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 A097843 First differences of Chebyshev polynomials S(n,123) = A049670(n+1) with Diophantine property. 7
 1, 122, 15005, 1845493, 226980634, 27916772489, 3433536035513, 422297015595610, 51939099382224517, 6388086926998019981, 785682752921374233146, 96632590522402032656977, 11885022951502528642575025, 1461761190444288621004071098, 179784741401695997854858170029 (list; graph; refs; listen; history; text; internal format)
 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) = (1/2)*((123/2 - (55/2)*sqrt(5))^n + (123/2 + (55/2)*sqrt(5))^n) + (11/50)*sqrt(5)*((123/2 + (55/2)*sqrt(5))^n - (123/2 - (55/2)*sqrt(5))^n), with n >= 0. - Paolo P. Lava, Dec 12 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:=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

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Last modified July 8 21:52 EDT 2020. Contains 335537 sequences. (Running on oeis4.)