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 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
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Sep 10 2004
STATUS
approved