A059832 -id:A059832 - cp's OEIS frontend

A138054 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).

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

Offset: 1

Roger L. Bagula and Gary W. Adamson, May 02 2008

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.

			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},

Cf. A059832, A105256.

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}];

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).

A305389 A ternary tribonacci sequence: define the morphism f: 1 -> 2, 2 -> 3, 3 -> 1,2,3; let S[k] be result of applying f k times to 1, for k =- 0,1,2,...; sequence gives limit S[3k] as k -> oo.

Original entry on oeis.org

1, 2, 3, 2, 3, 1, 2, 3, 3, 1, 2, 3, 2, 3, 1, 2, 3, 2, 3, 1, 2, 3, 3, 1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 2, 3, 1, 2, 3, 3, 1, 2, 3, 2, 3, 1, 2, 3, 3, 1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 2, 3, 1, 2, 3, 3, 1, 2, 3, 2, 3, 1, 2, 3, 2, 3, 1, 2, 3, 3, 1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 2, 3, 1, 2, 3, 3

Offset: 0

N. J. A. Sloane, Jun 21 2018

nonn

N. J. A. Sloane, Table of n, a(n) for n = 0..25280

The three sequence A305389, A305390, A305391 together give the limiting forms of the rows of A059832.

See A316324, A316325, A316326 for indices of 1's, 2's, 3's respectively.

A305390 A ternary tribonacci sequence: define the morphism f: 1 -> 2, 2 -> 3, 3 -> 1,2,3; let S[k] be result of applying f k times to 1, for k =- 0,1,2,...; sequence gives limit S[3k+1] as k -> oo.

Original entry on oeis.org

2, 3, 1, 2, 3, 3, 1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 2, 3, 1, 2, 3, 3, 1, 2, 3, 2, 3, 1, 2, 3, 3, 1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 2, 3, 1, 2, 3, 3, 1, 2, 3, 2, 3, 1, 2, 3, 2, 3, 1, 2, 3, 3, 1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 2, 3, 1, 2, 3, 3, 1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 2, 3

Offset: 0

N. J. A. Sloane, Jun 21 2018

nonn

N. J. A. Sloane, Table of n, a(n) for n = 0..25280

The three sequence A305389, A305390, A305391 together give the limiting forms of the rows of A059832.

Used by A107793.

A305391 A ternary tribonacci sequence: define the morphism f: 1 -> 2, 2 -> 3, 3 -> 1,2,3; let S[k] be result of applying f k times to 1, for k =- 0,1,2,...; sequence gives limit S[3k+2] as k -> oo.

Original entry on oeis.org

3, 1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 2, 3, 1, 2, 3, 3, 1, 2, 3, 2, 3, 1, 2, 3, 2, 3, 1, 2, 3, 3, 1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 2, 3, 1, 2, 3, 3, 1, 2, 3, 2, 3, 1, 2, 3, 1, 2, 3, 2, 3, 1, 2, 3, 3, 1, 2, 3, 2, 3, 1, 2, 3, 2, 3, 1, 2, 3, 3, 1, 2, 3, 2, 3, 1, 2, 3, 1, 2

Offset: 0

N. J. A. Sloane, Jun 21 2018

nonn

N. J. A. Sloane, Table of n, a(n) for n = 0..13744

The three sequence A305389, A305390, A305391 together give the limiting forms of the rows of A059832.

A107793 Differences between successive indices of 1's in the ternary tribonacci sequence A305390.

Original entry on oeis.org

4, 5, 3, 5, 4, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5

Offset: 0

Roger L. Bagula, Jun 11 2005

nonn

Average value is 4.38095...

Conjecture (N. J. A. Sloane, Jun 22 2018) This is a disguised form of A275925. More precisely, if we replace the 5's by 6's and the 4's by 5's, and ignore the first two terms, we appear to get a sequence which is a shifted version of A275925.

Cf. A000045, A000213, A000931.

Cf. also A059832, A275925, A305389, A305390, A305391.

# From N. J. A. Sloane, Jun 22 2018. The value 16 can be replaced (in two places) by any number congruent to 1 mod 3.
with(ListTools); S := Array(0..30);
psi:=proc(T) Flatten(subs( {1=[2], 2=[3], 3=[1,2,3]}, T)); end;
S[0]:=[1];
for n from 1 to 16 do S[n]:=psi(S[n-1]): od:
# Get differences between indices of 1's in S:
Bag:=proc(S) local i,a; global DIFF; a:=[];
for i from 1 to nops(S) do if S[i]=1 then a:=[op(a),i]; fi; od:
DIFF(a); end;
Bag(S[16]);

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]] pp = p[13] a = Flatten[Table[If[pp[[j]] == 1, j, {}], {j, 1, Length[pp]}]] b = Table[a[[n]] - a[[n - 1]], {n, 2, Length[a]}]

Edited (and checked) by N. J. A. Sloane, Jun 21 2018 (the original version did not make it clear that this is based on only one of the three tribonacci sequences A305389, A305390, A305391).

A059835 Form triangle as follows: start with three single digits: 0, 1, 2. Each succeeding row is a concatenation of the previous three rows.

Original entry on oeis.org

0, 1, 2, 0, 1, 2, 1, 2, 0, 1, 2, 2, 0, 1, 2, 1, 2, 0, 1, 2, 0, 1, 2, 1, 2, 0, 1, 2, 2, 0, 1, 2, 1, 2, 0, 1, 2, 1, 2, 0, 1, 2, 2, 0, 1, 2, 1, 2, 0, 1, 2, 0, 1, 2, 1, 2, 0, 1, 2, 2, 0, 1, 2, 1, 2, 0, 1, 2, 2, 0, 1, 2, 1, 2, 0, 1, 2, 0, 1, 2, 1, 2, 0, 1, 2, 2, 0, 1, 2, 1, 2, 0, 1, 2, 1, 2, 0, 1, 2, 2, 0, 1, 2, 1, 2

Offset: 0

Jason Earls, Feb 25 2001

Trajectory of 0 under the morphism 0 -> 1, 1-> 2, 2 -> 012. - Robert G. Wilson v, May 20 2014

The sequence of row lengths is A000213. - Michael Somos, May 22 2014

			Triangle begins:
  0
  1
  2
  0 1 2
  1 2 0 1 2
  2 0 1 2 1 2 0 1 2
  ...

C. Pickover, Wonders of Numbers, Oxford University Press, NY, 2001, p. 273.

C. A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Zentralblatt review
Index entries for sequences that are fixed points of mappings

Cf. A059832.

T:= proc(n) option remember;
      `if`(n<3, n, seq(T(i), i=n-3..n-1))
    end:
seq(T(n), n=0..10);  # Alois P. Heinz, May 22 2014

NestList[ Flatten[# /. {0 -> {1}, 1 -> {2}, 2 -> {0, 1, 2}}] &, {0}, 8] // Flatten (* Robert G. Wilson v, May 20 2014 *)

a(n) = A059832(n) - 1. - Sean A. Irvine, Oct 11 2022

More terms from Larry Reeves (larryr(AT)acm.org), Feb 26 2001

cp's OEIS Frontend

A138054 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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A305389 A ternary tribonacci sequence: define the morphism f: 1 -> 2, 2 -> 3, 3 -> 1,2,3; let S[k] be result of applying f k times to 1, for k =- 0,1,2,...; sequence gives limit S[3k] as k -> oo.

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A305390 A ternary tribonacci sequence: define the morphism f: 1 -> 2, 2 -> 3, 3 -> 1,2,3; let S[k] be result of applying f k times to 1, for k =- 0,1,2,...; sequence gives limit S[3k+1] as k -> oo.

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A305391 A ternary tribonacci sequence: define the morphism f: 1 -> 2, 2 -> 3, 3 -> 1,2,3; let S[k] be result of applying f k times to 1, for k =- 0,1,2,...; sequence gives limit S[3k+2] as k -> oo.

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A107793 Differences between successive indices of 1's in the ternary tribonacci sequence A305390.

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A059835 Form triangle as follows: start with three single digits: 0, 1, 2. Each succeeding row is a concatenation of the previous three rows.

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