cp's OEIS Frontend

Let M denotes the 4 X 4 matrix = row by row (1,1,1,1)(1,1,1,0)(1,1,0,0)(1,0,0,0) and A(n) the vector (x(n),y(n),z(n),t(n))=M^n*A where A is the vector (1,1,1,1) then a(n)=x(n). - Benoit Cloitre, Apr 02 2002

In general, the g.f. for p glass plates is A(x) = F_{p-1}(-x)/F_p(x) where F_p(x) = Sum_{k=0,p} (-1)^[(k+1)/2]*C([(p+k)/2],k)*x^k. - Paul D. Hanna, Feb 06 2006

a(n)/a(n-1) tends to 2.879385..., the longest diagonal of a nonagon with edge 1; or: sin(4*Pi/9)/sin(Pi/9). The sequence is the INVERT transform of (1, 3, 3, 5, 8, 13, 21, 34, 55, 89, 144, ...). - Gary W. Adamson, Jul 16 2015

The 3-wave sequence with initial values a, b, c is formed by the following construction:

a.......a+b+c............3a+5b+6c...

..b...b+c...a+2b+2c..2a+4b+5c...

....c..........a+2b+3c...

This sequence is related to the hendecagon or 11-gon, see A120747.

Sequence of perfect distributions for a cascade merge with six tapes according to Knuth. - Michael Somos, Feb 07 2004

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.

See A187506 for supporting theory. Define the matrix

U_3 = (0 0 0 1)

(0 0 1 1)

(0 1 1 1)

(1 1 1 1). Let r>=0 and M=(m_(i,j))=(U_3)^r, i,j=1,2,3,4. Let A_r be the r-th "block" defined by A_r={a(3*r-1),a(3*r),a(3*r+1),a(3*r+2)} with a(-1)=1. Note that A_r-2*A_(r-1)-3*A_(r-2)+A_(r-3)+A_(r-4)={0,0,0,0}, for r>=4. Let p={p_1,p_2,p_3,p_4}={-1,0,1,2} and n=3*r+p_i. Then a(n)=a(3*r+p_i)=m_(i,1), where M=(m_(i,j))=(U_3)^r was defined above. Hence the block A_r corresponds component-wise to the first column of M, and a(3*r+p_i)=m_(i,1) gives the quantity of H_(9,1,0) tiles that should appear in a subdivided H_(9,i,r) tile.

Since a(3*r+2)=a(3*(r+1)-1) for all r, this sequence arises by concatenation of first-column entries m_(2,1), m_(3,1) and m_(4,1) of M=(U_3)^r.

This sequence is a nontrivial extension of both A038197 and A187506.

Theory. (Start)

1. Definitions. Let T_(9,j,0) denote the rhombus with sides of unit length (=1), interior angles given by the pair (j*Pi/9,(9-j)*Pi/9) and Area(T_(9,j,0))=sin(j*Pi/9), j in {1,2,3,4}. Associated with T_(9,j,0) are its angle coefficients (j, 9-j) in which one coefficient is even while the other is odd. A half-tile is created by cutting T_(9,j,0) along a line extending between its two corners with even angle coefficient; let H_(9,j,0) denote this half-tile. Similarly, a T_(9,j,r) tile is a linearly scaled version of T_(9,j,0) with sides of length Q^r and Area(T_(9,j,r))=Q^(2*r)*sin(j*Pi/9), r>=0 an integer, where Q is the positive, constant square root Q=sqrt(x^3-2*x) with x=2*cos(Pi/9); likewise let H_(9,j,r) denote the corresponding half-tile. Often H_(9,i,r) (i in {1,2,3,4}) can be subdivided into an integral number of each equivalence class H_(9,j,0). But regardless of whether or not H_(9,j,r) subdivides, in theory such a proposed subdivision for each j can be represented by the matrix M=(m_(i,j)), i,j=1,2,3,4, in which the entry m_(i,j) gives the quantity of H_(9,j,0) tiles that should be present in a subdivided H_(9,i,r) tile. The number Q^(2*r) (the square of the scaling factor) is an eigenvalue of M=(U_3)^r, where

U_3=

(0 0 0 1)

(0 0 1 1)

(0 1 1 1)

(1 1 1 1).

2. The sequence. Let r>=0, and let D_r be the r-th "block" defined by D_r={a(3*r-1),a(3*r),a(3*r+1),a(3*r+2)} with a(-1)=0. Note that D_r-2*D_(r-1)-3*D_(r-2)+D_(r-3)+D_(r-4)={0,0,0,0}, for r>=4. Let p={p_1,p_2,p_3,p_4}={-1,0,1,2} and n=3*r+p_i. Then a(n)=a(3*r+p_i)=m_(i,4), where M=(m_(i,j))=(U_3)^r was defined above. Hence the block D_r corresponds component-wise to the fourth column of M, and a(3*r+p_i)=m_(i,4) gives the quantity of H_(9,4,0) tiles that should appear in a subdivided H_(9,i,r) tile. (End)

Combining blocks A_r, B_r, C_r and D_r, from A187503, A187504, A187505 and this sequence, respectively, as matrix columns [A_r,B_r,C_r,D_r] generates the matrix (U_3)^r, and a negative index (-1)*r yields the corresponding inverse [A_(-r),B_(-r),C_(-r),D(-r)]=(U_3)^(-r) of (U_3)^r. Therefore, the four sequences need not be causal.

Since U_3 is symmetric, so is M=(U_3)^r, so the block D_r also corresponds to the fourth row of M. Therefore, alternatively, for j=1,2,3,4, a(3r+p_j)=m_(4,j) gives the quantity of H_(9,j,0) tiles that should be present in a H_(9,4,r) tile.

Since a(3*r+2)=a(3*(r+1)-1) for all r>0, this sequence arises by concatenation of fourth-column entries m_(2,4), m_(3,4) and m_(4,4) (or fourth-row entries m_(4,2), m_(4,3) and m_(4,4)) from successive matrices M=(U_2)^r.

This sequence is a nontrivial extension of A038197.