%I M1701 #216 Apr 20 2023 13:18:54

%S 2,6,34,198,1154,6726,39202,228486,1331714,7761798,45239074,263672646,

%T 1536796802,8957108166,52205852194,304278004998,1773462177794,

%U 10336495061766,60245508192802,351136554095046,2046573816377474,11928306344169798,69523264248641314

%N a(n) = 6*a(n-1) - a(n-2), with a(0) = 2, a(1) = 6.

%C Two times Chebyshev polynomials of the first kind evaluated at 3.

%C Also 2(a(2*n)-2) and a(2*n+1)-2 are perfect squares. - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003

%C Chebyshev polynomials of the first kind evaluated at 3, then multiplied by 2. - _Michael Somos_, Apr 07 2003

%C Also gives solutions > 2 to the equation x^2 - 3 = floor(x*r*floor(x/r)) where r=sqrt(2). - _Benoit Cloitre_, Feb 14 2004

%C Output of Lu and Wu's formula for the number of perfect matchings of an m X n Klein bottle where m and n are both even specializes to this sequence for m=2. - _Sarah-Marie Belcastro_, Jul 04 2009

%C It appears that for prime P = 8*n +- 3, that a((P-1)/2) == -6 (mod P) and for all composites C = 8*n +- 3, there is at least one i < (C-1)/2 such that a(i) == -6 (mod P). Only a few of the primes P of the form 8*n +-3, e.g., 29, had such an i less than (P-1)/2. As for primes P = 8*n +- 1, it seems that the sum of the two adjacent terms, a((P-1)/2) and a((P+1)/2), is congruent to 8 (mod P). - _Kenneth J Ramsey_, Feb 14 2012 and Mar 05 2012

%C For n >= 1, a(n) is also the curvature of circles (rounded to the nearest integer) successively inscribed toward angle 90 degree of tangent lines, starting with a unit circle. The expansion factor is 5.828427... or 1/(3 - 2*sqrt(2)), which is also 3 + 2*sqrt(2) or A156035. See illustration in links. - _Kival Ngaokrajang_, Sep 04 2013

%C Except for the first term, positive values of x (or y) satisfying x^2 - 6*x*y + y^2 + 32 = 0. - _Colin Barker_, Feb 08 2014

%D A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 198.

%D Jay Kappraff, Beyond Measure, A Guided Tour Through Nature, Myth and Number, World Scientific, 2002; pp. 480-481.

%D Thomas Koshy, Fibonacci and Lucas Numbers with Applications, 2001, Wiley, pp. 77-79.

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%H Vincenzo Librandi, <a href="/A003499/b003499.txt">Table of n, a(n) for n = 0..300</a>

%H Seyed Hassan Alavi, Ashraf Daneshkhah, and Cheryl E. Praeger, <a href="https://arxiv.org/abs/2004.04535">Symmetries of biplanes</a>, arXiv:2004.04535 [math.GR], 2020. See Lemma 7.9 p. 21.

%H Peter Bala, <a href="/A174500/a174500_2.pdf">Some simple continued fraction expansions for an infinite product, Part 1</a>

%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 P. Bhadouria, D. Jhala, and B. Singh, <a href="http://dx.doi.org/10.22436/jmcs.08.01.07">Binomial Transforms of the k-Lucas Sequences and its Properties</a>, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92; sequence K_3.

%H S. Falcon, <a href="http://dx.doi.org/10.4236/am.2014.515216">Relationships between Some k-Fibonacci Sequences</a>, Applied Mathematics, 2014, 5, 2226-2234.

%H Refik Keskin and Olcay Karaatli, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL15/Karaatli/karaatli5.html">Some New Properties of Balancing Numbers and Square Triangular Numbers</a>, Journal of Integer Sequences, Vol. 15 (2012), #12.1.4.

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H W. Lu and F. Y. Wu, <a href="https://arxiv.org/abs/cond-mat/0110035">Close-packed dimers on nonorientable surfaces</a>, arXiv:cond-mat/0110035 [cond-mat.stat-mech], 2001-2002; Physics Letters A, 293(2002), 235-246. [From Sarah-Marie Belcastro, Jul 04 2009]

%H Kival Ngaokrajang, <a href="/A003499/a003499.pdf">Illustration of initial terms</a>

%H Simon Plouffe, <a href="https://arxiv.org/abs/0911.4975">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.

%H Simon Plouffe, <a href="/A000051/a000051_2.pdf">1031 Generating Functions</a>, Appendix to Thesis, Montreal, 1992

%H Jeffrey Shallit, <a href="http://www.jstor.org/stable/2690344">An interesting continued fraction</a>, Math. Mag., 48 (1975), 207-211.

%H Jeffrey Shallit, <a href="/A005248/a005248_1.pdf">An interesting continued fraction</a>, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]

%H Soumeya M. Tebtoub, Hacène Belbachir, and László Németh, <a href="https://hal.archives-ouvertes.fr/hal-02918958/document#page=18">Integer sequences and ellipse chains inside a hyperbola</a>, Proceedings of the 1st International Conference on Algebras, Graphs and Ordered Sets (ALGOS 2020), hal-02918958 [math.cs], 17-18.

%H <a href="/index/Rea#recur1">Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)</a>

%H <a href="/index/Tu#2wis">Index entries for two-way infinite sequences</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6,-1).

%F G.f.: (2-6*x)/(1 - 6*x + x^2).

%F a(n) = (3+2*sqrt(2))^n + (3-2*sqrt(2))^n = 2*A001541(n).

%F For all sequence elements n, 2*n^2 - 8 is a perfect square. Limit_{n->infinity} a(n)/a(n-1) = 3 + 2*sqrt(2). - _Gregory V. Richardson_, Oct 06 2002

%F a(2*n)+2 is a perfect square, 2(a(2*n+1)+2) is a perfect square. a(n), a(n-1) and A077445(n), n > 0, satisfy the Diophantine equation x^2 + y^2 - 3*z^2 = -8. - Mario Catalani (mario.catalani(AT)unito.it), Mar 24 2003

%F a(n+1) is the trace of n-th power of matrix {{6, -1}, {1, 0}}. - _Artur Jasinski_, Apr 22 2008

%F a(n) = Product_{r=1..n} (4*sin^2((4*r-1)*Pi/(4*n)) + 4). [Lu/Wu] - _Sarah-Marie Belcastro_, Jul 04 2009

%F a(n) = (1 + sqrt(2))^(2*n) + (1 + sqrt(2))^(-2*n). - _Gerson Washiski Barbosa_, Sep 19 2010

%F For n > 0, a(n) = A001653(n) + A001653(n+1). - _Charlie Marion_, Dec 27 2011

%F For n > 0, a(n) = b(4*n)/b(2*n) where b(n) is the Pell sequence, A000129. - _Kenneth J Ramsey_, Feb 14 2012

%F From _Peter Bala_, Jan 06 2013: (Start)

%F Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 3 - 2*sqrt(2). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.16585 37786 96882 80543 ... = 2 + 1/(6 + 1/(34 + 1/(198 + ...))). Cf. A174501.

%F Also F(-alpha) = 0.83251219269380007634 ... has the continued fraction representation 1 - 1/(6 - 1/(34 - 1/(198 - ...))) and the simple continued fraction expansion 1/(1 + 1/((6-2) + 1/(1 + 1/((34-2) + 1/(1 + 1/((198-2) + 1/(1 + ...))))))). Cf. A174501 and A003500.

%F F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((6^2-4) + 1/(1 + 1/((34^2-4) + 1/(1 + 1/((198^2-4) + 1/(1 + ...))))))).

%F (End)

%F G.f.: G(0), where G(k) = 1 + 1/(1 - x*(8*k-9)/( x*(8*k-1) - 3/G(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Aug 12 2013

%F Inverse binomial transform of A228568 [Bhadouria]. - _R. J. Mathar_, Nov 10 2013

%F From _Peter Bala_, Oct 16 2019: (Start)

%F 4*Sum_{n >= 1} 1/(a(n) - 8/a(n)) = 1.

%F 8*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 4/a(n)) = 1.

%F Series acceleration formulas for sum of reciprocals:

%F Sum_{n >= 1} 1/a(n) = 1/4 - 8*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 8)) and

%F Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/8 + 4*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 4)).

%F Sum_{n >= 1} 1/a(n) = ( (theta_3(3-2*sqrt(2)))^2 - 1 )/4 and

%F Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - (theta_3(2*sqrt(2)-3))^2 )/4, where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A067902.

%F (End)

%F E.g.f.: 2*exp(3*x)*cosh(2*sqrt(2)*x). - _Stefano Spezia_, Oct 18 2019

%F a(2*n)+2 = a(n)^2. - _Greg Dresden_ and Shraya Pal, Jun 29 2021

%p A003499:=-2*(-1+3*z)/(1-6*z+z**2); # conjectured by _Simon Plouffe_ in his 1992 dissertation

%t a[0]=2; a[1]=6; a[n_]:= 6a[n-1] -a[n-2]; Table[a[n], {n,0,25}] (* _Robert G. Wilson v_, Jan 30 2004 *)

%t Table[Tr[MatrixPower[{{6, -1}, {1, 0}}, n]], {n, 25}] (* _Artur Jasinski_, Apr 22 2008 *)

%t LinearRecurrence[{6, -1}, {2, 6}, 25] (* _Vladimir Joseph Stephan Orlovsky_, Feb 26 2012 *)

%t CoefficientList[Series[(2-6x)/(1-6x+x^2), {x,0,25}], x] (* _Vincenzo Librandi_, Jun 07 2013 *)

%t (* From _Eric W. Weisstein_, Apr 17 2018 *)

%t Table[(3-2Sqrt[2])^n + (3+2Sqrt[2])^n, {n,0,25}]//Expand

%t Table[(1+Sqrt[2])^(2n) + (1-Sqrt[2])^(2n), {n,0,25}]//FullSimplify

%t Join[{2}, Table[Fibonacci[4n, 2]/Fibonacci[2n, 2], {n, 25}]]

%t 2*ChebyshevT[Range[0, 25], 3] (* End *)

%o (PARI) a(n)=2*real((3+quadgen(32))^n)

%o (PARI) a(n)=2*subst(poltchebi(abs(n)),x,3)

%o (PARI) a(n)=if(n<0,a(-n),polsym(1-6*x+x^2,n)[n+1])

%o (Sage) [lucas_number2(n,6,1) for n in range(37)] # _Zerinvary Lajos_, Jun 25 2008

%o (Magma) I:=[2,6]; [n le 2 select I[n] else 6*Self(n-1) -Self(n-2): n in [1..25]]; // _G. C. Greubel_, Jan 16 2020

%o (Magma) R<x>:=PowerSeriesRing(Integers(), 25); Coefficients(R!( (2-6*x)/(1 - 6*x + x^2) )); // _Marius A. Burtea_, Jan 16 2020

%o (GAP) a:=[2,6];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]; od; a; # _G. C. Greubel_, Jan 16 2020

%Y A081555(n) = 1 + a(n).

%Y Bisection of A002203.

%Y First row of array A103999.

%Y Row 1 * 2 of array A188645. A174501.

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_