

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).


0



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
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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



