%I #48 Dec 31 2023 11:28:12

%S 1,122,15005,1845493,226980634,27916772489,3433536035513,

%T 422297015595610,51939099382224517,6388086926998019981,

%U 785682752921374233146,96632590522402032656977,11885022951502528642575025,1461761190444288621004071098,179784741401695997854858170029

%N First differences of Chebyshev polynomials S(n,123) = A049670(n+1) with Diophantine property.

%C (11*b(n))^2 - 5*(5*a(n))^2 = -4 with b(n)=A097842(n) give all positive solutions of this Pell equation.

%H Colin Barker, <a href="/A097843/b097843.txt">Table of n, a(n) for n = 0..478</a>

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H Giovanni Lucca, <a href="http://forumgeom.fau.edu/FG2019volume19/FG201902index.html">Integer Sequences and Circle Chains Inside a Hyperbola</a>, Forum Geometricorum (2019) Vol. 19, 11-16.

%H H. C. Williams and R. K. Guy, <a href="http://dx.doi.org/10.1142/S1793042111004587">Some fourth-order linear divisibility sequences</a>, Intl. J. Number Theory 7 (5) (2011) 1255-1277.

%H H. C. Williams and R. K. Guy, <a href="http://www.emis.de/journals/INTEGERS/papers/a17self/a17self.Abstract.html ">Some Monoapparitic Fourth Order Linear Divisibility Sequences</a> Integers, Volume 12A (2012) The John Selfridge Memorial Volume.

%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (123,-1).

%F a(n) = ((-1)^n)*S(2*n, 11*i) with the imaginary unit i and the S(n, x) = U(n, x/2) Chebyshev polynomials.

%F G.f.: (1-x)/(1-123*x+x^2).

%F a(n) = S(n, 123) - S(n-1, 123) = T(2*n+1, 5*sqrt(5)/2)/(5*sqrt(5)/2), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x) and T(n, x) Chebyshev's polynomials of the first kind, A053120.

%F a(n) = 123*a(n-1) - a(n-2) for n > 1, a(0)=1, a(1)=122. - _Philippe Deléham_, Nov 18 2008

%F a(n) = (F(10*(n+1)) - F(10*n))/F(10), with F=A000045 (Fibonacci). F(10*n)/F(10) = A049670. - _Wolfdieter Lang_, Oct 11 2012

%F a(n) = (1/5)*F(10*n + 5). Sum_{n >= 1} 1/(a(n) - 1/a(n)) = 1/11^2. Compare with A001519 and A007805. - _Peter Bala_, Nov 29 2013

%F From _Peter Bala_, Mar 23 2015: (Start)

%F a(n) = A049666(2*n + 1).

%F a(n) = ( Fibonacci(10*n + 10 - 2*k) - Fibonacci(10*n + 2*k) )/( Fibonacci(10 - 2*k) - Fibonacci(2*k) ), for k an arbitrary integer.

%F a(n) = ( Fibonacci(10*n + 10 - 2*k - 1) + Fibonacci(10*n + 2*k + 1) )/( Fibonacci(10 - 2*k - 1) + Fibonacci(2*k + 1) ), for k an arbitrary integer.

%F The aerated sequence (b(n))n>=1 = [1, 0, 122, 0, 15005, 0, 1845493, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -125, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)

%e All positive solutions of Pell equation x^2 - 125*y^2 = -4 are (11 = 11*1,1), (1364 = 11*124,122), (167761 = 11*15251,15005), (20633239 = 11*1875749,1845493), ...

%t LinearRecurrence[{123,-1}, {1,122}, 20] (* _G. C. Greubel_, Jan 14 2019 *)

%o (PARI) Vec((1-x)/(1-123*x+x^2) + O(x^30)) \\ _Colin Barker_, Jun 15 2015

%o (Magma) m:=20; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-x)/(1-123*x+x^2) )); // _G. C. Greubel_, Jan 14 2019

%o (Sage) ((1-x)/(1-123*x+x^2)).series(x, 20).coefficients(x, sparse=False) # _G. C. Greubel_, Jan 14 2019

%o (GAP) a:=[1,122];; for n in [3..20] do a[n]:=123*a[n-1]-a[n-2]; od; a; # _G. C. Greubel_, Jan 14 2019

%K nonn,easy

%O 0,2

%A _Wolfdieter Lang_, Sep 10 2004