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).

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

1,1

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# ] &)]

Links

Table of n, a(n) for n=1..80.

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).

Example

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}

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}];

Crossrefs

Cf. A103684.

Sequence in context: A084426 A255049 A187564 * A257257 A257251 A068555

Adjacent sequences: A138058 A138059 A138060 * A138062 A138063 A138064

Keyword

nonn,uned,tabf

Author

Roger L. Bagula, May 02 2008

Status

approved