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 A092499 Chebyshev polynomials S(n-1,21) with Diophantine property. 3
 0, 1, 21, 440, 9219, 193159, 4047120, 84796361, 1776676461, 37225409320, 779956919259, 16341869895119, 342399310878240, 7174043658547921, 150312517518628101, 3149388824232642200, 65986852791366858099 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS Sequence R_21: Starts with 0,1,21 and satisfies A*C=B^2-1 for successive A,B,C. The natural numbers a(n)=n satisfy the recurrence a(n-1)*a(n+1)=a(n)^2-1. Let R_r denote the sequence starting with 0,1,r and with this recurrence. We see that R_2 = "the natural numbers" and we find R_3 = A001906. These R_r form a "family" of sequences, which coincides with the m-family (r=m-2, n -> n+1) provided by Wolfdieter Lang (see A078368). This sequence R_21 is strongly related to A041833, which gives the denominators in the continued fraction of sqrt(437). All positive integer solutions of Pell equation b(n)^2 - 437*a(n)^2 = +4 together with b(n)=A097777(n), n>=0. For n>=2, a(n) equals the permanent of the (n-1)X(n-1) tridiagonal matrix with 21's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011 For n>=1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,20}. Milan Janjic, Jan 25 2015 LINKS Indranil Ghosh, Table of n, a(n) for n = 0..756 Tanya Khovanova, Recursive Sequences Index entries for linear recurrences with constant coefficients, signature (21,-1). FORMULA a(0)=0, a(1)=1, a(2)=21 and a(n-1)*a(n+1)=a(n)^2-1 a(n)=S(n-1, 21)=U(n-1, 21/2) with S(n, x)=U(n, x/2) Chebyshev's polynomials of the 2nd kind, A049310. S(-1, x)= 0 = U(-1, x). a(n)=S(2*n-1, sqrt(23))/sqrt(23), n>=1. a(n)=21*a(n-1)-a(n-2), n >= 1; a(0)=0, a(1)=1. a(n)=(ap^n-am^n)/(ap-am) with ap := (21+sqrt(437))/2 and am := (21-sqrt(437))/2. G.f.: x/(1-21*x+x^2). a(n+1) = Sum_{k, 0<=k<=n} A101950(n,k)*20^k. - Philippe Deléham, Feb 10 2012 Product {n >= 1} (1 + 1/a(n)) = 1/19*(19 + sqrt(437)). - Peter Bala, Dec 23 2012 Product {n >= 2} (1 - 1/a(n)) = 1/42*(19 + sqrt(437)). - Peter Bala, Dec 23 2012 EXAMPLE a(3)=440 because a(1)*440=a(2)^2-1 MATHEMATICA LinearRecurrence[{21, -1}, {0, 1}, 30] (* Harvey P. Dale, Apr 23 2015 *) PROG (Sage) [lucas_number1(n, 21, 1) for n in xrange(0, 20)] # Zerinvary Lajos, Jun 25 2008 CROSSREFS Cf. R_3=A001906, R_4=A001353, R_5=A004254, R_6=A001109, R_7=A004187, R_8=A001090, R_9=A018913, R_10=A004189, R_11=A004190, R_12=A004191, R_13=A078362, R_14=A007655, R_15=A078364, R_16=A077412, R_17=A078366, R_18=A049660, R_19=A078368, R_20=A075843, R_21=this, sequence, R_22=A077421. See also A041219 and A041917. Sequence in context: A297692 A218840 A171326 * A207776 A207273 A207766 Adjacent sequences:  A092496 A092497 A092498 * A092500 A092501 A092502 KEYWORD easy,nonn AUTHOR Rainer Rosenthal, Apr 05 2004 EXTENSIONS Extension, Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004 Corrected by T. D. Noe, Nov 07 2006 STATUS approved

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Last modified October 16 23:12 EDT 2018. Contains 316275 sequences. (Running on oeis4.)