

A003499


a(0) = 2, a(1) = 6; for n >= 2, a(n) = 6*a(n1)  a(n2).
(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^23 = 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.  SarahMarie Belcastro (smbelcas(AT)toroidalsnark.net), Jul 04 2009
It appears that for prime P = 8*n +/ 3, that a((P1)/2) equals 6 Mod P and for all composites C = 8*n +/ 3, there is at least one i < (C1)/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 (P1)/2. As for primes P = 8*n +/ 1, it seems that the sum of the two adjacent terms, a((P1)/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.
Jay Kappraff, Beyond Measure, A Guided Tour Through Nature, Myth and Number, World Scientific, 2002; p. 480481.
Thomas Koshy, Fibonacci and Lucas Numbers with Applications, 2001, Wiley, p. 7779.
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 kLucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 8192; sequence K_3.
S. Falcon, Relationships between Some kFibonacci Sequences, Applied Mathematics, 2014, 5, 22262234.
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, Closepacked dimers on nonorientable surfaces, arXiv:condmat/0110035 [condmat.statmech], 20012002; Physics Letters A, 293(2002), 235246. [From SarahMarie 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), 207211.
J. Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207211. [Annotated scanned copy]
Index entries for recurrences a(n) = k*a(n  1) +/ a(n  2)
Index entries for twoway infinite sequences
Index entries for linear recurrences with constant coefficients, signature (6,1).


FORMULA

G.f.: (26*x)/(16*x+x^2).
a(n) = (3+2*sqrt(2))^n+(32*sqrt(2))^n = 2*A001541(n).
For all sequence elements n, 2*n^2  8 is a perfect square. Lim a(n)/a(n1) = 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(n1) and A077445(n), n>0, satisfy the Diophantine equation x^2+y^23z^2=8.  Mario Catalani (mario.catalani(AT)unito.it), Mar 24 2003
a(n+1)=Trace of nth power of matrix {{6, 1}, {1, 0}}.  Artur Jasinski, Apr 22 2008
\prod_{r=1}^{n}(4\sin^2((4r1)\pi/(4n))+4) (Lu/Wu).  SarahMarie 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/((62) + 1/(1 + 1/((342) + 1/(1 + 1/((1982) + 1/(1 + ...))))))). Cf. A174501 and A003500.
F(alpha)*F(alpha) has the simple continued fraction expansion 1/(1 + 1/((6^24) + 1/(1 + 1/((34^24) + 1/(1 + 1/((198^24) + 1/(1 + ...))))))).
(End)
G.f.: G(0), where G(k)= 1 + 1/(1  x*(8*k9)/( x*(8*k1)  3/G(k+1) )); (continued fraction).  Sergei N. Gladkovskii, Aug 12 2013
Inverse Binomial transform of A228568 [Bhadoura].


MAPLE

A003499:=2*(1+3*z)/(16*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(16*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



