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.

A298678 Start with the hexagonal tile of the Shield tiling and recursively apply the substitution rule. a(n) is the number of hexagonal tiles after n iterations.

Original entry on oeis.org

1, 0, 7, 12, 73, 216, 919, 3204, 12409, 45408, 171271, 635580, 2379241, 8865000, 33113527, 123523572, 461111833, 1720661616, 6422058919, 23966525484, 89446140169, 333813840888, 1245817611991, 4649439829860, 17351975261881, 64758394108800, 241681735391047
Offset: 0

Views

Author

Felix Fröhlich, Jan 24 2018

Keywords

Comments

The following substitution rules apply to the tiles:
triangle with 6 markings -> 1 hexagon
triangle with 4 markings -> 1 square, 2 triangles with 4 markings
square -> 1 square, 4 triangles with 6 markings
hexagon -> 7 triangles with 6 markings, 3 triangles with 4 markings, 3 squares
For n > 0, a(n) is also the number of triangles with 6 markings after n iterations when starting with the hexagon.
a(n) is also the number of triangles with 6 markings after n iterations when starting with the triangle with 6 markings.
a(n) is also the number of hexagons after n iterations when starting with the triangle with 6 markings.

Links

Crossrefs

Cf. A298679, A298680, A298681, A298682, A298683.

Programs

  • PARI
    /* The function substitute() takes as argument a 4-element vector, where the first, second, third and fourth elements respectively are the number of triangles with 6 markings, the number of triangles with 4 markings, the number of squares and the number of hexagons that are to be substituted. The function returns a vector w, where the first, second, third and fourth elements respectively are the number of triangles with 6 markings, the number of triangles with 4 markings, the number of squares and the number of hexagons resulting from the substitution. */
substitute(v) = my(w=vector(4)); for(k=1, #v, while(v[1] > 0, w[4]++; v[1]--); while(v[2] > 0, w[3]++; w[2]=w[2]+2; v[2]--); while(v[3] > 0, w[3]++; w[1]=w[1]+4; v[3]--); while(v[4] > 0, w[1]=w[1]+7; w[2]=w[2]+3; w[3]=w[3]+3; v[4]--)); w
terms(n) = my(v=[0, 0, 0, 1], i=0); while(1, print1(v[4], ", "); i++; if(i==n, break, v=substitute(v)))
  • PARI
    Vec((1-2*x)/((1+2*x)*(1-4*x+x^2)) + O(x^40)) \\ Colin Barker, Jan 25 2018

Formula

G.f.: (1-2*x)/((1+2*x)*(1-4*x+x^2)). - Joerg Arndt, Jan 25 2018
13*a(n) = A077235(n) + 8*(-2)^n. - Bruno Berselli, Jan 25 2018
From Colin Barker, Jan 25 2018: (Start)
a(n) = (1/26)*((-1)^n*2^(4+n) + (5-2*sqrt(3))*(2-sqrt(3))^n + (2+sqrt(3))^n*(5+2*sqrt(3))).
a(n) = 2*a(n-1) + 7*a(n-2) - 2*a(n-3) for n>2.
(End)

Extensions

More terms from Colin Barker, Jan 25 2018

A298679 Start with the hexagonal tile of the Shield tiling and recursively apply the substitution rule. a(n) is the number of square tiles after n iterations.

Original entry on oeis.org

0, 3, 6, 33, 102, 423, 1494, 5745, 21102, 79431, 295086, 1103985, 4114710, 15367143, 57329286, 213999153, 798569022, 2980473543, 11122931934, 41512040625, 154923657702, 578185735911, 2157812994486, 8053078824945, 30054477139470, 112164880064583
Offset: 0

Views

Author

Felix Fröhlich, Jan 24 2018

Keywords

Comments

The following substitution rules apply to the tiles:
triangle with 6 markings -> 1 hexagon
triangle with 4 markings -> 1 square, 2 triangles with 4 markings
square -> 1 square, 4 triangles with 6 markings
hexagon -> 7 triangles with 6 markings, 3 triangles with 4 markings, 3 squares
a(n) is also the number of triangles with 4 markings after n+1 iterations when starting with the hexagonal tile.
a(n) is also the number of square tiles after n+1 iterations when starting with the hexagonal tile.

Links

Crossrefs

Cf. A298678, A298680, A298681, A298682, A298683.

Programs

  • Mathematica
    CoefficientList[Series[ 3x/((1+2x)(1-4x+x^2)) ,{x,0,40}],x] (* or *) LinearRecurrence[{2,7,-2},{0,3,6},40] (* Harvey P. Dale, Mar 02 2022 *)
  • PARI
    /* The function substitute() takes as argument a 4-element vector, where the first, second, third and fourth elements respectively are the number of triangles with 6 markings, the number of triangles with 4 markings, the number of squares and the number of hexagons that are to be substituted. The function returns a vector w, where the first, second, third and fourth elements respectively are the number of triangles with 6 markings, the number of triangles with 4 markings, the number of squares and the number of hexagons resulting from the substitution. */
substitute(v) = my(w=vector(4)); for(k=1, #v, while(v[1] > 0, w[4]++; v[1]--); while(v[2] > 0, w[3]++; w[2]=w[2]+2; v[2]--); while(v[3] > 0, w[3]++; w[1]=w[1]+4; v[3]--); while(v[4] > 0, w[1]=w[1]+7; w[2]=w[2]+3; w[3]=w[3]+3; v[4]--)); w
terms(n) = my(v=[0, 0, 0, 1], i=0); while(1, print1(v[3], ", "); i++; if(i==n, break, v=substitute(v)))
  • PARI
    concat(0, Vec(3*x / ((1 + 2*x)*(1 - 4*x + x^2)) + O(x^40))) \\ Colin Barker, Jan 25 2018

Formula

From Colin Barker, Jan 25 2018: (Start)
G.f.: 3*x / ((1 + 2*x)*(1 - 4*x + x^2)).
a(n) = (1/26)*(-3*(-1)^n*2^(2+n) + (6-5*sqrt(3))*(2-sqrt(3))^n + (2+sqrt(3))^n*(6+5*sqrt(3))).
a(n) = 2*a(n-1) + 7*a(n-2) - 2*a(n-3) for n>2.
(End)

Extensions

More terms from Colin Barker, Jan 25 2018

A298680 Start with the triangle with 4 markings of the Shield tiling and recursively apply the substitution rule. a(n) is the number of triangles with 6 markings after n iterations.

Original entry on oeis.org

0, 0, 4, 12, 56, 192, 756, 2748, 10408, 38544, 144452, 537900, 2009880, 7496160, 27985684, 104424732, 389756936, 1454515632, 5428480356, 20259056268, 75608443768, 282173320704, 1053087635252, 3930171627900, 14667610061160, 54740246247120, 204293419666564
Offset: 0

Views

Author

Felix Fröhlich, Jan 24 2018

Keywords

Comments

The following substitution rules apply to the tiles:
triangle with 6 markings -> 1 hexagon
triangle with 4 markings -> 1 square, 2 triangles with 4 markings
square -> 1 square, 4 triangles with 6 markings
hexagon -> 7 triangles with 6 markings, 3 triangles with 4 markings, 3 squares
a(n) is also the number of hexagonal tiles after n+1 iterations when starting with the triangle with 4 markings.

Links

Crossrefs

Cf. A298678, A298679, A298681, A298682, A298683.

Programs

  • PARI
    /* The function substitute() takes as argument a 4-element vector, where the first, second, third and fourth elements respectively are the number of triangles with 6 markings, the number of triangles with 4 markings, the number of squares and the number of hexagons that are to be substituted. The function returns a vector w, where the first, second, third and fourth elements respectively are the number of triangles with 6 markings, the number of triangles with 4 markings, the number of squares and the number of hexagons resulting from the substitution. */
substitute(v) = my(w=vector(4)); for(k=1, #v, while(v[1] > 0, w[4]++; v[1]--); while(v[2] > 0, w[3]++; w[2]=w[2]+2; v[2]--); while(v[3] > 0, w[3]++; w[1]=w[1]+4; v[3]--); while(v[4] > 0, w[1]=w[1]+7; w[2]=w[2]+3; w[3]=w[3]+3; v[4]--)); w
terms(n) = my(v=[0, 1, 0, 0], i=0); while(1, print1(v[1], ", "); i++; if(i==n, break, v=substitute(v)))
  • PARI
    concat(vector(2), Vec(4*x^2 / ((1 - x)*(1 + 2*x)*(1 - 4*x + x^2)) + O(x^40))) \\ Colin Barker, Jan 25 2018

Formula

From Colin Barker, Jan 25 2018: (Start)
G.f.: 4*x^2 / ((1 - x)*(1 + 2*x)*(1 - 4*x + x^2)).
a(n) = (1/39)*(-26 + (-1)^n*2^(3+n) - (2-sqrt(3))^n*(-9+sqrt(3)) + (2+sqrt(3))^n*(9+sqrt(3))).
a(n) = 3*a(n-1) + 5*a(n-2) - 9*a(n-3) + 2*a(n-4) for n>3.
(End)

Extensions

More terms from Colin Barker, Jan 25 2018

A298681 Start with the square tile of the Shield tiling and recursively apply the substitution rule. a(n) is the number of triangles with 6 markings after n iterations.

Original entry on oeis.org

0, 4, 4, 32, 80, 372, 1236, 4912, 17728, 67364, 248996, 934080, 3476400, 12993364, 48453364, 180907472, 675001760, 2519449092, 9402095556, 35090331232, 130956433168, 488740993844, 1823996357396, 6807266805360, 25405026124800, 94812927172324, 353846503607524
Offset: 0

Views

Author

Felix Fröhlich, Jan 24 2018

Keywords

Comments

The following substitution rules apply to the tiles:
triangle with 6 markings -> 1 hexagon
triangle with 4 markings -> 1 square, 2 triangles with 4 markings
square -> 1 square, 4 triangles with 6 markings
hexagon -> 7 triangles with 6 markings, 3 triangles with 4 markings, 3 squares

Links

Crossrefs

Cf. A298678, A298679, A298680, A298682, A298683.

Programs

  • PARI
    /* The function substitute() takes as argument a 4-element vector, where the first, second, third and fourth elements respectively are the number of triangles with 6 markings, the number of triangles with 4 markings, the number of squares and the number of hexagons that are to be substituted. The function returns a vector w, where the first, second, third and fourth elements respectively are the number of triangles with 6 markings, the number of triangles with 4 markings, the number of squares and the number of hexagons resulting from the substitution. */
substitute(v) = my(w=vector(4)); for(k=1, #v, while(v[1] > 0, w[4]++; v[1]--); while(v[2] > 0, w[3]++; w[2]=w[2]+2; v[2]--); while(v[3] > 0, w[3]++; w[1]=w[1]+4; v[3]--); while(v[4] > 0, w[1]=w[1]+7; w[2]=w[2]+3; w[3]=w[3]+3; v[4]--)); w
terms(n) = my(v=[0, 0, 1, 0], i=0); while(1, print1(v[1], ", "); i++; if(i==n, break, v=substitute(v)))
  • PARI
    concat(0, Vec(4*x*(1 - 2*x) / ((1 - x)*(1 + 2*x)*(1 - 4*x + x^2)) + O(x^40))) \\ Colin Barker, Jan 25 2018

Formula

From Colin Barker, Jan 25 2018: (Start)
G.f.: 4*x*(1 - 2*x) / ((1 - x)*(1 + 2*x)*(1 - 4*x + x^2)).
a(n) = (1/39)*(26 + (-1)^(1+n)*2^(5+n) + (3-9*sqrt(3))*(2-sqrt(3))^n + (2+sqrt(3))^n*(3+9*sqrt(3))).
a(n) = 3*a(n-1) + 5*a(n-2) - 9*a(n-3) + 2*a(n-4) for n>3.
(End)

Extensions

More terms from Colin Barker, Jan 25 2018

A298682 Start with the triangle with 4 markings of the Shield tiling and recursively apply the substitution rule. a(n) is the number of triangles with 4 markings after n iterations.

Original entry on oeis.org

1, 2, 4, 8, 28, 92, 352, 1280, 4828, 17900, 67024, 249680, 932716, 3479132, 12987904, 48464288, 180885628, 675045452, 2519361712, 9402270320, 35089981708, 130957132220, 488739595744, 1823999153600, 6807261212956, 25405037309612, 94812904802704
Offset: 0

Views

Author

Felix Fröhlich, Jan 24 2018

Keywords

Comments

The following substitution rules apply to the tiles:
triangle with 6 markings -> 1 hexagon
triangle with 4 markings -> 1 square, 2 triangles with 4 markings
square -> 1 square, 4 triangles with 6 markings
hexagon -> 7 triangles with 6 markings, 3 triangles with 4 markings, 3 squares
a(n) is also one more than the number of squares after n iterations when starting with the triangle with 4 markings.

Links

Crossrefs

Cf. A298678, A298679, A298680, A298681, A298683.

Programs

  • Mathematica
    LinearRecurrence[{3,5,-9,2},{1,2,4,8},40] (* Harvey P. Dale, Aug 20 2024 *)
  • PARI
    /* The function substitute() takes as argument a 4-element vector, where the first, second, third and fourth elements respectively are the number of triangles with 6 markings, the number of triangles with 4 markings, the number of squares and the number of hexagons that are to be substituted. The function returns a vector w, where the first, second, third and fourth elements respectively are the number of triangles with 6 markings, the number of triangles with 4 markings, the number of squares and the number of hexagons resulting from the substitution. */
substitute(v) = my(w=vector(4)); for(k=1, #v, while(v[1] > 0, w[4]++; v[1]--); while(v[2] > 0, w[3]++; w[2]=w[2]+2; v[2]--); while(v[3] > 0, w[3]++; w[1]=w[1]+4; v[3]--); while(v[4] > 0, w[1]=w[1]+7; w[2]=w[2]+3; w[3]=w[3]+3; v[4]--)); w
terms(n) = my(v=[0, 1, 0, 0], i=0); while(1, print1(v[2], ", "); i++; if(i==n, break, v=substitute(v)))
  • PARI
    Vec((1 + x)*(1 - 2*x - 5*x^2) / ((1 - x)*(1 + 2*x)*(1 - 4*x + x^2)) + O(x^40)) \\ Colin Barker, Jan 25 2018

Formula

From Colin Barker, Jan 25 2018: (Start)
G.f.: (1 + x)*(1 - 2*x - 5*x^2) / ((1 - x)*(1 + 2*x)*(1 - 4*x + x^2)).
a(n) = (1/13)*(26 + (-2)^n + (2+sqrt(3))^n*(-7+5*sqrt(3)) - (2-sqrt(3))^n*(7+5*sqrt(3))).
a(n) = 3*a(n-1) + 5*a(n-2) - 9*a(n-3) + 2*a(n-4) for n>3.
(End)

Extensions

More terms from Colin Barker, Jan 25 2018
Showing 1-5 of 5 results.

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