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

Author

Roger L. Bagula, Jun 11 2005

Keywords

• nonn

Comments

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.

Crossrefs

Cf. A000045, A000213, A000931.

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

Programs

Maple

# 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]);

Mathematica

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]}]

Extensions

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).