cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A052931 Expansion of 1/(1 - 3*x^2 - x^3).

Original entry on oeis.org

1, 0, 3, 1, 9, 6, 28, 27, 90, 109, 297, 417, 1000, 1548, 3417, 5644, 11799, 20349, 41041, 72846, 143472, 259579, 503262, 922209, 1769365, 3269889, 6230304, 11579032, 21960801, 40967400, 77461435, 144863001, 273351705, 512050438, 964918116, 1809503019
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

Let A be the tridiagonal unit-primitive matrix (see [Jeffery]) A = A_{9,1} = [0,1,0,0; 1,0,1,0; 0,1,0,1; 0,0,1,1]. Then a(n)=[A^n](2,3). - _L. Edson Jeffery, Mar 19 2011
From Wolfdieter Lang, Oct 02 2013: (Start)
This sequence a(n) appears in the formula for the nonnegative powers of the algebraic number rho(9) := 2*cos(Pi/9) of degree 3, the ratio of the smallest diagonal/side in the regular 9-gon, in terms of the power basis of the algebraic number field Q(rho(9)) (see A187360, n=9).
rho(9)^n = A(n)*1 + B(n)*rho(9) + C(n)*rho(9)^2, with A(0) = 1, A(1) = 0, A(n) = B(n-2), n >= 2, B(0) = 0, B(n) = a(n-1), n >= 1, C(0) = 0, C(n) = B(n-1), n >= 1. (End)

Examples

			From _Wolfdieter Lang_, Oct 02 2013: (Start)
In the 9-gon (enneagon), powers of rho(9) = 2*cos(pi/9):
rho(9)^5 = A(5)*1 + B(5)*rho(9) + C(5)*rho(9)^2, with A(5) = B(3) = a(2) = 3, B(5) = a(4) = 9 and C(5) = B(4) = a(3) = 1:
  rho(9)^5 = 3 + 9*rho(9) + rho(9)^2. (End)

Links

Crossrefs

Cf. A214699.

Programs

  • GAP
    a:=[1,0,3];; for n in [4..40] do a[n]:=3*a[n-2]+a[n-3]; od; a; # G. C. Greubel, Oct 17 2019
  • Magma
    R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( 1/(1-3*x^2-x^3) )); // G. C. Greubel, Oct 17 2019
  • Maple
    spec := [S,{S=Sequence(Prod(Z,Union(Z,Z,Z,Prod(Z,Z))))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);
seq(coeff(series(1/(1-3*x^2-x^3), x, n+1), x, n), n = 0..40); # G. C. Greubel, Oct 17 2019
  • Mathematica
    CoefficientList[Series[1/(1-3x^2-x^3),{x,0,40}],x] (* or *) LinearRecurrence[{0,3,1},{1,0,3},40] (* Vladimir Joseph Stephan Orlovsky, Jan 28 2012 *)
  • PARI
    x='x+O('x^40); Vec(1/(1-3*x^2-x^3)) \\ Altug Alkan, Feb 20 2018
  • Sage
    def A052931_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P(1/(1-3*x^2-x^3)).list()
A052931_list(40) # G. C. Greubel, Oct 17 2019

Formula

G.f.: 1/(1-3*x^2-x^3).
a(n) = 3*a(n-2) + a(n-3), with a(0)=1, a(1)=0, a(2)=3.
a(n) = Sum_{alpha=RootOf(-1+3*z^2+z^3)} (1/9)*(-1 +5*alpha +2*alpha^2) * alpha^(-1-n).
a(n) = Sum_{k=0..floor(n/2)} binomial(k, n-2k)3^(3k-n). - Paul Barry, Oct 04 2004
a(n) = A187497(3*(n+1)). - L. Edson Jeffery, Mar 19 2011.
3*a(n) = abs(A214699(n+1)). - Roman Witula, Oct 06 2012

Extensions

More terms from James Sellers, Jun 06 2000

A187498 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} = {-3,0,1,2}, n=3*r+p_i, and define a(-3)=0. Then a(n)=a(3*r+p_i) gives the quantity of H_(9,4,0) tiles in a subdivided H_(9,i,r) tile after linear scaling by the factor Q^r, where Q=sqrt(2*cos(Pi/9)).

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 1, 1, 2, 1, 3, 3, 4, 4, 6, 5, 10, 10, 14, 15, 20, 20, 34, 35, 48, 55, 69, 75, 117, 124, 165, 199, 241, 274, 406, 440, 571, 714, 846, 988, 1417, 1560, 1988, 2548, 2977, 3536, 4965, 5525, 6953, 9061, 10490, 12597, 17443, 19551
Offset: 0

Views

Author

L. Edson Jeffery, Mar 16 2011

Keywords

Comments

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(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_1)^r, where
U_1=
(0 1 0 0)
(1 0 1 0)
(0 1 0 1)
(0 0 1 1).
2. The sequence. Let r>=0, and let D_r be the r-th "block" defined by D_r={a(3*r-3),a(3*r),a(3*r+1),a(3*r+2)} with a(-3)=0. Note that D_r-D_(r-1)-3*D_(r-2)+2*D_(r-3)+D_(r-4)={0,0,0,0}, for r>=4, with initial conditions {D_k}={{0,0,0,1},{0,0,1,1},{0,1,1,2},{1,1,3,3}}, k=0,1,2,3. Let p={p_1,p_2,p_3,p_4}={-3,0,1,2} and n=3*r+p_i. Then a(n)=a(3*r+p_i)=m_(i,4), where M=(m_(i,j))=(U_1)^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 A187495, A187496, A187497 and this sequence, respectively, as matrix columns [A_r,B_r,C_r,D_r] generates the matrix (U_1)^r, and a negative index (-1)*r yields the corresponding inverse [A_(-r),B_(-r),C_(-r),D_(-r)]=(U_1)^(-r) of (U_1)^r.. Therefore the four sequences need not be causal.
Since U_1 is symmetric, so is M=(U_1)^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)=a(3*(r+1)-3) for all r, 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_1)^r.

References

  • L. E. Jeffery, Unit-primitive matrices and rhombus substitution tilings, (in preparation).

Links

Crossrefs

Cf. A187495, A187496, A187497.

Programs

  • Maple
    A052931 := proc(n) if n < 0 then 0; else coeftayl(1/(1-3*x^2-x^3),x=0,n) ; end if; end proc:
A052931a := proc(n) if n mod 3 = 0 then A052931(n/3) ; else 0 ; end if; end proc:
A057078 := proc(n) op(1+(n mod 3),[1,0,-1]) ; end proc:
A187498 := proc(n) -A057078(n) +A052931a(n) +2*A052931a(n-2) +A052931a(n-3) +3*A052931a(n-4) +2*A052931a(n-5) +A052931a(n-6) +3*A052931a(n-7) -A052931a(n-8) ; %/3 ; end proc:
seq(A187498(n),n=0..20) ; # R. J. Mathar, Mar 22 2011
  • Mathematica
    CoefficientList[Series[-x^2*(1 + x)*(x^6 + 3*x^4 + 2*x^2 + 1)/((1 + x + x^2)*(x^9 + 3*x^6 - 1)), {x, 0, 1000}], x] (* G. C. Greubel, Sep 23 2017 *)
  • PARI
    x='x+O('x^50); Vec(-x^2*(1+x)*(x^6+3*x^4+2*x^2+1)/((1+x+x^2)*(x^9+3*x^6-1))) \\ G. C. Greubel, Sep 23 2017

Formula

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

A188048 Expansion of (1 - x^2)/(1 - 3*x^2 - x^3).

Original entry on oeis.org

1, 0, 2, 1, 6, 5, 19, 21, 62, 82, 207, 308, 703, 1131, 2417, 4096, 8382, 14705, 29242, 52497, 102431, 186733, 359790, 662630, 1266103, 2347680, 4460939, 8309143, 15730497, 29388368, 55500634, 103895601, 195890270, 367187437, 691566411, 1297452581
Offset: 0

Views

Author

L. Edson Jeffery, Mar 19 2011

Keywords

Comments

Sequence is related to rhombus substitution tilings.

Links

Crossrefs

Cf. A052931.

Programs

  • Magma
    I:=[1,0,2,1]; [n le 4 select I[n] else Self(n-1)+3*Self(n-2)-2*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jun 22 2015
  • Maple
    F:= gfun:-rectoproc({a(n)=3*a(n-2)+a(n-3),a(0)=1,a(1)=0,a(2)=2},a(n),remember):
map(F, [$0..100]); # Robert Israel, Jun 21 2015
  • Mathematica
    CoefficientList[Series[(1-x^2)/(1-3x^2-x^3),{x,0,40}],x]  (* Harvey P. Dale, Mar 31 2011 *)
LinearRecurrence[{0,3,1}, {1,0,2}, 50] (* Roman Witula, Aug 20 2012 *)
  • PARI
    abs(polsym(1-3*x+x^3,66)/3) /* Joerg Arndt, Aug 19 2012 */

Formula

G.f.: (1 - x^2)/(1 - 3*x^2 - x^3).
a(n) = 3*a(n-2)+a(n-3), for n>=3, with a(0)=1, a(1)=0, a(2)=2.
a(n) = a(n-1)+3*a(n-2)-2*a(n-3)-a(n-4), for n>=4, with {a(k)}={1,0,2,1}, k=0,1,2,3.
a(n) = A187497(3*n+1).
a(n) = m_(3,3), where (m_(i,j)) = (U_1)^n, i,j=1,2,3,4 and U_1 is the tridiagonal unit-primitive matrix [0, 1, 0, 0; 1, 0, 1, 0; 0, 1, 0, 1; 0, 0, 1, 1].
3*(-1)^n*a(n) = A215664(n). - Roman Witula, Aug 20 2012
a(2n) = A094831(n); a(2n+1) = A094834(n). - John Blythe Dobson, Jun 20 2015
a(n) = A052931(n)-A052931(n-2). - R. J. Mathar, Nov 03 2020
a(n) = (2^n/3)*(cos^n(Pi/9) + cos^n(5*Pi/9) + cos^n(7*Pi/9)). - Greg Dresden, Sep 24 2022

A187495 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} = {-3,0,1,2}, n=3*r+p_i, and define a(-3)=1. Then a(n)=a(3*r+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(2*cos(Pi/9)).

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 1, 0, 2, 0, 1, 0, 3, 1, 5, 1, 4, 1, 9, 5, 14, 6, 14, 7, 28, 20, 42, 27, 48, 34, 90, 75, 132, 109, 165, 143, 297, 274, 429, 417, 571, 560, 1000, 988, 1429, 1548, 1988, 2108, 3417, 3536, 4846, 5644, 6953, 7752, 11799, 12597
Offset: 0

Views

Author

L. Edson Jeffery, Mar 17 2011

Keywords

Comments

(Start) See A187498 for supporting theory. Define the matrix
U_1=
(0 1 0 0)
(1 0 1 0)
(0 1 0 1)
(0 0 1 1).
Let r>=0, and let A_r be the r-th "block" defined by A_r={a(3*r-3),a(3*r),a(3*r+1),a(3*r+2)} with a(-3)=1. Note that A_r-A_(r-1)-3*A_(r-2)+2*A_(r-3)+A_(r-4)={0,0,0,0}, for r>=4, with initial conditions {A_k}={{1,0,0,0},{0,1,0,0},{1,0,1,0},{0,2,0,1}}, k=0,1,2,3. Let p={p_1,p_2,p_3,p_4}={-3,0,1,2}, n=3*r+p_i and M=(m_(i,j))=(U_1)^r, i,j=1,2,3,4. Then A_r corresponds component-wise to the first column of M, and a(n)=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. (End)
Since a(3*r)=a(3*(r+1)-3) for all r, this sequence arises by concatenation of first-column entries m_(2,1), m_(3,1) and m_(4,1) from successive matrices M=(U_1)^r.
This sequence is a nontrivial extension of A187496.

References

  • L. E. Jeffery, Unit-primitive matrices and rhombus substitution tilings, (in preparation).

Links

Crossrefs

Cf. A187496, A187497, A187498.

Programs

  • Magma
    I:=[0,0,0,1,0,0,0,1,0,2,0,1]; [n le 12 select I[n] else Self(n-3) + 3*Self(n-6) - 2*Self(n-9) - Self(n-12): n in [1..50]]; // G. C. Greubel, Apr 20 2018
  • Mathematica
    LinearRecurrence[{0,0,1,0,0,3,0,0,-2,0,0,-1}, {0,0,0,1,0,0,0,1,0,2,0,1}, 50] (* G. C. Greubel, Apr 20 2018 *)
  • PARI
    x='x+O('x^50); concat([0,0,0], Vec(x^3*(1-x^3+x^4-x^6-x^7+x^8)/(1-x^3-3*x^6+2*x^9+x^12))) \\ G. C. Greubel, Apr 20 2018

Formula

Recurrence: a(n) = a(n-3) +3*a(n-6) -2*a(n-9) -a(n-12), for n>=12, with initial conditions {a(m)}={0,0,0,1,0,0,0,1,0,2,0,1}, m=0,1,...,11.
G.f.: x^3*(1-x^3+x^4-x^6-x^7+x^8)/(1-x^3-3*x^6+2*x^9+x^12).

A187496 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} = {-3,0,1,2}, n=3*r+p_i, and define a(-3)=0. Then a(n)=a(3*r+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(2*cos(Pi/9)).

Original entry on oeis.org

1, 0, 0, 0, 1, 0, 2, 0, 1, 0, 3, 1, 5, 1, 4, 1, 9, 5, 14, 6, 14, 7, 28, 20, 42, 27, 48, 34, 90, 75, 132, 109, 165, 143, 297, 274, 429, 417, 571, 560, 1000, 988, 1429, 1548, 1988, 2108, 3417, 3536, 4846, 5644, 6953, 7752, 11799, 12597, 16645
Offset: 0

Views

Author

L. Edson Jeffery, Mar 17 2011

Keywords

Comments

(Start) See A187498 for supporting theory. Define the matrix
U_1=
(0 1 0 0)
(1 0 1 0)
(0 1 0 1)
(0 0 1 1).
Let r>=0, and let B_r be the r-th "block" defined by B_r={a(3*r-3),a(3*r),a(3*r+1),a(3*r+2)} with a(-3)=0. Note that B_r-B_(r-1)-3*B_(r-2)+2*B_(r-3)+B_(r-4)={0,0,0,0}, for r>=4, with initial conditions {B_k}={{0,1,0,0},{1,0,1,0},{0,2,0,1},{2,0,3,1}}, k=0,1,2,3. Let p={p_1,p_2,p_3,p_4}={-3,0,1,2}, n=3*r+p_i and M=(m_(i,j))=(U_1)^r, i,j=1,2,3,4. Then B_r corresponds component-wise to the second column of M, and a(n)=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. (End)
Since a(3*r)=a(3*(r+1)-3) for all r, this sequence arises by concatenation of second-column entries m_(2,2), m_(3,2) and m_(4,2) from successive matrices M=(U_1)^r.

References

  • L. E. Jeffery, Unit-primitive matrices and rhombus substitution tilings, (in preparation).

Links

Crossrefs

Cf. A187495, A187497, A187498.

Programs

  • Magma
    I:=[1,0,0,0,1,0,2,0,1,0,3,1]; [n le 12 select I[n] else Self(n-3) + 3*Self(n-6) - 2*Self(n-9) - Self(n-12): n in [1..50]]; // G. C. Greubel, Apr 20 2018
  • Mathematica
    LinearRecurrence[{0,0,1,0,0,3,0,0,-2,0,0,-1}, {1,0,0,0,1,0,2,0,1,0,3,1}, 50] (* G. C. Greubel, Apr 20 2018 *)
  • PARI
    x='x+O('x^50); Vec((1-x^3+x^4-x^6-x^7+x^8)/(1-x^3-3*x^6 +2*x^9 +x^12)) \\ G. C. Greubel, Apr 20 2018

Formula

Recurrence: a(n) = a(n-3) +3*a(n-6) -2*a(n-9) -a(n-12), for n>=12, with initial conditions {a(m)}={1,0,0,0,1,0,2,0,1,0,3,1}, m=0,1,...,11.
G.f.: (1-x^3+x^4-x^6-x^7+x^8)/(1-x^3-3*x^6+2*x^9+x^12).
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