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A003500 a(n) = 4*a(n-1) - a(n-2).
(Formerly M1278)
28
2, 4, 14, 52, 194, 724, 2702, 10084, 37634, 140452, 524174, 1956244, 7300802, 27246964, 101687054, 379501252, 1416317954, 5285770564, 19726764302, 73621286644, 274758382274, 1025412242452, 3826890587534, 14282150107684 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

COMMENTS

a(n) gives values of x satisfying x^2 - 3*y^2 = 4; corresponding y values are given by 2*A001353(n).

If M is any given term of the sequence, then the next one is 2M + sqrt(3M^2 - 12). - Lekraj Beedassy, Feb 18 2002

For n > 0, the three numbers a(n)-1, a(n), and a(n)+1 form a Fleenor-Heronian triangle, i.e., a Heronian triangle with consecutive sides, whose area A(n) may be obtained from the relation [4*A(n)]^2 = 3([a(2n)]^2 - 4); or A(n) = 3*A001353(2n)/2 and whose semiperimeter is 3*a[n]/2. The sequence is symmetrical about a[0], i.e., a[-n] = a[n].

For n > 0, a(n) + 2 is the number of dimer tilings of a 2n X 2 Klein bottle (cf. A103999).

Tsumura shows that, for prime p, a(p) is composite (contrary to a conjecture of Juricevic). - Charles R Greathouse IV, Apr 13 2010

Except for the first term, positive values of x (or y) satisfying x^2 - 4xy + y^2 + 12 = 0. - Colin Barker, Feb 04 2014

Except for the first term, positive values of x (or y) satisfying x^2 - 14xy + y^2 + 192 = 0. - Colin Barker, Feb 16 2014

REFERENCES

R. A. Beauregard and E. R. Suryanarayan, The Brahmagupta Triangles, The College Mathematics Journal 29(1) 13-7 1998 MAA.

B. C. Berndt, Ramanujan's Notebooks Part IV, Springer-Verlag, see p. 82.

Michael P. Cohen, Generating Heronian Triangles With Consecutive Integer Sides. Journal of Recreational Mathematics, vol. 30 no. 2 1999-2000 p. 123.

L. E. Dickson, History of The Theory of Numbers, Vol. 2 pp. 197;198;200;201. Chelsea NY.

Charles R. Fleenor, Heronian Triangles with Consecutive Integer Sides, Journal of Recreational Mathematics, Volume 28, no. 2 (1996-7) 113-115.

H. W. Gould, A triangle with integral sides and area, Fib. Quart., 11 (1973), 27-39.

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).

V. D. To, "Finding All Fleenor-Heronian Triangles", Journal of Recreational Mathematics vol. 32 no.4 2003-4 pp. 298-301 Baywood NY.

LINKS

T. D. Noe, Table of n, a(n) for n=0..200

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

K. S. Brown's Mathpages, Some Properties of the Lucas Sequence(2, 4, 14, 52, 194, ...)

Tanya Khovanova, Recursive Sequences

E. Keith Lloyd, The Standard Deviation of 1, 2,..., n: Pell's Equation and Rational Triangles, Math. Gaz. vol 81 (1997), 231-243.

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.

Yu Tsumura, On compositeness of special types of integers

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

Index entries for sequences related to linear recurrences with constant coefficients, signature (4,-1).

FORMULA

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

a(n) = trace of (n+1)st power of the 2 X 2 matrix [1 2 / 1 3]. - Gary W. Adamson, Jun 30 2003

From the addition formula, a(n+m) = a(n)*a(m) - a(m-n), it is easy to derive multiplication formulae, such as: a(2n) = (a(n))^2 - 2, a(3n) = (a(n))^3 - 3*(a(n)), a(4n) = (a(n))^4 - 4*(a(n))^2 + 2, a(5n) = (a(n))^5 - 5*(a(n))^3 + 5*(a(n)), a(6n) = (a(n))^6 - 6*(a(n))^4 + 9*(a(n))^2 - 2, etc. The absolute values of the coefficients in the expansions are given by the triangle A034807. - John Blythe Dobson, Nov 04 2007

G.f.: -2*(-1+2*x)/(1-4*x+x^2). a(n)=2*A001353(n+1)-4*A001353(n). - R. J. Mathar, Nov 16 2007

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 = 2 - sqrt(3). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.24561 99455 06551 88869 ... = 2 + 1/(4 + 1/(14 + 1/(52 + ...))). Cf. A174500.

Also F(-alpha) = 0.74544 81786 39692 68884 ... has the continued fraction representation 1 - 1/(4 - 1/(14 - 1/(52 - ...))) and the simple continued fraction expansion 1/(1 + 1/((4-2) + 1/(1 + 1/((14-2) + 1/(1 + 1/((52-2) + 1/(1 + ...))))))).

F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((4^2-4) + 1/(1 + 1/((14^2-4) + 1/(1 + 1/((52^2-4) + 1/(1 + ...))))))).

(End)

a(2^n) = A003010(n). - John Blythe Dobson, Mar 10 2014

MAPLE

A003500 := proc(n) option remember; if n <= 1 then 2*n+2 else 4*A003500(n-1)-A003500(n-2); fi; end;

A003500:=-2*(-1+2*z)/(1-4*z+z**2); # [Conjectured by Simon Plouffe in his 1992 dissertation.]

MATHEMATICA

a[0] = 2; a[1] = 4; a[n_] := a[n] = 4a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 23}]

LinearRecurrence[{4, -1}, {2, 4}, 30] (* Harvey P. Dale, Aug 20 2011 *)

PROG

(Sage) [lucas_number2(n, 4, 1) for n in xrange(0, 24)]# [Zerinvary Lajos, May 14 2009]

(Haskell)

a003500 n = a003500_list !! n

a003500_list = 2 : 4 : zipWith (-)

   (map (* 4) $ tail a003500_list) a003500_list

-- Reinhard Zumkeller, Dec 17 2011

CROSSREFS

Equals A001353(n+1) - A001353(n-1), also A001835(n) + A001835(n+1), also 2*A001075(n).

Cf. A001570, A006051, A048788, A002530, A011945. A174500.

Sequence in context: A092665 A046650 A055727 * A129876 A038055 A006385

Adjacent sequences:  A003497 A003498 A003499 * A003501 A003502 A003503

KEYWORD

nonn,easy,nice

AUTHOR

N. J. A. Sloane

EXTENSIONS

More terms from James A. Sellers, May 03 2000

Additional comments from Lekraj Beedassy, Feb 14 2002

STATUS

approved

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Last modified October 21 09:55 EDT 2014. Contains 248377 sequences.