A305390 -id:A305390 - cp's OEIS frontend

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

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).

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.

A059832 A ternary tribonacci triangle: form the triangle as follows: start with 3 single values: 1, 2, 3. Each succeeding row is a concatenation of the previous 3 rows.

Original entry on oeis.org

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, 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

Offset: 0

Jason Earls, Feb 25 2001

Alternatively, define a morphism f: 1 -> 2, 2 -> 3, 3 -> 1,2,3; let S(0)=1, S(k) = f(S(k-1)) for k>0; then sequence is the concatenation S(0) S(1) S(2) S(3) ...

			Rows 0, 1, 2, ..., 8, ... of the triangle are:
0, [1]
1, [2]
2, [3]
3, [1, 2, 3]
4, [2, 3, 1, 2, 3]
5, [3, 1, 2, 3, 2, 3, 1, 2, 3]
6, [1, 2, 3, 2, 3, 1, 2, 3, 3, 1, 2, 3, 2, 3, 1, 2, 3]
7, [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]
8, [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]
...

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

N. J. A. Sloane, Table of n, a(n) for n = 0..30121 (Roes 0 through 17, flattened.)
C. A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Zentralblatt review

Cf. A059835. Row sums A001590, row lengths A000213.

Rows 0,3,6,9,12,... converge to A305389, rows 1,4,7,10,... converge to A305390, and rows 2,5,8,11,... converge to A305391.

# To get successive rows of A059832
S:=Array(0..100);
S[0]:=[1];
S[1]:=[2];
S[2]:=[3];
for n from 3 to 12 do
S[n]:=[op(S[n-3]),op(S[n-2]), op(S[n-1])];
lprint(S[n]);
od: # N. J. A. Sloane, Jul 04 2018

a(n) = A059825(n) + 1. - Sean A. Irvine, Oct 11 2022

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

Entry revised by N. J. A. Sloane, Jun 21 2018

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.

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

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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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A059832 A ternary tribonacci triangle: form the triangle as follows: start with 3 single values: 1, 2, 3. Each succeeding row is a concatenation of the previous 3 rows.

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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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