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A003500
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a(n) = 4*a(n-1) - a(n-2).
(Formerly M1278)
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24
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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)
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OFFSET
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0,1
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COMMENTS
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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, a(n)-1, a(n), 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 = 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).
The terms whose index is a power of 2 form A003010. - John Blythe Dobson (j.dobson(AT)uwinnipeg.ca), Oct 28 2007
Tsumura shows that, for prime p, a(p) is composite (contrary to a conjecture of Juricevic). [From Charles R Greathouse IV, Apr 13 2010]
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REFERENCES
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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.
E. K. Lloyd, "The standard deviation of 1, 2, .., n, Pell's equation and rational triangles", preprint.
_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.
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LINKS
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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
_Simon Plouffe_, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
_Simon Plouffe_, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 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).
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FORMULA
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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 (j.dobson(AT)uwinnipeg.ca), 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)
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MAPLE
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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.]
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MATHEMATICA
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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] (* From Harvey P. Dale, Aug 20 2011 *)
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PROG
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(Sage) [lucas_number2(n, 4, 1) for n in xrange(0, 24)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), 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
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CROSSREFS
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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
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KEYWORD
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nonn,easy,nice
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AUTHOR
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N. J. A. Sloane.
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EXTENSIONS
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More terms from James A. Sellers, May 03 2000
Additional comments from Lekraj Beedassy, Feb 14 2002
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STATUS
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approved
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