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.

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A105105 Triangle read by rows, based on the morphism f: 1-> 2->3 3->{6,5,4} 4->5 5->6 6->{3,2,1}. First row is 1. If current row is a,b,c,..., then the next row is a,b,c,...,f(a),f(b),f(c),...

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 6, 5, 4, 1, 2, 2, 3, 2, 3, 3, 6, 5, 4, 2, 3, 3, 6, 5, 4, 3, 6, 5, 4, 6, 5, 4, 3, 2, 1, 6, 5, 1, 2, 2, 3, 2, 3, 3, 6, 5, 4, 2, 3, 3, 6, 5, 4, 3, 6, 5, 4, 6, 5, 4, 3, 2, 1, 6, 5, 2, 3, 3, 6, 5, 4, 3, 6, 5, 4, 6, 5, 4, 3, 2, 1, 6, 5, 3, 6, 5, 4, 6, 5, 4, 3, 2, 1, 6, 5, 6, 5
Offset: 0

Views

Author

Roger L. Bagula, Apr 07 2005

Keywords

Comments

Six-symbol substitution of the bi-Kenyon type.

Examples

			Triangle begins:
1,
1, 2,
1, 2, 2, 3,
1, 2, 2, 3, 2, 3, 3, 6, 5, 4,
1, 2, 2, 3, 2, 3, 3, 6, 5, 4, 2, 3, 3, 6, 5, 4, 3, 6, 5, 4, 6, 5, 4, 3, 2, 1, 6, 5,
...

Links

Crossrefs

Cf. A103684.

Programs

  • Mathematica
    s[n_] := n /. {1 -> 2, 2 -> 3, 3 -> {6, 5, 4}, 4 -> 5, 5 -> 6, 6 -> {3, 2, 1}}; t[a_] := Join[a, Flatten[s /@ a]]; Flatten[ NestList[t, {1}, 5]]
NestList[ Flatten[ Join[ #, # /. {1 -> {2}, 2 ->{3}, 3 ->{6,5,4}, 4 ->{5}, 5-> {6}, 6-> {3,2,1}} ]] &, {1}, 5] // Flatten (* Robert G. Wilson v, Jun 05 2014 *)

A105113 Triangle read by rows, based on the morphism f: 1->2, 2->3, 3->{3,5,5,5,4}, 4->5, 5->6, 6->{6,2,2,2,1}. First row is 1. If current row is a,b,c,..., then the next row is a,b,c,...,f(a),f(b),f(c),...

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 3, 5, 5, 5, 4, 1, 2, 2, 3, 2, 3, 3, 3, 5, 5, 5, 4, 2, 3, 3, 3, 5, 5, 5, 4, 3, 3, 5, 5, 5, 4, 3, 5, 5, 5, 4, 3, 5, 5, 5, 4, 6, 6, 6, 5, 1, 2, 2, 3, 2, 3, 3, 3, 5, 5, 5, 4, 2, 3, 3, 3, 5, 5, 5, 4, 3, 3, 5, 5, 5, 4, 3, 5, 5, 5, 4, 3, 5, 5, 5, 4, 6, 6, 6, 5, 2, 3, 3, 3, 5, 5
Offset: 0

Views

Author

Roger L. Bagula, Apr 07 2005

Keywords

Comments

This substitution with the polynomial that goes with it gives a new tile, not predicted in the Kenyon paper.
q=3 version of bi-Kenyon 6-symbol substitution.

Links

Crossrefs

Cf. A103684, A105112, A105111.

Programs

  • Mathematica
    s[n_] := n /. {1 -> 2, 2 -> 3, 3 -> {3, 5, 5, 5, 4}, 4 -> 5, 5 -> 6, 6 -> {6, 2, 2, 2, 1}}; t[a_] := Join[a, Flatten[s /@ a]] p[0] = {1}; p[1] = t[{1}]; Flatten[ NestList[t, {1}, 5]]

A119647 Fixed point of the morphism 1->{1,2}, 2->{1,3}, 3->{1}.

Original entry on oeis.org

1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3
Offset: 1

Views

Author

Robert G. Wilson v, Jul 27 2006

Keywords

Crossrefs

Cf. A103684.

Programs

  • Mathematica
    Nest[Flatten[Join[ #, # /. {1 -> {1, 2}, 2 -> {1, 3}}]] &, {1}, 5] // Flatten

Extensions

Name and data corrected by Michel Dekking, Dec 02 2019

A138061 This sequence is a triangular sequence formed by the substitution: ( French sideways graph) 1->1,2;2->3;3->4;4->1; as a Markov style substitution form. The result is the differential polynomial coefficient form. ( first zero omitted).

Original entry on oeis.org

2, 2, 6, 2, 6, 12, 2, 6, 12, 4, 2, 6, 12, 4, 5, 12, 2, 6, 12, 4, 5, 12, 7, 16, 27, 2, 6, 12, 4, 5, 12, 7, 16, 27, 10, 22, 36, 52, 2, 6, 12, 4, 5, 12, 7, 16, 27, 10, 22, 36, 52, 14, 30, 48, 68, 18, 2, 6, 12, 4, 5, 12, 7, 16, 27, 10, 22, 36, 52, 14, 30, 48, 68, 18, 19, 40, 63, 88, 23, 24
Offset: 1

Views

Author

Roger L. Bagula, May 02 2008

Keywords

Comments

Row sums are:
{0, 2, 8, 20, 24, 41, 91, 211, 389, 696, 1307}
This uses the French sideways graph method as in:
A103684:the morphism f: 1->{1,2}, 2->{1,3}, 3->{3}.
These sequences in the polynomial form were created to see what the
fractal implicit pictures would look like and not for the sequences:
Clear[a, s, p, t, m, n, t, p, k]
(* substitution *)
s[1] = {1, 2}; s[2] = {3}; s[3] = {4}; s[4] = {1};
t[a_] := Flatten[s /(AT) a];
p[0] = {1}; p[1] = t[p[0]];
p[n_] := t[p[n - 1]];
a = Table[p[n], {n, 0, 12}];
k = Table[D[Apply[Plus, Table[
a[[n]][[m]]*x^(m - 1), {m, 1, Length[a[[n]]]}]], x], {n, 3, 13}];
Clear[x, y, a, b, f, z, p];
nr = k /. x -> z;
p[z_] = Apply[Times, nr];
z = x + I*y;
f[x_, y_] = Re[1/(p[z])];
ContourPlot[ f[x, y], {x, -1.61,1.61}, {y, -1.61, 1.61}, PlotPoints -> {300, 300}, ImageSize ->600, ColorFunction -> (Hue[2# ] &)]

Examples

			First zero omitted:
{2},
{2, 6},
{2, 6, 12},
{2, 6, 12, 4},
{2, 6, 12, 4, 5, 12},
{2, 6, 12, 4, 5, 12, 7, 16, 27},
{2, 6, 12, 4, 5, 12, 7, 16, 27, 10, 22, 36, 52},
{2, 6, 12, 4, 5, 12, 7, 16, 27, 10, 22, 36, 52, 14, 30, 48, 68, 18},
{2, 6, 12, 4, 5, 12, 7, 16, 27, 10, 22, 36, 52, 14, 30, 48, 68, 18, 19, 40, 63, 88, 23, 24, 50},
{2, 6, 12, 4, 5, 12, 7, 16, 27, 10, 22, 36, 52, 14, 30, 48, 68, 18, 19, 40, 63, 88, 23, 24, 50, 26, 54, 84, 116, 30, 31, 64, 33, 68, 105}

Crossrefs

Cf. A103684.

Programs

  • Mathematica
    Clear[a, s, p, t, m, n] (* substitution *) s[1] = {1, 2}; s[2] = {3}; s[3] = {4}; s[4] = {1}; t[a_] := Flatten[s /@ a]; p[0] = {1}; p[1] = t[p[0]]; p[n_] := t[p[n - 1]]; a = Table[p[n], {n, 0, 10}]; Flatten[a]; b = Table[CoefficientList[D[Apply[Plus, Table[a[[n]][[m]]*x^(m - 1), {m, 1, Length[a[[n]]]}]], x], x], {n, 1, 11}]; Flatten[b] Table[Apply[Plus, CoefficientList[D[Apply[Plus, Table[a[[n]][[m]]*x^(m - 1), {m, 1, Length[a[[n]]]}]], x], x]], {n, 1, 11}];

Formula

( French sideways graph) 1->1,2;2->3;3->4;4->1; Substitution->p(x,n); out_n,m=Coefficients(dp(x,n)/dx).
Previous Showing 11-14 of 14 results.

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