%I #63 Dec 03 2023 10:43:38
%S 0,1,20,399,7960,158801,3168060,63202399,1260879920,25154396001,
%T 501827040100,10011386405999,199725901079880,3984506635191601,
%U 79490406802752140,1585823629419851199,31636982181594271840
%N Numbers k such that 99*k^2 + 1 is a square.
%C From _Wolfdieter Lang_, Nov 08 2002: (Start)
%C Chebyshev's polynomials U(n,x) evaluated at x=10.
%C The a(n) give all (unsigned, integer) solutions of Pell equation b(n)^2 - 99*a(n)^2 = +1 with b(n)= A001085(n). (End)
%C For n>=2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 20's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imagianry 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,...,19}. - _Milan Janjic_, Jan 25 2015
%D A. H. Beiler, "The Pellian", ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
%D L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
%D Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.
%H Vincenzo Librandi, <a href="/A075843/b075843.txt">Table of n, a(n) for n = 0..700</a>
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H J. J. O'Connor and E. F. Robertson, <a href="http://www-gap.dcs.st-and.ac.uk/~history/HistTopics/Pell.html">Pell's Equation</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PellEquation.html">Pell Equation.</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 (20,-1).
%F a(n) = ((10+3*sqrt(11))^n - (10-3*sqrt(11))^n) / (6*sqrt(11)).
%F a(n) = 20*a(n-1) - a(n-2), n>=1, a(0)=0, a(1)=1.
%F a(n) = S(n-1, 20), with S(n, x) := U(n, x/2), Chebyshev's polynomials of the second kind. S(-1, x) := 0. See A049310.
%F G.f.: x/(1 - 20*x + x^2).
%F a(n) = sqrt((A001085(n)^2 - 1)/99).
%F Lim_{n->inf.} a(n)/a(n-1) = 10 + 3*sqrt(11).
%F a(n+1) = Sum_{k=0..n} A101950(n,k)*19^k. - _Philippe Deléham_, Feb 10 2012
%F Product_{n>=1} (1 + 1/a(n)) = 1/3*(3 + sqrt(11)). - _Peter Bala_, Dec 23 2012
%F Product_{n>=2} (1 - 1/a(n)) = 3/20*(3 + sqrt(11)). - _Peter Bala_, Dec 23 2012
%p seq( simplify(ChebyshevU(n-1, 10)), n=0..20); # _G. C. Greubel_, Dec 22 2019
%t Table[GegenbauerC[n-1, 1, 10], {n,0,20}] (* _Vladimir Joseph Stephan Orlovsky_, Sep 11 2008 *)
%t CoefficientList[Series[x/(1-20x+x^2), {x,0,20}], x] (* _Vincenzo Librandi_, Dec 24 2012 *)
%t ChebyshevU[Range[22] -2, 10] (* _G. C. Greubel_, Dec 22 2019 *)
%t LinearRecurrence[{20,-1},{0,1},20] (* _Harvey P. Dale_, Dec 03 2023 *)
%o (Sage) [lucas_number1(n,20,1) for n in range(0,20)] # _Zerinvary Lajos_, Jun 25 2008
%o (Sage) [chebyshev_U(n-1,10) for n in (0..20)] # _G. C. Greubel_, Dec 22 2019
%o (Magma) I:=[0,1]; [n le 2 select I[n] else 20*Self(n-1)-Self(n-2): n in [1..20]]; // _Vincenzo Librandi_, Dec 24 2012
%o (PARI) vector( 22, n, polchebyshev(n-2, 2, 10) ) \\ _G. C. Greubel_, Dec 22 2019
%o (GAP) m:=10;; a:=[0,1];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # _G. C. Greubel_, Dec 22 2019
%Y Cf. A001084.
%Y Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), A001090 (m=4), A004189 (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), this sequence (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
%Y Cf. A323182.
%K nonn,easy
%O 0,3
%A _Gregory V. Richardson_, Oct 14 2002