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We have a(n)=C(n;-1), A121449(n)=A(n;-1), A085810(n+1)=-B(n+1;-1), where A(n;d), B(n;d), and C(n;d), n in N, d in C, are so-called quasi-Fibonacci numbers defined and discussed in the comments to A121449 and in Witula-Slota-Warzynski's paper. It follows from formulas (3.47-49) in this paper that the value of A(n;1/3), B(n;1/3) and C(n;1/3) could be obtained from special convolution type identities for sequences a(n), A121449, and A085810.

Essentially a duplicate of A077998: a(n) = A077998(n-1). - Joerg Arndt, Aug 14 2015

a(n) appears in the formula for the nonnegative powers of sigma, the ratio of the smaller diagonal in the heptagon to the side length s=2*sin(Pi/7), when expressed in the basis <1,rho,sigma>, with rho = 2*cos(Pi/7), the ratio of the smaller heptagon diagonal to the side length, as follows. sigma^n = a(n-1)*1 + B(n)*rho + a(n)*sigma, n>=0, with B(n)=A006054(n). Put a(-1):= 1. See the Steinbach reference, and a comment under A052547.

a(n-1) is the top left entry of the n-th power of the 3X3 matrix [0, 1, 0; 1, 1, 1; 0, 1, 1] or of the 3X3 matrix [0, 0, 1; 0, 1, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014

[Empirical] a(base,n) = a(base-1,n) + A005191(n+1) for base >= 2*floor(n/2) + 1.

The Ramanujan-type sequence number 7 for the argument 2Pi/7 (see also A214683, A215112, A006053, A006054, A215076, A215100, A120757, A215560 for the numbers: 1, 1a, 2, 2a, 3-6 respectively). The sequence a(n) is one of the three special sequences (the remaining two are A215560 and A215572) connected with the following recurrence relation:

(c(1)^4/c(2))^(n/3) + (c(2)^4/c(4))^(n/3) + (c(4)^4/c(1))^(n/3) = at(n) + bt(n)*7^(1/3) + ct(n)*49^(1/3), where c(j):=2*cos(2*Pi*j/7), and the sequences at(n), bt(n), and ct(n) satisfy the following system of recurrence equations: at(n)=7*bt(n-2)+at(n-3),

bt(n)=7*ct(n-2)+bt(n-3), ct(n)=at(n-2)+ct(n-3), with at(0)=3, at(1)=at(2)=bt(0)=bt(1)=bt(2)=ct(0)=ct(1)=0, ct(2)=2 - for details see the Witula's first paper (see also A215560). It follows that a(n)=bt(3*n+1), at(3*n+1)=ct(3*n+1)=0, which implies the first formula below.

We note that all numbers a(n) are divided by 7.

a(n) = A052534(n+1) - 1.

The hendecagon is an 11-sided polygon. The preferred word in the OEIS is 11-gon.

The lengths of the diagonals of the regular 11-gon are r[k] = sin(k*Pi/11)/sin(Pi/11), 1 <= k <= 5, where r[1] = 1 is the length of the edge.

The value of limit(a(n)/a(n-1),n=infinity) equals the longest diagonal r[5].

The a(n) equal the matrix elements M^n[1,2], where M = Matrix([[1,1,1,1,1], [1,1,1,1,0], [1,1,1,0,0], [1,1,0,0,0], [1,0,0,0,0]]). The characteristic polynomial of M is (x^5 - 3x^4 - 3x^3 + 4x^2 + x - 1) with roots x1 = -r[4]/r[3], x2 = -r[2]/r[4], x3 = r[1]/r[2], x4 = r[3]/r[5] and x5 = r[5]/r[1].

Note that M^4*[1,0,0,0,0] = [55, 50, 41, 29, 15] which are all terms of the 5-wave sequence A038201. This is also the case for the terms of M^n*[1,0,0,0,0], n>=1.

Linear hydrocarbons are molecules made of carbon (C) and hydrogen (H) atoms organized without cycles.

a(n) <= A002986(n) because molecules can be acyclic but not linear (i.e., including carbon atoms bonded with more than two other carbons).

From Petros Hadjicostas, Nov 16 2019: (Start)

We prove Vaclav Kotesovec's conjectures from the Formula section. Let M = [[0,0,1], [0,1,1], [1,1,1]], X(n) = M^(n-2), and Y(n) = M^(floor(n/2)-2) = X(floor(n/2)) (with negative powers indicating matrix inverses). Let also, t_1 = [1,1,1]^T, t_2 = [1,2,2]^T, and t_3 = [1,2,3]^T. In addition, define b(n) = (1/2)*(t_1^T X(n) t_1) and c(n) = (1/2)*(t_3^T Y(n) t_1) if n is even and = (1/2)*(t_2^T Y(n) t_1) if n is odd.

We have a(n) = b(n) + c(n) for n >= 1. Since the characteristic polynomial of Vaclav Kotesovec's recurrence is x^9 - 2*x^8 - 3*x^7 + 5*x^6 + x^5 + 2*x^3 - 3*x^2 - x + 1 = g(x)*g(x^2), where g(x) = x^3 - 2*x^2 - x + 1, to prove his first conjecture, it suffices to show that b(n) - 2*b(n-1) - b(n-2) + b(n-3) = 0 (whose characteristic polynomial is g(x)) and c(n) - 2*c(n-2) - c(n-4) + c(n-6) = 0 (whose characteristic polynomial is g(x^2)).

Note that 2*b(n) = A006356(n-1) for n >= 1. (See the comments by L. Edson Jeffery and R. J. Mathar in the documentation of that sequence.) Also, 2*c(2*n) = A006356(n) and 2*c(2*n-1) = A006054(n+1) for n >= 1.

Properties of the polynomial g(x) = x^3 - 2*x^2 - x + 1 and its roots were studied by Witula et al. (2006) (see Corollary 2.4). This means that a(n) can essentially be expressed in terms of exp(I*2*Pi/7), but we omit the discussion. See also the comments for sequence A006054.

The characteristic polynomial of matrix M is g(x). By the Cayley-Hamilton theorem, 0 = g(M) = M^3 - 2*M^2 - M + I_3, and thus, for n >= 5, X(n) - 2*X(n-1) - X(n-2) + X(n-3) = M^(n-2) - 2*M^(n-3) - M^(n-4) + M^(n-5) = 0. Pre-multiplying by (1/2)*t_1^T and post-multiplying by t_1, we get that b(n) - 2*b(n-1) - b(n-2) + b(n-3) = 0 for n >= 5.

Similarly, for n >= 10, Y(n) - 2*Y(n-2) - Y(n-4) + Y(n-6) = X(floor(n/2)) - 2*X(floor((n-2)/2)) - X(floor((n-4)/2)) + X(floor((n-6)/2)) = X(floor(n/2)) - 2*X(floor(n/2) - 1) - X(floor(n/2) - 2) + X(floor(n/2) - 3) = 0. Pre-multiplying by (1/2)*t_3^T (when n is even) or by (1/2)*t_2^T (when n is odd), and post-multiplying by t_1, we get c(n) - 2*c(n-2) - c(n-4) + c(n-6) = 0 for n >= 10.

Since the characteristic polynomial of Vaclav Kotesovec's recurrence is g(x)*g(x^2), which is a polynomial of degree 9, the denominator of the g.f. of the sequence (a(n): n >= 1) should be x^9*g(1/x)*g(1/x^2) = (1 - 2*x - x^2 + x^3)*(1 - 2*x^2 - x^4 + x^6), as Vaclav Kotesovec conjectured below. The numerator of Vaclav Kotesovec's g.f. can be easily derived using the initial conditions (from a(1) = 1 to a(9) = 409). (End)

Modified versions of the generating function for D(0)={1,2,5,11,...}=A006054(m+2), m=0,1,2,..., are related to rhombus substitution tilings (see A187068, A187069 and A187070). The columns of the triangle have generating functions 1/(1-x), 2*x/(1-x)^2, x^2*(5-x)/(1-x)^3, x^3*(11-2*x-x^2)/(1-x)^4, x^4*(25-6*x-3*x^2)/(1-x)^5, ..., for which the sum of the signed coefficients in the n-th numerator equals 2^n. The diagonals {1,2,5,...}, {1,4,14,...}, ..., are generated by successive series expansion of F(n+1,x), n=0,1,..., where F(n,x)=1/(1-2*x-x^2+x^3)^n. For example, the second diagonal is {T{1,0},T{2,1},...}={1,4,14,...}=A189426, for which successive partial sums give A189427 (excluding the zero terms). Moreover, the diagonals correspond to successive convolutions of A006054 (= the first diagonal) with itself.

Second of a series of sequences of partial sums of (nonzero) diagonals of triangle A188106 whose diagonals correspond to successive convolutions of A006054 with itself, where the first such sequence of partial sums is given by A077850. For n=1,2,..., this series of sequences is generated by successive series expansion of 1/((1-x)*(1-2*x-x^2+x^3)^n), for which A077850 corresponds to n=1 and A189427 corresponds to n=2.

a(n)=Sum_{k=0..n} A189426(k), where A189426={0,0,1,4,14,42,119,322,...} is the convolution of A006054={0,0,1,2,5,11,25,56,126,...} with itself. Also, a(n+2)=Sum_{k=0..n} A188106{n+k+1,k}, n=0,1,2,....

We have a(n)=B(n;3), where B(n;d), n=1,2,..., d \in C, denote one of the quasi-Fibonacci numbers defined in the comments to A121449 and in the Witula-Slota-Warzynski paper. Its conjugate sequences A(n;3) and C(n;3) are discussed in A121458 and A215484 respectively. Similarly as in A121458 we deduce that each of the following elements a(3*n), a(3*n+1), a(3*n+2) is divided by 3*7^n for every n=0,1,... .