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a(n)/a(n-1) tends to 2.801... = 1 + 2*cos(Pi/7).

A(n) := a(n+1)*(-1)^(n+1) appears in the following formula for the nonpositive powers of rho*sigma, where rho:=2*cos(Pi/7) and sigma:=sin(3*Pi/7)/sin(Pi/7) = rho^2-1 are the ratios of the smaller and larger diagonal length to the side length in a regular 7-gon (heptagon). See the Steinbach reference where the basis <1,rho,sigma> is used in an extension of the rational field. (rho*sigma)^(-n) = C(n) + B(n)*rho + A(n)*sigma, n >= 0, with C(n)= A085810(n)*(-1)^n, and B(n)= A181880(n-2)*(-1)^n. For the nonnegative powers see A120757(n), |A122600(n-1)| and A181879(n), respectively. See also a comment under A052547.

This sequence is constructible as a spiral tiling of similar trapezoids, as follows: start with an isosceles trapezoid with side lengths 3,1,4,1. Each new trapezoid is rotated and scaled so one leg fills all unoccupied space on the short base of the previous trapezoid. a(n) is given by the length of the n-th trapezoid's legs. This process is identical to the recursion relation added by R. J. Mathar in the Formula section. See the Links section for an illustration. - Andrew B. Hudson, Jun 19 2019

The g.f. is the transform of the g.f. of A006053 under the Chebyshev mapping G(x)-> (1/(1+x^2))G(x/(1+x^2)). The denominator of the g.f. is a parameterization of the Alexander polynomial of 7_1. It is also the 14th cyclotomic polynomial.

Riordan array (1/(1-x-2*x^2+x^3), x/(1-x-2*x^2+x^3)).

Subtriangle of triangle given by (0, 1, 2, -5/2, 1/10, 2/5, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Diagonal sums are A125691(n).

Row sums are A001654(n+1).

Mirror image of triangle in A188107.

Let u(k), v(k), w(k) be defined by u(1)=1, v(1)=0, w(1)=0 and u(k+1)=u(k)+v(k)+w(k), v(k+1)=u(k)+v(k), w(k+1)=u(k); then {u(n)} = 1,1,3,6,14,31,... (A006356 with an extra initial 1), {v(n)} = 0,1,2,5,11,25,... (this sequence with its initial 0 deleted) and {w(n)} = {u(n)} prefixed by an extra 0 = A077998 with an extra initial 0. - Benoit Cloitre, Apr 05 2002. Also u(k)^2+v(k)^2+w(k)^2 = u(2k). - Gary W. Adamson, Dec 23 2003

Form the graph with matrix A=[1, 1, 1; 1, 0, 0; 1, 0, 1]. Then A006054 counts walks of length n between the vertex of degree 1 and the vertex of degree 3. - Paul Barry, Oct 02 2004

Form the digraph with matrix [1,1,0; 1,0,1; 1,1,1]. A006054(n) counts walks of length n between the vertices with loops. - Paul Barry, Oct 15 2004

Nonzero terms = INVERT transform of (1, 1, 2, 2, 3, 3, ...). Example: 56 = (1, 1, 2, 5, 11, 25) dot (3, 3, 2, 2, 1, 1) = (3 + 3 + 4 + 10 + 11 + 25). - Gary W. Adamson, Apr 20 2009

-a(n+1) appears in the formula for the nonpositive powers of rho:= 2*cos(Pi/7), 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 sigma:=rho^2-1, the ratio of the larger heptagon diagonal to the side length, as follows. rho^(-n) = C(n)*1 + C(n-1)*rho - a(n+1)*sigma, n >= 0, with C(n)=A077998(n), C(-1):=0. See the Steinbach reference, and a comment under A052547.

If, with the above notations, the power basis of the field Q(rho) is taken one has for nonpositive powers of rho, rho^(-n) = a(n+2)*1 + A077998(n-1)*rho - a(n+1)*rho^2. For nonnegative powers see A006053. See also the Steinbach reference. - Wolfdieter Lang, May 06 2011

a(n) appears also in the nonnegative powers of sigma,(defined in the above comment, where also the basis is given). See a comment in A106803.

The sequence b(n):=(-1)^(n+1)*a(n) forms the negative part (i.e., with nonpositive indices) of the sequence (-1)^n*A006053(n+1). In this way we obtain what we shall call the Ramanujan-type sequence number 2a for the argument 2*Pi/7 (see the comment to Witula's formula in A006053). We have b(n) = -2*b(n-1) + b(n-2) + b(n-3) and b(n) * 49^(1/3) = (c(1)/c(4))^(1/3) * (c(1))^(-n) + (c(2)/c(1))^(1/3) * (c(2))^(-n) + (c(4)/c(2))^(1/3) * (c(4))^(-n) = (c(2)/c(1))^(1/3) * (c(1))^(-n+1) + (c(4)/c(2))^(1/3) * (c(2))^(-n+1) + (c(1)/c(4))^(1/3) * (c(4))^(-n+1), where c(j) := 2*cos(2*Pi*j/7) (for the proof, see the comments to A215112). - Roman Witula, Aug 06 2012

(1, 1, 2, 5, 11, 25, 56, ...) * (1, 0, 1, 0, 1, ...) = the variant of A006356: (1, 1, 3, 6, 14, 31, ...). - Gary W. Adamson, May 15 2013

The limit of a(n+1)/a(n) for n -> infinity is, for all generic sequences with this recurrence of signature (2,1,-1), sigma = rho^2-1, approximately 2.246979603, the length ratio (largest diagonal)/side in the regular heptagon (7-gon). For rho = 2*cos(Pi/7) and sigma see a comment above, and the P. Steinbach reference. Proof: a(n+1)/a(n) = 2 + 1/(a(n)/a(n-1)) - 1/((a(n)/a(n-1))*(a(n-1)/a(n-2))), leading in the limit to sigma^3 -2*sigma^2 - sigma + 1, which is solved by sigma = rho^2-1, due to C(7, rho) = 0 , with the minimal polynomial C(7, x) = x^3 - x^2 - 2*x + 1 of rho (see A187360). - Wolfdieter Lang, Nov 07 2013

Numbers of straight-chain aliphatic amino acids involving single, double or triple bonds (allowing adjacent double bonds) when cis/trans isomerism is neglected. - Stefan Schuster, Apr 19 2018

Let A(r,n) be the total number of ordered arrangements of an n+r tiling of r red squares and white tiles of total length n, where the individual tile lengths can range from 1 to n. A(r,0) corresponds to a tiling of r red squares only, and so A(r,0) = 1. Also, A(r,n)=0 for n<0. Let A_1(r,n) = Sum_{j=0..n} A(r,j). Then the expansion of 1/(1 - 2*x - x^2 + x^3) is A_1(0,n) + A_1(1,n-2) + A_1(n-4) + ... = a(n) without the initial two 0's. In general, the expansion of 1/(1 - 2*x -x^k + x^(k+1)) is equal to Sum_{j>=0} A_1(j, n-j*k). - Gregory L. Simay, May 25 2018

For n>1, a(n) is the number of ways to tile a strip of length n-1 with one color of squares and dominos, two colors of trominos and quadrominos, 3 colors of 5-minos and 6-minos, and so on. - Greg Dresden and Zhiyu Zhang, Jun 26 2025

Form the graph with matrix A = [0,1,1; 1,0,0; 1,0,1] (P_3 with a loop at an extremity). Then A028495 counts closed walks of length n at the degree 3 vertex. - Paul Barry, Oct 02 2004

Equals INVERT transform of (1, 1, 0, 1, 0, 1, 0, 1, ...). - Gary W. Adamson, Apr 28 2009

From Johannes W. Meijer, May 29 2010: (Start)

a(n) is the number of ways White can force checkmate in exactly (n+1) moves, n>=0, ignoring the fifty-move and the triple repetition rules, in the following chess position: White Ka1, Ra8, Bc1, Nb8, pawns a6, a7, b2, c6, d2, f6 and h6; Black Kc8, pawns b3, c7, d3, f7 and h7. (After Noam D. Elkies, see link; diagram 5).

Counts all paths of length n, n>=0, starting at the initial node on the path graph P_6, see the second Maple program. (End)

a(n) is the number of length n-1 binary words such that each maximal block of 1's has odd length. a(4) = 6 because we have: 000, 001, 010, 100, 101, 111. - Geoffrey Critzer, Nov 17 2012

a(n) is the number of compositions of n where increments can only appear at every second position, starting with the second and third part, see example. Also, a(n) is the number of compositions of n where there is no fall between every second pair of parts, starting with the first and second part; see example. - Joerg Arndt, May 21 2013

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

Range of row n of the circular Pascal array of order 7. - Shaun V. Ault, Jun 05 2014

a(n) is the number of compositions of n into parts from {1,2,4,6,8,10,...}. Example: a(4)= 6 because we have 4, 22, 211, 121, 112, and 1111. - Emeric Deutsch, Aug 17 2016

In general, a(n,m) = (2^n/(m+1))*Sum_{r=1..m} (1-(-1)^r)*cos(Pi*r/(m+1))^n*(1+cos(Pi*r/(m+1))) gives the number of paths of length n starting at the initial node on the path graph P_m. Here we have m=6. - Herbert Kociemba, Sep 15 2020

a(n-1) is the number of triangular dcc-polyominoes having area n (see Baril et al. at page 11). - Stefano Spezia, Oct 14 2023

a(n) is the number of permutations p of [n] with p(j)Alois P. Heinz, Mar 29 2024

Form the graph with matrix A=[0,1,1;1,0,0;1,0,1] (P_3 with a loop at an extremity). Then A052547 counts closed walks of length n at the degree 2 vertex. - Paul Barry, Oct 02 2004

The characteristic polynomial x^3 - x^2 - 2*x + 1 generates a 3 step recursion: a(0)=1,a(1)=0,a(2)=2, for n>2 a(n)=a(n-1)+2*a(n-2)-a(n-3) so we can also prepend the term 1,0 to a(n) and get the same sequence, i.e. start with a(0)=1,a(1)=0,a(2)=1. - Lambert Klasen (lambert.klasen(AT)gmx.net), Jan 30 2005

The length of the diagonals (including the side) of a regular 7-gon (heptagon) inscribed in a circle of radius r=1 are d_1=2*sin(Pi/7) (the side length), d_2=2*cos(Pi/7)*d_1, and d_3=2*sin(3*Pi/7). The two ratios are rho := R_2 = d_2/d_1 = 2*cos(Pi/7) approximately 1.801937736, and sigma:= R_3 = d_3/d_1 = S(2,rho) = rho^2-1, approximately 2.246979604. See A049310 for Chebyshev S-polynomials. See the Steinbach reference where the basis <1,rho,sigma> has been considered for an extension of the rational field Q, which is there called Q(rho). This rho is the largest zero of S(6,x). For nonnegative powers of rho one has rho^n = C(n)*1 + B(n)*rho + A(n)*sigma, with B(n)=a(n-1), a(-1):=0, a(-2):=1, A(n)=B(n+1)-B(n-1)= A006053(n), and C(n)=B(n-1)=a(n-2), n>=0. For negative powers see A106803 and -A006054. For nonnegative and negative powers of sigma see A006054, A106803 and a(n), -A006053, respectively.

a(n) appears also in the formula for the nonpositive powers of sigma (see the comment above for the definition and the Steinbach basis) as sigma^(-n) = a(n)*1 - A006053(n+1)*rho - a(n-1)*sigma, n>=0. Put a(-1):=0. 1/sigma=sigma-rho, the smallest positive zero of S(6,x) (see A049310 for Chebyshev S-polynomials). - Wolfdieter Lang, Dec 01 2010

The pair A094648 and the alternating sequence A033304 when joined form a two-sided sequence defined by the recurrence formula x(n+3) + x(n+2) - 2x(n+1) - x(n) = 0, n in Z, x(-1)=-2, x(0)=3, x(1)=-1 - for details see Witula's comments to A033304. - Roman Witula, Jul 25 2012

From Roman Witula, Aug 09 2012: (Start)

There exist two interesting subsequences b(n) and c(n) of the given above sequence x(n) defined by the following relations: b(n)=a(2^n) and c(n)=x(-2^n). These subsequences satisfy the following system of recurrence equations:

b(n+1)=b(n)^2-2*c(n), and c(n+1)=c(n)^2-2*b(n),

which easily follow from the general identity: x(n)^2=x(2*n)-2*x(-n), n in Z. We note that b(0)=-1, b(1)=5, b(2)=13, b(3)=117, c(0)=-2, c(1)=6, c(2)=26, c(3)=650. From the above system we deduce that all b(n) are odd, whereas all c(n) are even. Moreover we obtain c(n+1)-b(n+1)=(c(n)-b(n))*(b(n)+c(n)+2), which yields b(n+1)-c(n+1)=product{k=1,..,n}(b(k)+c(k)+2)=13*product{k=2,..,n}(b(k)+c(k)+2)=13^2*41*product{k=3,..,n}(b(k)+c(k)+2). It follows that b(n)-c(n) is divisible by 13^2*41 for every n=3,4,..., and after using the above system again each b(n) and c(n), for n=2,3,..., is divisible by 13. (End)

If we set W(n):=3*A077998(n)-A006054(n+1)-A006054(n), n=0,1,..., then a(n)=(W(n)^2-W(2*n))/2 and W(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 = (-1-c(1))^n + (-1-c(2))^n + (-1-c(4))^n, where c(j):=2*cos(2*Pi*j/7) - for the proof see Witula-Slota-Warzynski's paper. Moreover it follows from the comment at the top and from comments to A033304 that W(n+1)=A033304(n)=(-1)^(n+1)*x(-n-1). - Roman Witula, Aug 11 2012

The following trigonometric type identitities hold true: (1) -a(n-1)-a(n) = c(1)*c(2)^n + c(2)*c(4)^n + c(4)*c(1)^n and (2) a(n)-a(n+2) = c(4)*c(2)^(n+1) + c(1)*c(4)^(n+1) + c(2)*c(1)^(n+1), where a(-1)=-2 and c(j) is defined as above (see also the respective comment to A033304). For the proof see Remark 6 in Witula's paper. - Roman Witula, Aug 14 2012

It can be proved that A033304(n-1)*(-1)^n = (a(n)^2 - a(2*n))/2, n=1,2,... - Roman Witula, Sep 30 2012

With respect to the form of the trigonometric formulas describing a(n), we call this sequence the Berndt-type sequence number 19 for the argument 2*Pi/7. The A-numbers of other Berndt-type sequences numbers are given in below. - Roman Witula, Sep 30 2012

Ramanujan-type sequence number 1 for the argument 2Pi/7.

The discussed sequence is associated with the sequence A006053 (with respect to the similar trigonometric formulas describing both sequences). Indeed, we have 7^(1/3)*a(n) = (c(1)/c(2))^(1/3)*(2c(1))^n + (c(2)/c(4))^(1/3)*(2c(2))^n + (c(4)/c(1))^(1/3)*(2c(4))^n = (c(1)/c(2))^(1/3)*(2c(2))^(n+1) + (c(2)/c(4))^(1/3)*(2c(4))^(n+1) + (c(4)/c(1))^(1/3)*(2c(1))^(n+1), where c(j) := Cos(2Pi*j/7), which is "almost" the copy of the respective formula for A006053. From a(0), A006053(0) and a(1), A006053(1), (and again) A006053(0) we deduce the following attractive decompositions

x^3 - 7^(1/3)*x - 1 = (x - (c(1)/c(4))^(1/3))*(x - (c(2)/c(1))^(1/3))*(x - (c(4)/c(2))^(1/3)), and

x^3 - 49^(1/3)*x - 1 = (x - (c(1)/c(2))^(1/3)*2c(1))*(x - (c(2)/c(4))^(1/3)*2c(2))*(x - (c(4)/c(1))^(1/3)*2c(4)).

From Newton-Girard formulas applied to these polynomials we generate two new sequences of real numbers S(n) := (c(1)/c(4))^(n/3) + (c(2)/c(1))^(n/3) + (c(4)/c(2))^(n/3), and T(n) := ((c(1)/c(2))^(1/3)*2c(1))^n + ((c(2)/c(4))^(1/3)*2c(2))^n + ((c(4)/c(1))^(1/3)*2c(4))^n. In first Witula's paper it is proved that S(n) = as(n) + bs(n)*7^(1/3) + cs(n)*49^(1/3), where as(n+3) = as(n) + 7cs(n+1), bs(n+3) = bs(n) + as(n+1), cs(n+3) = cs(n) + bs(n+1), as(0)=3, as(1)=as(2)=bs(0)=bs(1)=0, bs(2)=2, cs(0)=cs(1)=cs(2)=0, and T(n) = at(n) + bt(n)*7^(1/3) + ct(n)*49^(1/3), where at(n+3) = at(n) + 7bt(n+1), bt(n+3) = bt(n) + 7ct(n+1), ct(n+3) = ct(n) + at(n+1), at(0)=3, at(1)=at(2)=bt(0)=bt(1)=bt(2)=ct(0)=ct(1)=0, ct(2)=2. All six sequences as(n),bs(n),...,ct(n) are created from integers and will be discussed in separate sequences .

All a(n) are divisible by 3.

The Ramanujan-type sequence number 1 for the argument 2*Pi/9 defined by the following identity:

3^(1/3)*a(n) = (c(1)/c(2))^(1/3)*c(1)^n + (c(2)/c(4))^(1/3)*c(2)^n + (c(4)/c(1))^(1/3)*c(4)^n = -( (c(1)/c(2))^(1/3)*c(2)^(n+1) + (c(2)/c(4))^(1/3)*c(4)^(n+1) + (c(4)/c(1))^(1/3)*c(1)^(n+1) ), where c(j) := 2*cos(2*Pi*j/9).

The definitions of other Ramanujan-type sequences, for the argument of 2*Pi/9 in one's, are given in the Crossrefs section.