cp's OEIS Frontend

First entry of v(n) gives 1,1,4,10,30,85 = A006357 prefixed with an initial 1, the second entry gives 0,1,3,9,26,... = A076264 prefixed with an initial 0.

A sequence derived from 9-gonal diagonal ratios.

a(n)/a(n-1) converges to D = 2.879385... = longest 9-Gon diagonal with edge = 1. E.g., a(7)/a(6) = 707/246 = 2.873983...(a(n)/a(n-1) of all 4 columns converge to 2.8739...). For each row, left to right, terms converge upon 9-Gon ratios: (2.879...):(2.53208...):(1.87938...):(1) Example: row 7 = 707 622 462 246, from A006357, A076264, A091024 and A006357(offset), respectively. The ratios 707/246, 622/246, 462/246 and 246/246 are: (2.8739...):(2.528...):(1.87804...):(1)

From L. Edson Jeffery, Mar 15 2011: (Start)

In fact, the above ratios (2.8739...):(2.528...):(1.87804...):(1) converge to Q_3(w):Q_2(w):Q_1(w):Q_0(w), where the polynomials Q_r(w) are defined by Q_r(w)=w*Q_(r-1)(w)-Q_(r-2)(w) (r>1), Q_0(w)=1, Q_1(w)=w, and w=2*cos(Pi/9).

Moreover, this sequence and a variant of its g.f. are related to rhombus substitution tilings showing 9-fold rotational symmetry (cf. A187503, A187504, A187505, A187506). (End)

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 B_r be the r-th "block" defined by B_r={a(3*r-1),a(3*r),a(3*r+1),a(3*r+2)} with a(-1)=0. Note that B_r-2*B_(r-1)-3*B_(r-2)+B_(r-3)+B_(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,2), where M=(m_(i,j))=(U_3)^r was defined above. Hence the block B_r corresponds component-wise to the second column of M, and a(3*r+p_i)=m_(i,2) gives the quantity of H_(9,2,0) 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 second-column entries m_(2,2), m_(3,2) and m_(4,2) of M=(U_3)^r.

(Start) 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).

2. Let r>=0 and M=(m_(i,j))=(U_3)^r, i,j=1,2,3,4. Let C_r be the r-th "block" defined by C_r={a(3*r-1),a(3*r),a(3*r+1),a(3*r+2)} with a(-1)=0. Note that C_r-2*C_(r-1)-3*C_(r-2)+C_(r-3)+C_(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,3), where M=(m_(i,j))=(U_3)^r was defined above. Hence the block C_r corresponds component-wise to the third column of M, and a(3*r+p_i)=m_(i,3) gives the quantity of H_(9,3,0) tiles that should appear in a subdivided H_(9,i,r) tile. (End)

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

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.