OFFSET
0,1
COMMENTS
The Berndt-type sequence number 5a for the argument 2Pi/9 defined by the first relation from the section "Formula". We see that a(n) is equal to the sum of the n-th negative powers of the c(j) := 2*cos(2*Pi*j/9), j=1,2,4 (the A215664(n) is equal to the respective n-th positive powers, further both sequences can be obtained from the two-sided recurrence relation: X(n+3) = 3*X(n+1) - X(n), n in Z, with X(-1) = X(0) = 3, and X(1) = 0).
From the last formula in Witula's comments to A215664 it follows that 2*(-1)^n*a(n) = A215664(n)^2 - A215664(2*n).
The following decomposition holds true: (X - c(1)^(-n))*(X - c(2)^(-n))*(X - c(4)^(-n)) = X^3 - a(n)*X^2 - (-1)^n*A215664(n)*X - (-1)^n.
For n >= 1, a(n) is the number of cyclic (0,1,2)-compositions of n that avoid the pattern 110 provided the positions of the parts of the composition on the circle are fixed. (Similar comments hold for the pattern 012 and for the pattern 001.) - Petros Hadjicostas, Sep 13 2017
See the Maple program by Edlin and Zeilberger for counting the q-ary cyclic compositions of n that avoid one or more patterns provided the positions of the parts of the composition are fixed on the circle. The program is located at D. Zeilberger's personal website (see links). For the sequence here, q=3 and the pattern is A=110. - Petros Hadjicostas, Sep 13 2017
LINKS
Michael De Vlieger, Table of n, a(n) for n = 0..1999
A. E. Edlin and D. Zeilberger, The Goulden-Jackson cluster method for cyclic words, Adv. Appl. Math. 25 (2000), 228-232.
A. E. Edlin and D. Zeilberger, Maple program; Local copy
Kai Wang, Fibonacci Numbers And Trigonometric Functions Outline, (2019).
Roman Witula, Ramanujan Type Trigonometric Formulae, Demonstratio Math. 45 (2012), 779-796.
Index entries for linear recurrences with constant coefficients, signature (3,0,-1).
FORMULA
a(n) = 3*A147704(n).
a(n) = c(1)^(-n) + c(2)^(-n) + c(4)^(-n) = (-c(1)*c(2))^n + (-c(1)*c(4))^n + (-c(2)*c(4))^n, where c(j) := 2*cos(2*Pi*j/9).
G.f.: Sum_{n>=0} a(n)*x^n = 3-3*x*(x^2-1)/(1-3*x+x^3) = 3*(1-2*x)/(1-3*x+x^3).
G.f. of Edlin and Zeilberger (2000): 1+Sum_{n>=1} a(n)*x^n = 1-3*x*(x^2-1)/(1-3*x+x^3) = (1-2*x^3)/(1-3*x+x^3). - Petros Hadjicostas, Sep 13 2017
a(n) = ceiling(r^n) for n >= 1, where r = 1/A130880 is the largest root of x^3 - 3*x^2 + 1. - Tamas Lengyel, Feb 20 2022
EXAMPLE
For n=3, we have a(3) = 3^3 - 3 = 24 ternary cyclic compositions of n=3 (with fixed positions on the circle for the parts) that avoid 110 because we have to exclude 110, 101, and 011. - Petros Hadjicostas, Sep 13 2017
MATHEMATICA
LinearRecurrence[{3, 0, -1}, {3, 3, 9}, 50].
PROG
(PARI) x='x+O('x^99); Vec(3*(1-2*x)/(1-3*x+x^3)) \\ Altug Alkan, Sep 13 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Roman Witula, Aug 25 2012
STATUS
approved