%I #51 Sep 08 2022 08:46:03

%S 3,7,35,175,931,5047,27587,151263,830403,4560871,25054435,137642127,

%T 756187747,4154438295,22824258947,125395430335,688917131651,

%U 3784882096583,20793986742179,114241312597615,627637106311971,3448212648805239,18944339609269571

%N a(n) = 7*(a(n-1) - a(n-2) - a(n-3)), with a(0)=3, a(1)=7, a(2)=35.

%C The Berndt-type sequence number 8 for the argument 2Pi/7 defined by trigonometric relations from "Formula" below.

%C We note that the following decompositions hold true: (X-cot(2*Pi/7)^n)*(X-cot(4*Pi/7)^n)*(X-cot(8*Pi/7)^n) = X^3 - sqrt(7)^(-n)*a(n)*X^2 + (-sqrt(7))^(-n)*B(n)*X

%C - (-sqrt(7))^(-n), and (X-tan(2*Pi/7)^n)*(X-tan(4*Pi/7)^n)*(X-tan(8*Pi/7)^n) = X^3 - B(n)*X^2 + (-1)^n*a(n)*X - (-sqrt(7))^n, where B(n) := tan(2*Pi/7)^n + tan(4*Pi/7)^n + tan(8*Pi/7)^n = (-sqrt(7) + 4*sin(2*Pi/7))^n + (-sqrt(7) + 4*sin(4*Pi/7))^n + (-sqrt(7) + 4*sin(8*Pi/7))^n. Moreover we have 2*(-1)^n*B(n) = 7^(-n/2)*(a(n)^2 - a(2*n)). For the proof of these decompositions see Witula-Slota's (Section 6) and Witula's (Remark 11) reference.

%C We note that the numbers a(n)*7^(-ceiling(n/3)) are all integers. Moreover from the recurrence relation: a(n+3)+7*a(n+1)=7*(a(n+2)-a(n)) it can be easily obtained the following summations formulas: 8*sum{k=1,..,n} a(2*k) = 7*(a(2*n+1)-2)-a(2*n+2), which also means that the result is divisible by 8 for every n=1,2,..., and 8*sum{k=1,..,n} a(2*k-1) = 7*(a(2*n)-2)-a(2*n+1), which implies that 7*(a(n)-2)-a(n+1) is divisible by 8 for each n=0,1,...

%D E Hetmaniok, P Lorenc, S Damian, et al., Periodic orbits of boundary logistic map and new kind of modified Chebyshev polynomials in R. Witula, D. Slota, W. Holubowski (eds.), Monograph on the Occasion of 100th Birthday Anniversary of Zygmunt Zahorski. Wydawnictwo Politechniki Slaskiej, Gliwice 2015, pp. 325-343.

%H G. C. Greubel, <a href="/A215575/b215575.txt">Table of n, a(n) for n = 0..1350</a>

%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, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/Witula/witula17.html">Ramanujan Type Trigonometric Formulas: The General Form for the Argument 2*Pi/7</a>, Journal of Integer Sequences, Vol. 12 (2009), Article 09.8.5.

%H R. Witula, P. Lorenc, M. Rozanski, M. Szweda, <a href="http://yadda.icm.edu.pl/yadda/element/bwmeta1.element.baztech-142e1ad8-f56c-4c6e-b6e6-f71d1f31e76e">Sums of the rational powers of roots of cubic polynomials</a>, Zeszyty Naukowe Politechniki Slaskiej, Seria: Matematyka Stosowana z. 4, Nr. kol. 1920, 2014.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7, -7, -7).

%F a(n) = (sqrt(7)^n)*(cot(2*Pi/7)^n + cot(4*Pi/7)^n + cot(8*Pi/7)^n) = (3 + 4*cos(2*Pi/7))^n + (3 + 4*cos(4*Pi/7))^n + (3 + 4*cos(8*Pi/7))^n = (-tan(2*Pi/7)*tan(4*Pi/7))^n + (-tan(2*Pi/7)*tan(8*Pi/7))^n + (-tan(4*Pi/7)*tan(8*Pi/7))^n.

%F G.f.: (3-14*x+7*x^2)/(1-7*x+7*x^2+7*x^3).

%F a(n) = x1^n + x2^n + x3^n, where x1, x2, x3 are the roots of the polynomial x^3 - 7*x^2 + 7*x + 7, that is, x1 = sqrt(7)/tan(Pi/7), x2 = sqrt(7)/tan(2*Pi/7), x3 = sqrt(7)/tan(4*Pi/7). - _Kai Wang_, Jul 19 2016

%e We have cot(2*Pi/7)^2 + cot(4*Pi/7)^2 + cot(8*Pi/7)^2 = 5, cot(2*Pi/7)^4 + cot(4*Pi/7)^4 + cot(8*Pi/7)^4 = 19, but cot(2*Pi/7)^6 + cot(4*Pi/7)^6 + cot(8*Pi/7)^6 = 563/7. Similarly the numbers sqrt(7)*(cot(2*Pi/7)^n + cot(4*Pi/7)^n + cot(8*Pi/7)^n) are integers for n=1,3,5,7 (equal to 7, 25, 103, 441, respectively), whereas for n=9 we obtain the rational value 13297/7.

%t LinearRecurrence[{7,-7,-7}, {3,7,35}, 50]

%o (PARI) polsym(x^3 - 7*x^2 + 7*x + 7, 30) \\ _Charles R Greathouse IV_, Jul 20 2016

%o (Magma) I:=[3,7,35]; [n le 3 select I[n] else 7*Self(n-1) - 7*Self(n-2) - 7*Self(n-3): n in [1..30]]; // _G. C. Greubel_, Apr 25 2017

%Y Cf. A215007, A215008, A215143, A215493, A215494, A215510, A215512, A215455.

%K nonn,easy

%O 0,1

%A _Roman Witula_, Aug 16 2012