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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A215575 a(n) = 7*(a(n-1) - a(n-2) - a(n-3)), with a(0)=3, a(1)=7, a(2)=35.

Original entry on oeis.org

3, 7, 35, 175, 931, 5047, 27587, 151263, 830403, 4560871, 25054435, 137642127, 756187747, 4154438295, 22824258947, 125395430335, 688917131651, 3784882096583, 20793986742179, 114241312597615, 627637106311971, 3448212648805239, 18944339609269571
Offset: 0

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Author

Roman Witula, Aug 16 2012

Keywords

Comments

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

Examples

			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.
		

References

  • 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.

Crossrefs

Programs

  • 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
  • Mathematica
    LinearRecurrence[{7,-7,-7}, {3,7,35}, 50]
  • PARI
    polsym(x^3 - 7*x^2 + 7*x + 7, 30) \\ Charles R Greathouse IV, Jul 20 2016
    

Formula

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