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A003499 a(n) = 6*a(n-1) - a(n-2), with a(0) = 2, a(1) = 6.
(Formerly M1701)
35
2, 6, 34, 198, 1154, 6726, 39202, 228486, 1331714, 7761798, 45239074, 263672646, 1536796802, 8957108166, 52205852194, 304278004998, 1773462177794, 10336495061766, 60245508192802, 351136554095046, 2046573816377474, 11928306344169798, 69523264248641314 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

COMMENTS

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

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

Chebyshev polynomials of the first kind evaluated at 3, then multiplied by 2. - Michael Somos, Apr 07 2003

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

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 (smbelcas(AT)toroidalsnark.net), Jul 04 2009

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

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

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

REFERENCES

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

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

Thomas Koshy, Fibonacci and Lucas Numbers with Applications, 2001, Wiley, p. 77-79.

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

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..300

Seyed Hassan Alavi, Ashraf Daneshkhah, Cheryl E Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See Lemma 7.9 p. 21.

P. Bala, Some simple continued fraction expansions for an infinite product, Part 1

P. Bhadouria, D. Jhala, B. Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92; sequence K_3.

S. Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234.

Refik Keskin and Olcay Karaatli, Some New Properties of Balancing Numbers and Square Triangular Numbers, Journal of Integer Sequences, Vol. 15 (2012), #12.1.4.

Tanya Khovanova, Recursive Sequences

W. Lu and F. Y. Wu, Close-packed dimers on nonorientable surfaces, arXiv:cond-mat/0110035 [cond-mat.stat-mech], 2001-2002; Physics Letters A, 293(2002), 235-246. [From Sarah-Marie Belcastro (smbelcas(AT)toroidalsnark.net), Jul 04 2009]

Kival Ngaokrajang, Illustration of initial terms

Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.

Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.

Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211.

Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]

Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)

Index entries for two-way infinite sequences

Index entries for linear recurrences with constant coefficients, signature (6,-1).

FORMULA

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

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

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

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

a(n+1) is the trace of n-th power of matrix {{6, -1}, {1, 0}}. - Artur Jasinski, Apr 22 2008

a(n) = Product_{r=1..n} (4*sin^2((4*r-1)*Pi/(4*n)) + 4). [Lu/Wu] - Sarah-Marie Belcastro (smbelcas(AT)toroidalsnark.net), Jul 04 2009

a(n) = (1 + sqrt(2))^(2*n) + (1 + sqrt(2))^(-2*n). - Gerson Washiski Barbosa, Sep 19 2010

For n > 0, a(n) = A001653(n) + A001653(n+1). - Charlie Marion, Dec 27 2011

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

From Peter Bala, Jan 06 2013: (Start)

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.

Also F(-alpha) = 0.83251 21926 93800 07634 ... 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(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 + ...))))))).

(End)

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

Inverse binomial transform of A228568 [Bhadouria]. - R. J. Mathar, Nov 10 2013

From Peter Bala, Oct 16 2019: (Start)

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

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

Series acceleration formulas for sum of reciprocals:

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

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

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

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.

(End)

E.g.f.: 2*exp(3*x)*cosh(2*sqrt(2)*x). - Stefano Spezia, Oct 18 2019

MAPLE

A003499:=-2*(-1+3*z)/(1-6*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation

MATHEMATICA

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 *)

Table[Tr[MatrixPower[{{6, -1}, {1, 0}}, n]], {n, 25}] (* Artur Jasinski, Apr 22 2008 *)

LinearRecurrence[{6, -1}, {2, 6}, 25] (* Vladimir Joseph Stephan Orlovsky, Feb 26 2012 *)

CoefficientList[Series[(2-6x)/(1-6x+x^2), {x, 0, 25}], x] (* Vincenzo Librandi, Jun 07 2013 *)

(* From Eric W. Weisstein, Apr 17 2018 *)

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

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

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

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

PROG

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

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

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

(Sage) [lucas_number2(n, 6, 1) for n in range(37)] # Zerinvary Lajos, Jun 25 2008

(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

(MAGMA) R<x>:=PowerSeriesRing(Integers(), 25); Coefficients(R!( (2-6*x)/(1 - 6*x + x^2) )); // Marius A. Burtea, Jan 16 2020

(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

CROSSREFS

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

Bisection of A002203.

First row of array A103999.

Row 1 * 2 of array A188645. A174501.

Sequence in context: A026951 A030233 A233396 * A279609 A253778 A018953

Adjacent sequences:  A003496 A003497 A003498 * A003500 A003501 A003502

KEYWORD

nonn,easy

AUTHOR

N. J. A. Sloane

STATUS

approved

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Last modified July 8 21:38 EDT 2020. Contains 335537 sequences. (Running on oeis4.)