A075092 Sum of generalized tribonacci numbers (A001644) and reflected generalized tribonacci numbers (A073145).

6, 0, 2, 12, 6, 20, 50, 56, 134, 264, 402, 836, 1542, 2652, 5154, 9392, 16902, 31824, 58082, 106172, 197126, 360932, 662994, 1223784, 2245766, 4130520, 7606770, 13976436, 25711622, 47310252, 86978370, 160002656, 294324230, 541249952

Offset: 0

Author

Mario Catalani (mario.catalani(AT)unito.it), Aug 31 2002

Keywords

• easy
• nonn

Comments

Conjecture: a(n) >= 0.

For n > 2, a(n) is the number of cyclic sequences (q1, q2, ..., qn) consisting of zeros, ones and twos such that each triple contains 0 and 1 at least once, provided the positions of the zeros and ones are fixed on a circle. For example, a(5)=20 because only the sequences (00101), (01001), (01010), (01011), (01012), (01021), (01101), (01201), (02101), (20101) and those obtained from them by exchanging 0 and 1 contain 0 and 1 in each triple (including triples q4, q5, q1 and q5, q1, q2). For n = 1, 2 the statement is still true provided we allow the sequence to wrap around itself on a circle. E.g., a(2) = 2 since only sequences 01 and 10 can be wrapped so one obtains (010) and (101), respectively. - Wojciech Florek, Nov 25 2021

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Wojciech Florek, A class of generalized Tribonacci sequences applied to counting problems, Appl. Math. Comput., 338 (2018), 809-821.
• W. O. J. Moser, Cyclic binary strings without long runs of like (alternating) bits, Fibonacci Quart. 31(1) (1993), 2-6.
• Index entries for linear recurrences with constant coefficients, signature (0,1,4,1,0,-1).

Crossrefs

Cf. A001644, A073145, A075091, A294627.

Programs

Magma

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (6-4*x^2-12*x^3-2*x^4)/(1-x^2-4*x^3-x^4+x^6) ));  // G. C. Greubel, Apr 13 2019

Mathematica

CoefficientList[Series[(6-4*x^2-12*x^3-2*x^4)/(1-x^2-4*x^3-x^4+x^6), {x, 0, 40}], x]

PARI

my(x='x+O('x^40)); Vec((6-4*x^2-12*x^3-2*x^4)/(1-x^2-4*x^3-x^4+x^6)) \\ G. C. Greubel, Apr 13 2019

Sage

((6-4*x^2-12*x^3-2*x^4)/(1-x^2-4*x^3-x^4+x^6)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 13 2019

Formula

a(n) = a(n-2) + 4*a(n-3) + a(n-4) - a(n-6), a(0)=6, a(1)=0, a(2)=2, a(3)=12, a(4)=6, a(5)=20.

G.f.: (6 - 4*x^2 - 12*x^3 - 2*x^4)/(1 - x^2 - 4*x^3 - x^4 + x^6).