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A138054
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Levels of substitution A059832 taken as polynomials p(x,n)]and coefficients of the differential polynomials returned as q(x,n) =dp(x,n)dx coefficients. ( first three zeros omitted).
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0
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2, 6, 3, 2, 6, 12, 1, 4, 9, 8, 15, 6, 14, 24, 2, 6, 6, 12, 5, 12, 21, 24, 9, 20, 33, 24, 39, 14, 30, 48, 3, 2, 6, 12, 15, 6, 14, 24, 18, 30, 11, 24, 39, 14, 30, 48, 34, 54, 19, 40, 63, 66, 23, 48, 75, 52, 81, 28, 58, 90, 1, 4, 9, 8, 15, 6, 14, 24, 9, 20, 33, 24, 39, 14, 30, 48, 51, 18
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| Row sums are: (with zeros) {0, 0, 0, 8, 23, 81, 305, 1027, 3514, 12002, 40658, ...}.
This sequence is as a result of my Pc Mandelbrot-Julia work.
I noticed that these substitution levels
increased like iteration polynomials,
so I converted the substitution levels to polynomials.
To get a good implicit plot I have been using the inverse of the differential in polynomials as a product.
So I used that kind of procedure to get the differentiation of a substitution.
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FORMULA
| p(x,n)=Sum[A059832[n,m]*t(m-1),{m,1,n}]; q(x,n)=dp(x,n)dx; out_n,m=Coefficients(q(x,n).
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EXAMPLE
| Three zeros then:
{2, 6},
{3, 2, 6, 12},
{1, 4, 9, 8, 15, 6, 14, 24},
{2, 6, 6, 12, 5, 12, 21, 24, 9, 20, 33, 24, 39, 14, 30, 48},
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MATHEMATICA
| Clear[a, s, p, t, m, n] (* substitution *) s[1] = {2}; s[2] = {3}; s[3] = {1, 2, 3}; t[a_] := Flatten[s /@ a]; p[0] = {1}; p[1] = t[p[0]]; p[n_] := t[p[n - 1]]; (*A059832*) 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}];
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CROSSREFS
| Cf. A059832, A105256.
Sequence in context: A128203 A144531 A086357 * A198418 A086360 A066098
Adjacent sequences: A138051 A138052 A138053 * A138055 A138056 A138057
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KEYWORD
| nonn,uned,tabf
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AUTHOR
| Roger L. Bagula and Gary Adamson (rlbagulatftn(AT)yahoo.com), May 02 2008
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