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%I #91 Sep 08 2022 08:45:13
%S 3,-1,5,-4,13,-16,38,-57,117,-193,370,-639,1186,-2094,3827,-6829,
%T 12389,-22220,40169,-72220,130338,-234609,423065,-761945,1373466,
%U -2474291,4459278,-8034394,14478659,-26088169,47011093,-84708772,152642789,-275049240
%N An accelerator sequence for Catalan's constant.
%C The pair A094648 and the alternating sequence A033304 when joined form a two-sided sequence defined by the recurrence formula x(n+3) + x(n+2) - 2x(n+1) - x(n) = 0, n in Z, x(-1)=-2, x(0)=3, x(1)=-1 - for details see Witula's comments to A033304. - _Roman Witula_, Jul 25 2012
%C From _Roman Witula_, Aug 09 2012: (Start)
%C There exist two interesting subsequences b(n) and c(n) of the given above sequence x(n) defined by the following relations: b(n)=a(2^n) and c(n)=x(-2^n). These subsequences satisfy the following system of recurrence equations:
%C b(n+1)=b(n)^2-2*c(n), and c(n+1)=c(n)^2-2*b(n),
%C which easily follow from the general identity: x(n)^2=x(2*n)-2*x(-n), n in Z. We note that b(0)=-1, b(1)=5, b(2)=13, b(3)=117, c(0)=-2, c(1)=6, c(2)=26, c(3)=650. From the above system we deduce that all b(n) are odd, whereas all c(n) are even. Moreover we obtain c(n+1)-b(n+1)=(c(n)-b(n))*(b(n)+c(n)+2), which yields b(n+1)-c(n+1)=product{k=1,..,n}(b(k)+c(k)+2)=13*product{k=2,..,n}(b(k)+c(k)+2)=13^2*41*product{k=3,..,n}(b(k)+c(k)+2). It follows that b(n)-c(n) is divisible by 13^2*41 for every n=3,4,..., and after using the above system again each b(n) and c(n), for n=2,3,..., is divisible by 13. (End)
%C If we set W(n):=3*A077998(n)-A006054(n+1)-A006054(n), n=0,1,..., then a(n)=(W(n)^2-W(2*n))/2 and W(n) = (-c(1))^(-n) + (-c(2))^(-n) + (-c(4))^(-n) = (-c(1)*c(2))^n + (-c(1)*c(4))^n + (-c(2)*c(4))^n = (-1-c(1))^n + (-1-c(2))^n + (-1-c(4))^n, where c(j):=2*cos(2*Pi*j/7) - for the proof see Witula-Slota-Warzynski's paper. Moreover it follows from the comment at the top and from comments to A033304 that W(n+1)=A033304(n)=(-1)^(n+1)*x(-n-1). - _Roman Witula_, Aug 11 2012
%C The following trigonometric type identitities hold true: (1) -a(n-1)-a(n) = c(1)*c(2)^n + c(2)*c(4)^n + c(4)*c(1)^n and (2) a(n)-a(n+2) = c(4)*c(2)^(n+1) + c(1)*c(4)^(n+1) + c(2)*c(1)^(n+1), where a(-1)=-2 and c(j) is defined as above (see also the respective comment to A033304). For the proof see Remark 6 in Witula's paper. - _Roman Witula_, Aug 14 2012
%C It can be proved that A033304(n-1)*(-1)^n = (a(n)^2 - a(2*n))/2, n=1,2,... - _Roman Witula_, Sep 30 2012
%C With respect to the form of the trigonometric formulas describing a(n), we call this sequence the Berndt-type sequence number 19 for the argument 2*Pi/7. The A-numbers of other Berndt-type sequences numbers are given in below. - _Roman Witula_, Sep 30 2012
%H G. C. Greubel, <a href="/A094648/b094648.txt">Table of n, a(n) for n = 0..3900</a>
%H A. Akbary and Q. Wang, <a href="http://dx.doi.org/10.1155/IJMMS.2005.2631">On some permutation polynomials over finite fields</a>, International Journal of Mathematics and Mathematical Sciences, 2005:16 (2005) 2631-2640.
%H A. Akbary and Q. Wang, <a href="http://dx.doi.org/10.1090/S0002-9939-05-08220-1">A generalized Lucas sequence and permutation binomials</a>, Proceeding of the American Mathematical Society, 134 (1) (2006), 15-22.
%H David M. Bradley, <a href="http://dx.doi.org/10.1023/A:1006945407723">A Class of Series Acceleration Formulae for Catalan's Constant</a>, The Ramanujan Journal, Vol. 3, Issue 2, 1999, pp. 159-173
%H David M. Bradley, <a href="http://arxiv.org/abs/0706.0356">A Class of Series Acceleration Formulae for Catalan's Constant</a>, arXiv:0706.0356 [math.CA], 2007.
%H Kai Wang, <a href="https://www.researchgate.net/publication/337943524_Fibonacci_Numbers_And_Trigonometric_Functions_Outline">Fibonacci Numbers And Trigonometric Functions Outline</a>, (2019).
%H Q. Wang, <a href="https://www.semanticscholar.org/paper/On-generalized-Lucas-sequences-Wang-Akbari/7e33b3b79703dc6790fca133e8c92cc0cafcfe4a">On generalized Lucas sequences</a>, Contemp. Math. 531 (2010) 127-141
%H Roman Witula, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/Witula/witula17.html">Ramanujan Type Trigonometric Formulas: The General Form for the Argument 2Pi/7</a>, J. Integer Seq., 12 (2009), Article 09.8.5.
%H Roman Witula and Damian Slota, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Slota/witula13.html">New Ramanujan-Type Formulas and Quasi-Fibonacci Numbers of Order 7</a>, Journal of Integer Sequences, Vol. 10 (2007), Article 07.5.6
%H Roman Witula, Damian Slota and Adam Warzynski, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL9/Slota/slota57.html">Quasi-Fibonacci Numbers of the Seventh Order</a>, J. Integer Seq., 9 (2006), Article 06.4.3.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (-1,2,1).
%F G.f.: (3+2*x-2*x^2)/(1+x-2*x^2-x^3);
%F a(n) = (2*sin(3*Pi/14))^n+(-2*sin(Pi/14))^n+(-2*cos(Pi/7))^n.
%F a(p) == -1 mod(p), p prime. - _Philippe Deléham_, Oct 03 2009
%F a(n) = (2*cos(2*Pi/7))^n + (2*cos(4*Pi/7))^n + (2*cos(8*Pi/7))^n, which is equivalent to the formula given above (for analogous sums with sines see A215493 and A215494). Moreover we have a(n+3) + a(n+2) - 2a(n+1) - a(n) = 0 - for the proof see Witula-Slota's paper. - _Roman Witula_, Jul 24 2012
%F a(n) = 3*(-1)^n*A006053(n+2) +2*A078038(n-1). - _R. J. Mathar_, Nov 03 2020
%e We have a(17) = a(19) + 50000, a(4) + a(5) = -3, 2*a(7) + a(8) = 3, and 2*a(9) + a(10) = a(5). - _Roman Witula_, Sep 14 2012
%t CoefficientList[ Series[(3 + 2x - 2x^2)/(1 + x - 2x^2 - x^3), {x, 0, 33}], x] (* _Robert G. Wilson v_, May 24 2004 *)
%t a[n_] := Round[(2Sin[3Pi/14])^n + (-2Sin[Pi/14])^n + (-2Cos[Pi/7])^n]; Table[ a[n], {n, 0, 33}] (* _Robert G. Wilson v_, May 24 2004 *)
%t LinearRecurrence[{-1,2,1}, {3,-1,5}, 50] (* _Roman Witula_, Aug 09 2012 *)
%o (Magma) I:=[3,-1,5]; [n le 3 select I[n] else -Self(n-1)+2*Self(n-2)+Self(n-3): n in [1..40]]; // _Vincenzo Librandi_, Jul 25 2015
%o (PARI) x='x+O('x^30); Vec((3+2*x-2*x^2)/(1+x-2*x^2-x^3)) \\ _G. C. Greubel_, May 09 2018
%Y Cf. A000032, A094649, A094650.
%Y Cf. A215007, A215008, A215143, A215493, A215494, A215510, A215512, A215575, A215694, A215695, A108716, A215794, A215828, A215817, A215877, A094429, A094430, A217274.
%K easy,sign
%O 0,1
%A _Paul Barry_, May 18 2004
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