A094648 An accelerator sequence for Catalan's constant.

3, -1, 5, -4, 13, -16, 38, -57, 117, -193, 370, -639, 1186, -2094, 3827, -6829, 12389, -22220, 40169, -72220, 130338, -234609, 423065, -761945, 1373466, -2474291, 4459278, -8034394, 14478659, -26088169, 47011093, -84708772, 152642789, -275049240

Offset: 0

Author

Paul Barry, May 18 2004

Keywords

• easy
• sign

Comments

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

From Roman Witula, Aug 09 2012: (Start)

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:

b(n+1)=b(n)^2-2*c(n), and c(n+1)=c(n)^2-2*b(n),

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)

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

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

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

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

Examples

			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
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..3900
• A. Akbary and Q. Wang, On some permutation polynomials over finite fields, International Journal of Mathematics and Mathematical Sciences, 2005:16 (2005) 2631-2640.
• A. Akbary and Q. Wang, A generalized Lucas sequence and permutation binomials, Proceeding of the American Mathematical Society, 134 (1) (2006), 15-22.
• David M. Bradley, A Class of Series Acceleration Formulae for Catalan's Constant, The Ramanujan Journal, Vol. 3, Issue 2, 1999, pp. 159-173
• David M. Bradley, A Class of Series Acceleration Formulae for Catalan's Constant, arXiv:0706.0356 [math.CA], 2007.
• Kai Wang, Fibonacci Numbers And Trigonometric Functions Outline, (2019).
• Q. Wang, On generalized Lucas sequences, Contemp. Math. 531 (2010) 127-141
• Roman Witula, Ramanujan Type Trigonometric Formulas: The General Form for the Argument 2Pi/7, J. Integer Seq., 12 (2009), Article 09.8.5.
• Roman Witula and Damian Slota, New Ramanujan-Type Formulas and Quasi-Fibonacci Numbers of Order 7, Journal of Integer Sequences, Vol. 10 (2007), Article 07.5.6
• Roman Witula, Damian Slota and Adam Warzynski, Quasi-Fibonacci Numbers of the Seventh Order, J. Integer Seq., 9 (2006), Article 06.4.3.
• Index entries for linear recurrences with constant coefficients, signature (-1,2,1).

Crossrefs

Cf. A000032, A094649, A094650.

Cf. A215007, A215008, A215143, A215493, A215494, A215510, A215512, A215575, A215694, A215695, A108716, A215794, A215828, A215817, A215877, A094429, A094430, A217274.

Programs

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

Mathematica

CoefficientList[ Series[(3 + 2x - 2x^2)/(1 + x - 2x^2 - x^3), {x, 0, 33}], x] (* Robert G. Wilson v, May 24 2004 *)
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 *)
LinearRecurrence[{-1,2,1}, {3,-1,5}, 50] (* Roman Witula, Aug 09 2012 *)

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

Formula

G.f.: (3+2*x-2*x^2)/(1+x-2*x^2-x^3);

a(n) = (2*sin(3*Pi/14))^n+(-2*sin(Pi/14))^n+(-2*cos(Pi/7))^n.

a(p) == -1 mod(p), p prime. - Philippe Deléham, Oct 03 2009

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

a(n) = 3*(-1)^n*A006053(n+2) +2*A078038(n-1). - R. J. Mathar, Nov 03 2020