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

%I A249864 #26 May 23 2024 04:31:31
%S A249864 1,13,289,6877,164641,3943693,94468609,2262942877,54207552961,
%T A249864 1298512176013,31105146481249,745106711887837,17848622331502561,
%U A249864 427553951736562573,10241820250907001409,245337182888490470557
%N A249864 A special solution of X(n)^2 - 120*Y(n)^2 = 7^(2*n), n >= 0. The present sequence gives the X values.
%C A249864 The member Y(n) = A249863(n-1) with A249863(-1) = 0.
%C A249864 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.
%C A249864 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).
%C A249864 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.
%C A249864 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.
%C A249864 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).
%H A249864 G. C. Greubel, <a href="/A249864/b249864.txt">Table of n, a(n) for n = 0..720</a>
%H A249864 <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials</a>
%H A249864 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (26,-49)
%F A249864 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.
%F A249864 O.g.f.: (1 - 13*x)/(1 - 26*x + (7*x)^2).
%F A249864 a(n) = 26*a(n-1) - 49*a(n-2), a(0) = 1, a(1) = 13.
%F A249864 a(n) = (r^n + s^n)/2 where r,s are the roots of x^2 - 26*x + 49. - _Robert Israel_, Nov 18 2014
%p A249864 f:= gfun:-rectoproc({a(n)=26*a(n-1)-49*a(n-2),a(0)=1,a(1)=13},a(n),remember):
%p A249864 seq(f(n),n=0..50); # _Robert Israel_, Nov 18 2014
%t A249864 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 *)
%o A249864 (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
%o A249864 (PARI) x='x+O('x^30); Vec((1 - 13*x)/(1 - 26*x + (7*x)^2)) \\ _G. C. Greubel_, Dec 20 2017
%Y A249864 Cf. A249457, A249458, A249863.
%K A249864 nonn,easy
%O A249864 0,2
%A A249864 _Wolfdieter Lang_, Nov 09 2014