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A003499 a(0) = 2, a(1) = 6; for n >= 2, a(n) = 6*a(n-1) - a(n-2).
(Formerly M1701)
28
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(2n)-2) and a(2n+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) equals -6 Mod P and for all composites C = 8*n +/- 3, there is at least one i < (C-1)/2 such that a(i) is equal to - 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 equal 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 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 - 6xy + y^2 + 32 = 0. - Colin Barker, Feb 08 2014

REFERENCES

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

S. Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234; http://www.scirp.org/journal/am; http://dx.doi.org/10.4236/am.2014.515216

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

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

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

W. Lu and F. Y. Wu, Close-packed dimers on nonorientable surfaces, Physics Letters A, 293(2002), 235-246. [From Sarah-Marie Belcastro (smbelcas(AT)toroidalsnark.net), Jul 04 2009]

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

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

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.

Tanya Khovanova, Recursive Sequences

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.

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 a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 06 2002

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

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

\prod_{r=1}^{n}(4\sin^2((4r-1)\pi/(4n))+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(4n)/b(2n) 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..inf} (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 [Bhadoura].

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, 19}] (* Robert G. Wilson v, Jan 30 2004 *)

Table[Tr[MatrixPower[{{6, -1}, {1, 0}}, n]], {n, 1, 100}] (* Artur Jasinski, Apr 22 2008 *)

LinearRecurrence[{6, -1}, {2, 6}, 70] (* Vladimir Joseph Stephan Orlovsky, Feb 26 2012 *)

CoefficientList[Series[(2 - 6 x) / (1 - 6 x + x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Jun 07 2013 *)

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

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 January 21 14:24 EST 2017. Contains 281109 sequences.