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A097842
Chebyshev polynomials S(n,123) + S(n-1,123) with Diophantine property.
5
1, 124, 15251, 1875749, 230701876, 28374454999, 3489827263001, 429220378894124, 52790616776714251, 6492816643156958749, 798563656491529211876, 98216836931814936101999, 12079872378956745611334001, 1485726085774747895257980124, 182732228677915034371120221251, 22474578401297774479752529233749
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
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.
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
Sequence in context: A289299 A035816 A206077 * A206189 A299828 A280905
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Sep 10 2004
STATUS
approved