Search: id:A003500
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%I A003500 M1278
%S A003500 2,4,14,52,194,724,2702,10084,37634,140452,524174,1956244,7300802,
%T A003500 27246964,101687054,379501252,1416317954,5285770564,19726764302,
%U A003500 73621286644,274758382274,1025412242452,3826890587534,14282150107684
%N A003500 a(n) = 4a(n-1) - a(n-2).
%C A003500 a(n) gives values of x satisfying x^2 - 3*y^2 = 4; corresponding y values 
               are given by 2*A001353(n).
%C A003500 If M is any given term of the sequence, then the next one is 2M + sqrt(3M^2 
               - 12). - Lekraj Beedassy (blekraj(AT)yahoo.com), Feb 18 2002
%C A003500 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 semipermeter = 3*a[n]/2. The sequence is symmetrical 
               about a[0], i.e.; a[ -n] = a[n].
%C A003500 For n>0, a(n)+2 is the number of dimer tilings of a 2n x 2 Klein bottle 
               (cf. A103999).
%C A003500 The terms whose index is a power of 2 form A003010. - John Blythe Dobson 
               (j.dobson(AT)uwinnipeg.ca), Oct 28 2007
%D A003500 R. A. Beauregard and E. R. Suryanarayan, The Brahmagupta Triangles, The 
               College Mathematics Journal 29(1) 13-7 1998 MAA.
%D A003500 B. C. Berndt, Ramanujan's Notebooks Part IV, Springer-Verlag, see p. 
               82.
%D A003500 Michael P. Cohen, Generating Heronian Triangles With Consecutive Integer 
               Sides. Journal of Recreational Mathematics, vol. 30 no. 2 1999-2000 
               p. 123.
%D A003500 L. E. Dickson, History of The Theory of Numbers, Vol. 2 pp. 197;198;200;
               201. Chelsea NY.
%D A003500 Charles R. Fleenor, Heronian Triangles with Consecutive Integer Sides, 
               Journal of Recreational Mathematics, Volume 28, no. 2 (1996-7) 113-115.
%D A003500 H. W. Gould, A triangle with integral sides and area, Fib. Quart., 11 
               (1973), 27-39.
%D A003500 E. K. Lloyd, "The standard deviation of 1, 2, .., n, Pell's equation 
               and rational triangles", preprint.
%D A003500 J. O. Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 
               207-211.
%D A003500 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, 
               Academic Press, 1995 (includes this sequence).
%D A003500 V. D. To, "Finding All Fleenor-Heronian Triangles", Journal of Recreational 
               Mathematics vol. 32 no.4 2003-4 pp. 298-301 Baywood NY.
%H A003500 T. D. Noe, <a href="b003500.txt">Table of n, a(n) for n=0..200</a>
%H A003500 <a href="Sindx_Rea.html#recLCC">Index entries for sequences related to 
               linear recurrences with constant coefficients</a>
%H A003500 S. Plouffe, <a href="http://www.lacim.uqam.ca/%7Eplouffe/articles/MasterThesis.pdf">
               Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures</
               a>, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 
               1992.
%H A003500 S. Plouffe, <a href="http://www.lacim.uqam.ca/%7Eplouffe/articles/FonctionsGeneratrices.pdf">
               1031 Generating Functions and Conjectures</a>, Universit\'{e} du 
               Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
%H A003500 K. S. Brown's Mathpages, <a href="http://www.mathpages.com/home/kmath480.htm">
               Some Properties of the Lucas Sequence(2, 4, 14, 52, 194, ...)</a>
%H A003500 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/
               RecursiveSequences.html">Recursive Sequences</a>
%H A003500 <a href="Sindx_Rea.html#recur1">Index entries for recurrences a(n) = 
               k*a(n - 1) +/- a(n - 2)</a>
%F A003500 a(n) = ( 2 + Sqrt(3) )^n + ( 2 - Sqrt(3) )^n.
%F A003500 a(n) = trace of (n+1)st power of the 2 X 2 matrix [1 2 / 1 3]. - Gary 
               W. Adamson (qntmpkt(AT)yahoo.com), Jun 30 2003
%F A003500 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
%F A003500 G.f.: -2*(-1+2*x)/(1-4*x+x^2). a(n)=2*A001353(n+1)-4*A001353(n). - R. 
               J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 16 2007
%p A003500 A003500 := proc(n) option remember; if n <= 1 then 2*n+2 else 4*A003500(n-1)-A003500(n-2); 
               fi; end;
%p A003500 A003500:=-2*(-1+2*z)/(1-4*z+z**2); [Conjectured by S. Plouffe in his 
               1992 dissertation.]
%t A003500 a[0] = 2; a[1] = 4; a[n_] := a[n] = 4a[n - 1] - a[n - 2]; Table[ a[n], 
               {n, 0, 23}]
%o A003500 (Other) sage: [lucas_number2(n,4,1) for n in xrange(0, 24)]# [From Zerinvary 
               Lajos (zerinvarylajos(AT)yahoo.com), May 14 2009]
%Y A003500 Equals A001353(n+1) - A001353(n-1), also A001835(n) + A001835(n+1), also 
               2*A001075(n).
%Y A003500 Cf. A001570, A006051, A048788, A002530, A011945.
%Y A003500 Sequence in context: A032222 A046650 A055727 this_sequence A129876 A038055 
               A006385
%Y A003500 Adjacent sequences: A003497 A003498 A003499 this_sequence A003501 A003502 
               A003503
%K A003500 nonn,easy,nice
%O A003500 0,1
%A A003500 N. J. A. Sloane (njas(AT)research.att.com).
%E A003500 More terms from James A. Sellers (sellersj(AT)math.psu.edu), May 03 2000
%E A003500 Additional comments from Lekraj Beedassy (blekraj(AT)yahoo.com), Feb 
               14 2002

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