A215828 a(n) = 7^(floor(n/3))*A(n), where A(n) = A(n-1) + A(n-2) + A(n-3)/7, with A(0)=3, A(1)=1, A(2)=3.

3, 1, 3, 31, 53, 87, 1011, 1673, 2771, 32119, 53189, 88079, 1020995, 1690737, 2799811, 32454831, 53744245, 88998887, 1031656755, 1708393209, 2829048851, 32793751175, 54305486341, 89928286367, 1042430160131, 1726233651041, 2858592097539, 33136210400191

Offset: 0

Author

Roman Witula, Aug 24 2012

Keywords

• nonn
• easy

Comments

The Berndt-type sequence number 13 for the argument 2Pi/7

defined by the relation ((-sqrt(7))^n)*A(n) = t(1)^n + t(2)^n + t(4)^n = (-sqrt(7) + 4*s(1))^n + (-sqrt(7) + 4*s(2))^n + (-sqrt(7) + 4*s(4))^n, where t(j) := tan(2*Pi*j/7) and s(j) := sin(2*Pi*j/7), and the fact that all numbers 7^(floor(n/3))*A(n) are integers. We note that ((-sqrt(7))^n)*A(n) = B(n), where B(n) is defined in the comments to A215575. For more details see also A108716, A215794, Witula-Slota's (Section 6) and Witula's (Remark 11) papers.

Examples

			We have A(3)=31/7, A(4)=53/7 and A(5)=87/7. On the other hand we have a(2)+a(3)+a(4)=a(5).
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Roman Witula, Ramanujan Type Trigonometric Formulas: The General Form for the Argument 2*Pi/7, Journal of Integer Sequences, Vol. 12 (2009), Article 09.8.5.
• R. Witula, P. Lorenc, M. Rozanski, M. Szweda, Sums of the rational powers of roots of cubic polynomials, Zeszyty Naukowe Politechniki Slaskiej, Seria: Matematyka Stosowana z. 4, Nr. kol. 1920, 2014.
• 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.
• Index entries for linear recurrences with constant coefficients, signature (0,0,31,0,0,25,0,0,1).

Crossrefs

Cf. A215575, A108716, A215794.

Programs

Magma

/* By definition: */ i:=28; I:=[3,1,3]; A:=[m le 3 select I[m] else Self(m-1)+Self(m-2)+Self(m-3)/7: m in [1..i]]; [7^(Floor((n-1)/3))*A[n]: n in [1..i]]; // Bruno Berselli, Oct 28 2012

Mathematica

CoefficientList[Series[(x^8 - 5 x^7 + 25 x^6 + 6 x^5 - 22 x^4 + 62 x^3 - 3 x^2 - x - 3)/(x^9 + 25 x^6 + 31 x^3 - 1), {x, 0, 30}], x] (* Vincenzo Librandi, Mar 19 2013 *)

Formula

G.f.: (x^8-5*x^7+25*x^6+6*x^5-22*x^4+62*x^3-3*x^2-x-3)/(x^9+25*x^6+31*x^3-1). [Colin Barker, Oct 28 2012]