A097843 First differences of Chebyshev polynomials S(n,123) = A049670(n+1) with Diophantine property.

1, 122, 15005, 1845493, 226980634, 27916772489, 3433536035513, 422297015595610, 51939099382224517, 6388086926998019981, 785682752921374233146, 96632590522402032656977, 11885022951502528642575025, 1461761190444288621004071098, 179784741401695997854858170029

Offset

0,2

Comments

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

Links

Colin Barker, Table of n, a(n) for n = 0..478

Tanya Khovanova, Recursive Sequences

Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.

H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.

H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.

Index entries for sequences related to Chebyshev polynomials.

Index entries for linear recurrences with constant coefficients, signature (123,-1).

Formula

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.

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

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.

a(n) = 123*a(n-1) - a(n-2) for n > 1, a(0)=1, a(1)=122. - Philippe Deléham, Nov 18 2008

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

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

From Peter Bala, Mar 23 2015: (Start)

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

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.

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.

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)

Example

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), ...

Mathematica

LinearRecurrence[{123, -1}, {1, 122}, 20] (* G. C. Greubel, Jan 14 2019 *)

Prog

(PARI) Vec((1-x)/(1-123*x+x^2) + O(x^30)) \\ Colin Barker, Jun 15 2015
(Magma) m:=20; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-x)/(1-123*x+x^2) )); // G. C. Greubel, Jan 14 2019
(Sage) ((1-x)/(1-123*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jan 14 2019
(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

Crossrefs

Sequence in context: A131970 A233096 A121916 * A223385 A241375 A275094

Adjacent sequences: A097840 A097841 A097842 * A097844 A097845 A097846

Keyword

nonn,easy

Author

Wolfdieter Lang, Sep 10 2004

Status

approved