(n)^2 = 9^n, with X(n) := 2*bhat(n) - 3^n for n >= 0. For fixed n there are infinitely many solutions of this Diophantine equation. Here we only need a special solution for each n, which has to be an odd positive integer X(n) and X(0) = 1. Such a solution is X(n) = 3^(n-1)*(3*S(n, 34/3) - 17*S(n-1, 34/3)) and Y(n) = 3^(n-1)*S(n-1, 34/3) given in A249863(n-1), with Chebyshev's S-polynomial S(n, x). The proof is easy (once the o.g.f. for one of the sequences X or Y has been guessed by superseeker). Inserting the given formulas one has to prove S(n, 34/3)^2 + S(n-1, 34/3)^2 = 1 + (34/3)*S(n, 34/3)*S(n-1,34/3) which reduces after use of the recurrence relation for S to the well known Cassini-Simson identity S(n-1, x)^2 = 1 + S(n, x)*S(n-2, x), with S(-2) = -1, n >= 0, here with x = 34/3.
%C The solution for bhat(n) is then (X(n) + 3^n)/2. This satisfies indeed the original recurrence with input due to the recurrence relation of S(n, 34/3). Therefore, A249457(n)/10 = bhat(n).
%H G. C. Greubel, <a href="/A249862/b249862.txt">Table of n, a(n) for n = 0..650</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (34,-9).
%F a(n) = 3^(n-1)*(3*S(n, 34/3) - 17*S(n-1, 34/3)), n >= 0, with the scaled Chebyshev S sequence 3^n*S(n, 34/3) given in A248163.
%F O.g.f.: (1 - 17*x)/(1 - 34*x + 9*x^2).
%F a(n) = 34*a(n-1) - 9*a(n-2), a(0) = 0, a(1) = 17.
%F E.g.f.: exp(17*x)*cosh(2*sqrt(70)*x). - _Stefano Spezia_, Mar 24 2023
%t CoefficientList[Series[(1 - 17 x) / (1 - 34 x + 9 x^2), {x, 0, 40}], x] (* _Vincenzo Librandi_, Nov 08 2014 *)
%t LinearRecurrence[{34,-9},{1,17},30] (* _Harvey P. Dale_, Dec 13 2016 *)
%o (Magma) I:=[1,17]; [n le 2 select I[n] else 34*Self(n-1) - 9*Self(n-2): n in [1..20]]; // _Vincenzo Librandi_, Nov 08 2014
%o (PARI) x='x+O('x^30); Vec((1 - 17*x)/(1 - 34*x + 9*x^2)) \\ _G. C. Greubel_, Dec 20 2017
%Y Cf. A249457, A240926, A248163.
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, Nov 07 2014
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