A187506 -id:A187506 - cp's OEIS frontend

A091024 Let v(0) be the column vector (1,0,0,0)'; for n>0, let v(n) = [1 1 1 1 / 1 1 1 0 / 1 1 0 0/ 1 0 0 0] v(n-1). Sequence gives third entry of v(n).

0, 1, 2, 7, 19, 56, 160, 462, 1329, 3828, 11021, 31735, 91376, 263108, 757588, 2181389, 6281058, 18085587, 52075371, 149945056, 431749580, 1243173370, 3579575053, 10306975580, 29677753369, 85453685055, 246054079584

Offset: 0

Gary W. Adamson, Dec 14 2003

nonn

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)

			A006357, A076264, a(n) and A006357 (offset) gives the 4 components of v(n) transposed:
1 0 0 0
1 1 1 1
4 3 2 1
10 9 7 4
30 26 19 10
85 75 56 30

Jay Kappraff, "Beyond Measure, A Guided Tour Through Nature, Myth and Number" (p. 497 gives the analogous case for the Heptagon).

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,3,-1,-1).

Cf. A006357, A076264.

a[n_] := (MatrixPower[{{1, 1, 1, 1}, {1, 1, 1, 0}, {1, 1, 0, 0}, {1, 0, 0, 0}}, n].{{1}, {0}, {0}, {0}})[[3, 1]]; Table[ a[n], {n, 0, 26}] (* Robert G. Wilson v, Feb 21 2005 *)
LinearRecurrence[{2,3,-1,-1},{0,1,2,7},30] (* Harvey P. Dale, Feb 18 2016 *)

Recurrence: a(n) = 2*a(n-1) + 3*a(n-2) - a(n-3) - a(n-4), with initial conditions {a(k)}={0,1,2,7}, k=0,1,2,3. - L. Edson Jeffery, Mar 15 2011
G.f.: x/(1 - 2*x - 3*x^2 + x^3 + x^4). - L. Edson Jeffery, Mar 15 2011
G.f.: Q(0)*x/(2+2*x) , where Q(k) = 1 + 1/(1 - x*(12*k-3 + x^2)/( x*(12*k+3 + x^2 ) - 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 12 2013

More terms from Robert G. Wilson v, Feb 21 2005

A187503 Let i be in {1,2,3,4} and let r >= 0 be an integer. Let p = {p_1, p_2, p_3, p_4} = {-1,0,1,2}, n=3r+p_i, and define a(-1)=1. Then a(n)=a(3r+p_i) gives the quantity of H_(9,1,0) tiles in a subdivided H_(9,i,r) tile after linear scaling by the factor Q^r, where Q=sqrt(x^3-2x) with x=2cos(Pi/9).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 3, 4, 7, 9, 10, 19, 26, 30, 56, 75, 85, 160, 216, 246, 462, 622, 707, 1329, 1791, 2037, 3828, 5157, 5864, 11021, 14849, 16886, 31735, 42756, 48620, 91376, 123111, 139997, 263108, 354484, 403104, 757588, 1020696, 1160693

Offset: 0

L. Edson Jeffery, Mar 15 2011

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.

Cf. A187504, A187505, A187506.

Recurrence: a(n)=2*a(n-3)+3*a(n-6)-a(n-9)-a(n-12), for n>=12, with initial conditions {a(k)}={0,0,0,0,0,1,1,1,1,2,3,4}, k=0,1,...,11.

G.f.: x^5(1+x+x^2-x^3+x^5-x^6)/(1-2*x^3-3*x^6+x^9+x^12).

A187504 Let i be in {1,2,3,4} and let r >= 0 be an integer. Let p = {p_1, p_2, p_3, p_4} = {-1,0,1,2}, n=3r+p_i, and define a(-1)=0. Then a(n)=a(3r+p_i) gives the quantity of H_(9,2,0) tiles in a subdivided H_(9,i,r) tile after linear scaling by the factor Q^r, where Q=sqrt(x^3-2x) with x=2cos(Pi/9).

Original entry on oeis.org

1, 0, 0, 0, 1, 1, 2, 2, 2, 4, 6, 7, 13, 17, 19, 36, 49, 56, 105, 141, 160, 301, 406, 462, 868, 1169, 1329, 2498, 3366, 3828, 7194, 9692, 11021, 20713, 27907, 31735, 59642, 80355, 91376, 171731, 231373, 263108, 494481, 666212, 757588

Offset: 0

L. Edson Jeffery, Mar 15 2011

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.

Cf. A187503, A187505, A187506.

Recurrence: a(n)=2*a(n-3)+3*a(n-6)-a(n-9)-a(n-12), for n>=12, with initial conditions {a(k)}={1,0,0,0,1,1,2,2,2,4,6,7}, k=0,1,...,11.

G.f.: (1-2*x^3+x^4+x^5-x^6+x^9-x^10)/(1-2*x^3-3*x^6+x^9+x^12).

A187505 Let i be in {1,2,3,4} and let r >= 0 be an integer. Let p = {p_1, p_2, p_3, p_4} = {-1,0,1,2}, n=3r+p_i, and define a(-1)=0. Then a(n)=a(3r+p_i) gives the quantity of H_(9,3,0) tiles in a subdivided H_(9,i,r) tile after linear scaling by the factor Q^r, where Q=sqrt(x^3-2x) with x=2cos(Pi/9).

Original entry on oeis.org

0, 1, 0, 1, 1, 1, 2, 3, 3, 6, 8, 9, 17, 23, 26, 49, 66, 75, 141, 190, 216, 406, 547, 622, 1169, 1575, 1791, 3366, 4535, 5157, 9692, 13058, 14849, 27907, 37599, 42756, 80355, 108262, 123111, 231373, 311728, 354484, 666212, 897585, 1020696

Offset: 0

L. Edson Jeffery, Mar 14 2011

(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.

Cf. A187503, A187504, A187506.

Recurrence: a(n)=2*a(n-3)+3*a(n-6)-a(n-9)-a(n-12), for n>=12, with initial conditions {a(k)}={0,1,0,1,1,1,2,3,3,6,8,9}, k=0,1,...,11.

G.f.: x*(1+x^2-x^3+x^4-2*x^6+x^7-x^8)/(1-2*x^3-3*x^6+x^9+x^12).

cp's OEIS Frontend

A091024 Let v(0) be the column vector (1,0,0,0)'; for n>0, let v(n) = [1 1 1 1 / 1 1 1 0 / 1 1 0 0/ 1 0 0 0] v(n-1). Sequence gives third entry of v(n).

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