A216605 Expansion of g.f. (6 + 5*x - 20*x^2 - 12*x^3 + 12*x^4 + 3*x^5)/(1 + x - 5*x^2 - 4*x^3 + 6*x^4 + 3*x^5 - x^6).

6, -1, 11, -4, 31, -16, 98, -64, 327, -256, 1126, -1024, 3958, -4083, 14116, -16189, 50887, -63768, 184958, -249547, 676626, -970771, 2488156, -3756867, 9188406, -14474916, 34049481, -55564474, 126540536, -212637571, 471398623, -811660849, 1759603367

Offset: 0

Author

Roman Witula, Sep 10 2012

Keywords

• sign
• easy

Comments

Previous name was: a(n) is equal to the rational part (considering of the ring Z(sqrt(13))) of the numbers A(n) = ((sqrt(13) - 1)/2)*A(n-1) + A(n-2) + ((3-sqrt(13))/2)*A(n-3), with A(0) = 6, A(1) = sqrt(13) - 1, and A(2) = 11 - sqrt(13).

The Berndt-type sequence number 1 for the argument 2*Pi/13 defined by the following relation: a(n) + b(n)*sqrt(13) = A(n) = 2*(c(1)^n + c(3)^n + c(4)^n), where c(j) := 2*cos(2*Pi*j/13), j=1..6. The b(n) = A216486(n), n=0,1,..., are all positive integers. We note that we also have a(n) - b(n)*sqrt(13) = B(n) = 2*(c(2)^n + c(5)^n + c(6)^n) and the following recurrence relation holds: B(n) = -((sqrt(13)+ 1)/2)*B(n-1) + B(n-2) + ((3+sqrt(13))/2)*B(n-3), with B(0) = 6, B(1) = -sqrt(13) - 1, and B(2) = 11 + sqrt(13).

We note that 4*a(n) - a(n+2) is divisible by 13 for every n = 0,1,... .

Examples

			We have 4*a(2*n-1)=a(2*n+1) for every n = 1,2,...,5 and a(13) - 4*a(11) = 13. Further we have c(1)^3 + c(3)^3 + c(4)^3 = 4*(c(1) + c(3) + c(4)) since A(3) = 4*sqrt(13) - 4, c(2)^3 + c(5)^3 + c(6)^3 = 4*(c(2) + c(5) + c(6)) since B(3) = - 4*sqrt(13) - 4, 2 + c(1)^4 + c(3)^4 + c(4)^4 = 3*(c(1)^2 + c(3)^2 + c(4)^2) and 2 + c(2)^4 + c(5)^4 + c(6)^4 = 3*(c(2)^2 + c(5)^2 + c(6)^2).
		

References

• R. Witula and D. Slota, Quasi-Fibonacci numbers of order 13, Thirteenth International Conference on Fibonacci Numbers and Their Applications, Congressus Numerantium, 201 (2010), 89-107.
• R. Witula, On some applications of formulas for sums of the unimodular complex numbers, Wyd. Pracowni Komputerowej Jacka Skalmierskiego, Gliwice 2011 (in Polish).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• A. Akbary and Q. Wang, A generalized Lucas sequence and permutation binomials, Proc. Am. Math. Soc. 134 (2006) 15-22. Sequence a(n) with l=13 is the unsigned version.
• Russell A. Gordon, Lucas Type Sequences and Sums of Binomial Coefficients, Integers (2023) Vol 23, Art. No. A84. See p. 21.
• R. Witula and D. Slota, Quasi-Fibonacci numbers of order 13, (abstract) see p. 15.
• Index entries for linear recurrences with constant coefficients, signature (-1,5,4,-6,-3,1).

Crossrefs

Cf. A094649, A189234, A216486.

Programs

Magma

I:=[6,-1,11,-4,31,-16]; [n le 6 select I[n] else -Self(n-1)+5*Self(n-2)+4*Self(n-3)-6*Self(n-4)-3*Self(n-5)+Self(n-6): n in [1..35]]; // Vincenzo Librandi, Aug 30 2017

Mathematica

LinearRecurrence[{-1, 5, 4, -6, -3, 1}, {6, -1, 11, -4, 31, -16}, 30]
CoefficientList[Series[(6 + 5 x - 20 x^2 - 12 x^3 + 12 x^4 + 3 x^5)/(1 + x - 5 x^2 - 4 x^3 + 6 x^4 + 3 x^5 - x^6), {x, 0, 33}], x] (* Vincenzo Librandi, Aug 30 2017 *)

Formula

G.f.: (6 + 5*x - 20*x^2 - 12*x^3 + 12*x^4 + 3*x^5)/(1 + x - 5*x^2 - 4*x^3 + 6*x^4 + 3*x^5 - x^6). - Bruno Berselli, Sep 11 2012

a(n) = -a(n-1) + 5*a(n-2) + 4*a(n-3) - 6*a(n-4) - 3*a(n-5) + a(n-6), which from the following decomposition can be generated (see Witula-Slota's and Witula's references for details): X^6 + X^5 - 5*X^4 - 4*X^3 + 6*X^2 + 3*X - 1 = ((X - c(1))*(X - c(3))*(X - c(4)))*((X - c(2))*(X - c(5))*(X - c(6))) = (X^3 + ((1 - sqrt(13))/2)*X^2 - X + (sqrt(13) - 3)/2)*(X^3 + ((1 + sqrt(13))/2)*X^2 - X - (sqrt(13) + 3)/2). - Roman Witula, Sep 11 2012

Extensions

New name using existing g.f. from Joerg Arndt, Feb 15 2024