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%I #46 Sep 01 2021 13:12:34
%S 0,1,21,440,9219,193159,4047120,84796361,1776676461,37225409320,
%T 779956919259,16341869895119,342399310878240,7174043658547921,
%U 150312517518628101,3149388824232642200,65986852791366858099
%N Chebyshev polynomials S(n-1,21) with Diophantine property.
%C Sequence R_21: Starts with 0,1,21 and satisfies A*C=B^2-1 for successive A,B,C.
%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).
%C All positive integer solutions of Pell equation b(n)^2 - 437*a(n)^2 = +4 together with b(n)=A097777(n), n>=0.
%C 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
%C 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
%H Indranil Ghosh, <a href="/A092499/b092499.txt">Table of n, a(n) for n = 0..756</a>
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (21,-1).
%F a(0)=0, a(1)=1, a(2)=21 and a(n-1)*a(n+1) = a(n)^2-1
%F 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).
%F a(n) = S(2*n-1, sqrt(23))/sqrt(23), n>=1.
%F a(n) = 21*a(n-1)-a(n-2), n >= 1; a(0)=0, a(1)=1.
%F a(n) = (ap^n-am^n)/(ap-am) with ap := (21+sqrt(437))/2 and am := (21-sqrt(437))/2.
%F G.f.: x/(1-21*x+x^2).
%F a(n+1) = Sum_{k, 0<=k<=n} A101950(n,k)*20^k. - _Philippe Deléham_, Feb 10 2012
%F Product {n >= 1} (1 + 1/a(n)) = 1/19*(19 + sqrt(437)). - _Peter Bala_, Dec 23 2012
%F Product {n >= 2} (1 - 1/a(n)) = 1/42*(19 + sqrt(437)). - _Peter Bala_, Dec 23 2012
%e a(3)=440 because a(1)*440 = a(2)^2-1.
%t LinearRecurrence[{21,-1},{0,1},30] (* _Harvey P. Dale_, Apr 23 2015 *)
%o (Sage) [lucas_number1(n,21,1) for n in range(0,20)] # _Zerinvary Lajos_, Jun 25 2008
%Y 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.
%K easy,nonn
%O 0,3
%A _Rainer Rosenthal_, Apr 05 2004
%E Extension, Chebyshev and Pell comments from _Wolfdieter Lang_, Aug 31 2004
%E Corrected by _T. D. Noe_, Nov 07 2006
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