OFFSET
0,2
COMMENTS
(11*a(n))^2 - 5*(5*b(n))^2 = -4 with b(n)=A097843(n) give all positive solutions of this Pell equation.
LINKS
Indranil Ghosh, Table of n, a(n) for n = 0..477
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 linear recurrences with constant coefficients, signature (123,-1).
FORMULA
a(n) = S(n, 123) + S(n-1, 123) = S(2*n, 5*sqrt(5)), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x)= 0 = U(-1, x). S(n, 123) = A049670(n+1).
a(n) = (-2/11)*i*((-1)^n)*T(2*n+1, 11*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1-123*x+x^2).
a(n) = 123*a(n-1) - a(n-2) for n > 1, a(0)=1, a(1)=124. - Philippe Deléham, Nov 18 2008
From Peter Bala, Mar 23 2015: (Start)
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, k != 2.
The aerated sequence (b(n))n>=1 = [1, 0, 124, 0, 15251, 0, 1875749, 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 = -121, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)
a(n) = Lucas(10*n + 5)/11. - Ehren Metcalfe, Jul 29 2017
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
CoefficientList[Series[(1+x)/(1-123x+x^2), {x, 0, 20}], x] (* Michael De Vlieger, Feb 08 2017 *)
LinearRecurrence[{123, -1}, {1, 124}, 20] (* G. C. Greubel, Jan 13 2019 *)
PROG
(PARI) a(n)=polchebyshev(n, 2, 123/2) + polchebyshev(n - 1, 2, 123/2); \\ Michel Marcus, Aug 04 2017
(PARI) my(x='x+O('x^20)); Vec((1+x)/(1-123*x+x^2)) \\ G. C. Greubel, Jan 13 2019
(Magma) m:=20; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1+x)/(1-123*x+x^2) )); // G. C. Greubel, Jan 13 2019
(Sage) ((1+x)/(1-123*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jan 13 2019
(GAP) a:=[1, 124];; for n in [3..20] do a[n]:=123*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 13 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Sep 10 2004
STATUS
approved