A249864 A special solution of X(n)^2 - 120*Y(n)^2 = 7^(2*n), n >= 0. The present sequence gives the X values.

1, 13, 289, 6877, 164641, 3943693, 94468609, 2262942877, 54207552961, 1298512176013, 31105146481249, 745106711887837, 17848622331502561, 427553951736562573, 10241820250907001409, 245337182888490470557

Offset: 0

Author

Wolfdieter Lang, Nov 09 2014

Keywords

• nonn
• easy

Comments

The member Y(n) = A249863(n-1) with A249863(-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 A249458. The curvatures (inverse radii) b(n) (for bend) of the circles in the smaller 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 II. Now the original radius of the large circle is R = 1 l.u. (length units) and the larger sagitta h is 7/5 l.u. The circle radii are R(n), n >= 0, starting with R(0) = 3/10. Then a rescaling is done by 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*7^(n-1)*b(n) and found to be: bhat(n) = (1/7)*(91*bhat(n-1) - 3*7^n + 2*sqrt(210)*sqrt((7*bhat(n-1) - 7^n)*bhat(n-1))), n >= 1, with input bhat(0) = 1. This looks like A249458(n)/10. We search therefore for a positive integer solution which will then be the unique solution.

Define Y(n) := sqrt((bhat(n)-7^(n))*bhat(n)/(30)). This means (for positive bhat(n)) bhat(n) = (7^n + sqrt(7^(2*n) + 120*Y(n)^2))/2. Isolating the root and squaring yields X(n)^2 - 120*Y(n)^2 = 7^(2*n), with X(n) := 2*bhat(n) - 7^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. This solution is X(n) = 7^(n-1)*(7*S(n, 26/7) - 13*S(n-1, 26/7)) and Y(n) = 7^(n-1)*S(n-1, 26/7) 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, e.g., by superseeker). Inserting the given formulas one has to prove S(n, 26/7)^2 + S(n-1, 26/7)^2 = 1 + (26/7)*S(n, 26/7)*S(n-1, 26/7) 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 = 26/7.

The solution for bhat(n) is then (X(n) + 7^n)/2. This satisfies indeed the original recurrence with input due to the recurrence relation of S(n, 26/7). Therefore, A249458(n)/10 = bhat(n).

Links

• G. C. Greubel, Table of n, a(n) for n = 0..720
• Index entries for sequences related to Chebyshev polynomials
• Index entries for linear recurrences with constant coefficients, signature (26,-49)

Crossrefs

Cf. A249457, A249458, A249863.

Programs

Magma

I:=[1,13]; [n le 2 select I[n] else 26*Self(n-1)-49*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Nov 09 2014

Maple

f:= gfun:-rectoproc({a(n)=26*a(n-1)-49*a(n-2),a(0)=1,a(1)=13},a(n),remember):
seq(f(n),n=0..50); # Robert Israel, Nov 18 2014

Mathematica

LinearRecurrence[{26, -49}, {1, 13} , 30] (* or *) CoefficientList[ Series[(1 - 13*x)/(1 - 26*x + (7*x)^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)

PARI

x='x+O('x^30); Vec((1 - 13*x)/(1 - 26*x + (7*x)^2)) \\ G. C. Greubel, Dec 20 2017

Formula

a(n) = (7^n)*(S(n, 26/7) - (13/7)*S(n-1, 26/7)), n >= 0, with the scaled Chebyshev S sequence 7^n*S(n, 26/7) given in A249863.

O.g.f.: (1 - 13*x)/(1 - 26*x + (7*x)^2).

a(n) = 26*a(n-1) - 49*a(n-2), a(0) = 1, a(1) = 13.

a(n) = (r^n + s^n)/2 where r,s are the roots of x^2 - 26*x + 49. - Robert Israel, Nov 18 2014