

A107793


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


2



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
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OFFSET

0,1


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.


LINKS

Table of n, a(n) for n=0..99.


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[n1]): 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]}]


CROSSREFS

Cf. A000045, A000213, A000931.
Cf. also A059832, A275925, A305389, A305390, A305391.
Sequence in context: A069197 A021692 A222627 * A275275 A196402 A268741
Adjacent sequences: A107790 A107791 A107792 * A107794 A107795 A107796


KEYWORD

nonn


AUTHOR

Roger L. Bagula, Jun 11 2005


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).


STATUS

approved



