A003500 - OEIS

%I M1278 #247 Aug 29 2024 06:05:19

%S 2,4,14,52,194,724,2702,10084,37634,140452,524174,1956244,7300802,

%T 27246964,101687054,379501252,1416317954,5285770564,19726764302,

%U 73621286644,274758382274,1025412242452,3826890587534,14282150107684,53301709843202,198924689265124

%N a(n) = 4*a(n-1) - a(n-2) with a(0) = 2, a(1) = 4.

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

%C If M is any given term of the sequence, then the next one is 2*M + sqrt(3*M^2 - 12). - _Lekraj Beedassy_, Feb 18 2002

%C 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(2*n)/2 and whose semiperimeter is 3*a[n]/2. The sequence is symmetrical about a[0], i.e., a[-n] = a[n].

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

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

%C Except for the first term, positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 12 = 0. - _Colin Barker_, Feb 04 2014

%C Except for the first term, positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 192 = 0. - _Colin Barker_, Feb 16 2014

%C A268281(n) - 1 is a member of this sequence iff A268281(n) is prime. - _Frank M Jackson_, Feb 27 2016

%C a(n) gives values of x satisfying 3*x^2 - 4*y^2 = 12; corresponding y values are given by A005320. - _Sture Sjöstedt_, Dec 19 2017

%C Middle side lengths of almost-equilateral Heronian triangles. - _Wesley Ivan Hurt_, May 20 2020

%C For all elements k of the sequence, 3*(k-2)*(k+2) is a square. - _Davide Rotondo_, Oct 25 2020

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

%D J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p.91.

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

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

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

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%D R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

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

%H T. D. Noe, <a href="/A003500/b003500.txt">Table of n, a(n) for n=0..200</a>

%H P. Bala, <a href="/A174500/a174500_2.pdf">Some simple continued fraction expansions for an infinite product, Part 1</a>

%H R. A. Beauregard and E. R. Suryanarayan, <a href="http://www.maa.org/sites/default/files/pdf/mathdl/CMJ/methodoflastresort.pdf">The Brahmagupta Triangles</a>, The College Mathematics Journal 29(1) 13-7 1998 MAA.

%H Hacène Belbachir, Soumeya Merwa Tebtoub and László Németh, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL23/Nemeth/nemeth7.html">Ellipse Chains and Associated Sequences</a>, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.

%H Daniel Birmajer, Juan B. Gil and Michael D. Weiner, <a href="http://arxiv.org/abs/1505.06339">Linear recurrence sequences with indices in arithmetic progression and their sums</a>, arXiv preprint arXiv:1505.06339 [math.NT], 2015.

%H K. S. Brown's Mathpages, <a href="http://www.mathpages.com/home/kmath480/kmath480.htm">Some Properties of the Lucas Sequence(2, 4, 14, 52, 194, ...)</a>

%H H. W. Gould, <a href="http://www.fq.math.ca/Scanned/11-1/gould.pdf">A triangle with integral sides and area</a>, Fib. Quart., 11 (1973), 27-39.

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H E. Keith Lloyd, <a href="http://www.jstor.org/stable/3619201">The Standard Deviation of 1, 2, ..., n: Pell's Equation and Rational Triangles</a>, Math. Gaz. vol 81 (1997), 231-243.

%H S. Northshield, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Northshield/north4.html">An Analogue of Stern's Sequence for Z[sqrt(2)]</a>, Journal of Integer Sequences, 18 (2015), #15.11.6.

%H Hideyuki Ohtskua, proposer, <a href="https://www.fq.math.ca/Problems/FQElemProbAug2024.pdf">Problem B-1351</a>, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 62, No. 3 (2024), p. 258.

%H Simon Plouffe, <a href="https://arxiv.org/abs/0911.4975">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.

%H Simon Plouffe, <a href="/A000051/a000051_2.pdf">1031 Generating Functions</a>, Appendix to Thesis, Montreal, 1992

%H Jeffrey Shallit, <a href="http://www.jstor.org/stable/2690344">An interesting continued fraction</a>, Math. Mag., 48 (1975), 207-211.

%H Jeffrey Shallit, <a href="/A005248/a005248_1.pdf">An interesting continued fraction</a>, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]

%H Yu Tsumura, <a href="http://arxiv.org/abs/1004.1244">On compositeness of special types of integers</a>, arXiv:1004.1244 [math.NT], 2010.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HeronianTriangle.html">Heronian Triangle</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Heronian_triangle">Heronian triangle</a>

%H A. V. Zarelua, <a href="https://doi.org/10.1007/s11006-006-0090-y">On Matrix Analogs of Fermat's Little Theorem</a>, Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.

%H <a href="/index/Rea#recur1">Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (4,-1).

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

%F a(n) = 2*A001075(n).

%F G.f.: 2*(1 - 2*x)/(1 - 4*x + x^2). _Simon Plouffe_ in his 1992 dissertation.

%F a(n) = A001835(n) + A001835(n+1).

%F a(n) = trace of n-th power of the 2 X 2 matrix [1 2 / 1 3]. - _Gary W. Adamson_, Jun 30 2003 [corrected by _Joerg Arndt_, Jun 18 2020]

%F From the addition formula, a(n+m) = a(n)*a(m) - a(m-n), it is easy to derive multiplication formulas, such as: a(2*n) = (a(n))^2 - 2, a(3*n) = (a(n))^3 - 3*(a(n)), a(4*n) = (a(n))^4 - 4*(a(n))^2 + 2, a(5*n) = (a(n))^5 - 5*(a(n))^3 + 5*(a(n)), a(6*n) = (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

%F a(n) = 2*A001353(n+1) - 4*A001353(n). - _R. J. Mathar_, Nov 16 2007

%F From _Peter Bala_, Jan 06 2013: (Start)

%F Let F(x) = Product_{n=0..infinity} (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.

%F 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 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 + ...))))))).

%F (End)

%F a(2^n) = A003010(n). - _John Blythe Dobson_, Mar 10 2014

%F a(n) = [x^n] ( (1 + 4*x + sqrt(1 + 8*x + 12*x^2))/2 )^n for n >= 1. - _Peter Bala_, Jun 23 2015

%F E.g.f.: 2*exp(2*x)*cosh(sqrt(3)*x). - _Ilya Gutkovskiy_, Apr 27 2016

%F a(n) = Sum_{k=0..floor(n/2)} (-1)^k*n*(n - k - 1)!/(k!*(n - 2*k)!)*4^(n - 2*k) for n >= 1. - _Peter Luschny_, May 10 2016

%F From _Peter Bala_, Oct 15 2019: (Start)

%F a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; -1, 4].

%F Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).

%F 2*Sum_{n >= 1} 1/( a(n) - 6/a(n) ) = 1.

%F 6*Sum_{n >= 1} (-1)^(n+1)/( a(n) + 2/a(n) ) = 1.

%F 8*Sum_{n >= 1} 1/( a(n) + 24/(a(n) - 12/(a(n))) ) = 1.

%F 8*Sum_{n >= 1} (-1)^(n+1)/( a(n) + 8/(a(n) + 4/(a(n))) ) = 1.

%F Series acceleration formulas for sums of reciprocals:

%F Sum_{n >= 1} 1/a(n) = 1/2 - 6*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 6)),

%F Sum_{n >= 1} 1/a(n) = 1/8 + 24*Sum_{n >= 1} 1/(a(n)*(a(n)^2 + 12)),

%F Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/6 + 2*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 2)) and

%F Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/8 + 8*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 12)).

%F Sum_{n >= 1} 1/a(n) = ( theta_3(2-sqrt(3))^2 - 1 )/4 = 0.34770 07561 66992 06261 .... See Borwein and Borwein, Proposition 3.5 (i), p.91.

%F Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - theta_3(sqrt(3)-2)^2 )/4. Cf. A003499 and A153415. (End)

%F a(n) = tan(Pi/12)^n + tan(5*Pi/12)^n. - _Greg Dresden_, Oct 01 2020

%F From _Wolfdieter Lang_, Sep 06 2021: (Start)

%F a(n) = S(n, 4) - S(n-2, 4) = 2*T(n, 2), for n >= 0, with S and T Chebyshev polynomials, with S(-1, x) = 0 and S(-2, x) = -1. S(n, 4) = A001353(n+1), for n >= -1, and T(n, 2) = A001075(n).

%F a(2*k) = A067902(k), a(2*k+1) = 4*A001570(k+1), for k >= 0. (End)

%F a(n) = sqrt(2 + 2*A011943(n+1)) = sqrt(2 + 2*A102344(n+1)), n>0. - _Ralf Steiner_, Sep 23 2021

%F Sum_{n>=1} arctan(3/a(n)^2) = Pi/6 - arctan(1/3) = A019673 - A105531 (Ohtskua, 2024). - _Amiram Eldar_, Aug 29 2024

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

%p end proc;

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

%t LinearRecurrence[{4,-1},{2,4},30] (* _Harvey P. Dale_, Aug 20 2011 *)

%t Table[Round@LucasL[2n, Sqrt[2]], {n, 0, 20}] (* _Vladimir Reshetnikov_, Sep 15 2016 *)

%o (Sage) [lucas_number2(n,4,1) for n in range(0, 24)] # _Zerinvary Lajos_, May 14 2009

%o (Haskell)

%o a003500 n = a003500_list !! n

%o a003500_list = 2 : 4 : zipWith (-)

%o    (map (* 4) $ tail a003500_list) a003500_list

%o -- _Reinhard Zumkeller_, Dec 17 2011

%o (PARI) x='x+O('x^99); Vec(-2*(-1+2*x)/(1-4*x+x^2)) \\ _Altug Alkan_, Apr 04 2016

%o (Magma) I:=[2,4]; [n le 2 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..30]]; // _Vincenzo Librandi_, Nov 14 2018

%Y Cf. A001075, A001353, A001835.

%Y Cf. A011945 (areas), A334277 (perimeters).

%Y Cf. this sequence (middle side lengths), A016064 (smallest side lengths), A335025 (largest side lengths).

%Y Cf. A001570, A002530, A005320, A006051, A048788, A174500, A268281.

%Y Cf. A011943, A102344, A019673, A105531.

%K nonn,easy,nice

%O 0,1

%A _N. J. A. Sloane_

%E More terms from _James A. Sellers_, May 03 2000

%E Additional comments from _Lekraj Beedassy_, Feb 14 2002