%I #121 Jun 06 2022 10:33:26
%S 1,9,161,2889,51841,930249,16692641,299537289,5374978561,96450076809,
%T 1730726404001,31056625195209,557288527109761,10000136862780489,
%U 179445175002939041,3220013013190122249,57780789062419261441
%N a(n) = 18*a(n-1) - a(n-2).
%C The primitive Heronian triangle 3*a(n) +- 2, 4*a(n) has the latter side cut into 2*a(n) +- 3 by the corresponding altitude and has area 10*a(n)*A060645(n). - _Lekraj Beedassy_, Jun 25 2002
%C Chebyshev polynomials T(n,x) evaluated at x=9.
%C {a(n)} gives all (unsigned, integer) solutions of Pell equation a(n)^2 - 80*b(n)^2 = +1 with b(n) = A049660(n), n >= 0.
%C {a(n)} gives all possible solutions for x in Pell equation x^2 - D*y^2 = 1 for D=5, D=20 and D=80. The corresponding values for y are A060645 (D=5), A207832 (D=20) and A049660 (D=80). - _Herbert Kociemba_, Jun 05 2022
%C Also gives solutions to the equation x^2 - 1 = floor(x*r*floor(x/r)) where r=sqrt(5). - _Benoit Cloitre_, Feb 14 2004
%C Appears to give all solutions > 1 to the equation: x^2 = ceiling(x*r*floor(x/r)) where r=sqrt(5). - _Benoit Cloitre_, Feb 24 2004
%C For all terms x of the sequence, 5*x^2 - 5 is a square, A004292(n)^2.
%C The a(n) are the x-values in the nonnegative integer solutions of x^2 - 5y^2 = 1, see A060645(n) for the corresponding y-values. - _Sture Sjöstedt_, Nov 29 2011
%C Rightmost digits alternate repeatedly: 1 and 9 in fact, a(2) = 18*9 - 1 == 1 (mod 10); a(3) = 18*1 - 9 == 9 (mod 10) therefore a(2n) == 1 (mod 10), a(2n+1) == 9 (mod 10). - _Carmine Suriano_, Oct 03 2013
%H Seiichi Manyama, <a href="/A023039/b023039.txt">Table of n, a(n) for n = 0..750</a> (terms 0..200 from Vincenzo Librandi)
%H Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL23/Nemeth/nemeth7.html">Ellipse Chains and Associated Sequences</a>, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
%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 (18,-1).
%F a(n) ~ (1/2)*(sqrt(5) + 2)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
%F Limit_{n->infinity} a(n)/a(n-1) = phi^6 = 9 + 4*sqrt(5). - _Gregory V. Richardson_, Oct 13 2002
%F a(n) = T(n, 9) = (S(n, 18) - S(n-2, 18))/2, with S(n, x) := U(n, x/2) and T(n, x), resp. U(n, x), are Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(-2, x) := -1, S(-1, x) := 0, S(n, 18)=A049660(n+1).
%F a(n) = sqrt(80*A049660(n)^2 + 1) (cf. Richardson comment).
%F a(n) = ((9 + 4*sqrt(5))^n + (9 - 4*sqrt(5))^n)/2.
%F G.f.: (1 - 9*x)/(1 - 18*x + x^2).
%F a(n) = cosh(2*n*arcsinh(2)). - _Herbert Kociemba_, Apr 24 2008
%F a(n) = A001077(2*n). - _Michael Somos_, Aug 11 2009
%F From _Johannes W. Meijer_, Jul 01 2010: (Start)
%F a(n) = 2*A167808(6*n+1) - A167808(6*n+3).
%F Limit_{k->infinity} a(n+k)/a(k) = a(n) + A060645(n)*sqrt(5).
%F Limit_{n->infinity} a(n)/A060645(n) = sqrt(5).
%F (End)
%F a(n) = (1/2)*A087215(n) = (1/2)*(sqrt(5) + 2)^(2*n) + (1/2)*(sqrt(5) - 2)^(2*n).
%F Sum_{n >= 1} 1/( a(n) - 5/a(n) ) = 1/8. Compare with A005248, A002878 and A075796. - _Peter Bala_, Nov 29 2013
%F a(n) = 2*A115032(n-1) - 1 = S(n, 18) - 9*S(n-1, 18), with A115032(-1) = 1, and see the above formula with S(n, 18) using its recurrence. - _Wolfdieter Lang_, Aug 22 2014
%F a(n) = A128052(3n). - _A.H.M. Smeets_, Oct 02 2017
%F a(n) = A049660(n+1) - 9*A049660(n). - _R. J. Mathar_, May 24 2018
%F a(n) = hypergeom([n, -n], [1/2], -4). - _Peter Luschny_, Jul 26 2020
%F a(n) = L(6*n)/2 for L(n) the Lucas sequence A000032(n). - _Greg Dresden_, Dec 07 2021
%F a(n) = cosh(6*n*arccsch(2)). - _Peter Luschny_, May 25 2022
%e G.f. = 1 + 9*x + 161*x^2 + 2889*x^3 + 51841*x4 + 930249*x^5 + 16692641*x^6 + ...
%p a := n -> hypergeom([n, -n], [1/2], -4):
%p seq(simplify(a(n)), n=0..16); # _Peter Luschny_, Jul 26 2020
%t LinearRecurrence[{18, -1}, {1, 9}, 50] (* _Sture Sjöstedt_, Nov 29 2011 *)
%t CoefficientList[Series[(1-9*x)/(1-18*x+x^2), {x, 0, 50}], x] (* _G. C. Greubel_, Dec 19 2017 *)
%o (PARI) {a(n) = fibonacci(6*n) / 2 + fibonacci(6*n - 1)}; /* _Michael Somos_, Aug 11 2009 */
%o (Magma) I:=[1, 9]; [n le 2 select I[n] else 18*Self(n-1)-Self(n-2): n in [1..20]]; // _Vincenzo Librandi_, Feb 13 2012
%o (PARI) x='x+O('x^30); Vec((1-9*x)/(1-18*x+x^2)) \\ _G. C. Greubel_, Dec 19 2017
%Y Cf. A001077, A115032.
%Y Row 2 of array A188645.
%Y Row 4 of A322790.
%K nonn,easy
%O 0,2
%A _David W. Wilson_
%E Chebyshev and Pell comments from _Wolfdieter Lang_, Nov 08 2002
%E _Sture Sjöstedt_'s comment corrected and reformulated by _Wolfdieter Lang_, Aug 24 2014