A249862 A special solution of X(n)^2 - 280*Y(n)^2 = 3^(2*n), n >= 0; here the X member.

1, 17, 569, 19193, 647441, 21840257, 736741769, 24852657833, 838359690401, 28280555553137, 953993651593049, 32181259154185433, 1085576868377967281, 36619982192463218657, 1235309202728347728809, 41670933053031653811593, 1405693940978521100034881

Offset: 0

Author

Wolfdieter Lang, Nov 07 2014

Keywords

• nonn
• easy

Comments

The member Y(n) = A248163(n-1) with A248163(-1) = 0.

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.

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).

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.

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.

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).

Links

• G. C. Greubel, Table of n, a(n) for n = 0..650
• Index entries for linear recurrences with constant coefficients, signature (34,-9).

Crossrefs

Cf. A249457, A240926, A248163.

Programs

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

Mathematica

CoefficientList[Series[(1 - 17 x) / (1 - 34 x + 9 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 08 2014 *)
LinearRecurrence[{34,-9},{1,17},30] (* Harvey P. Dale, Dec 13 2016 *)

PARI

x='x+O('x^30); Vec((1 - 17*x)/(1 - 34*x + 9*x^2)) \\ G. C. Greubel, Dec 20 2017

Formula

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.

O.g.f.: (1 - 17*x)/(1 - 34*x + 9*x^2).

a(n) = 34*a(n-1) - 9*a(n-2), a(0) = 0, a(1) = 17.

E.g.f.: exp(17*x)*cosh(2*sqrt(70)*x). - Stefano Spezia, Mar 24 2023