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

%I A107793 #21 Jun 22 2018 08:21:35
%S A107793 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,
%T A107793 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,
%U A107793 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
%N A107793 Differences between successive indices of 1's in the ternary tribonacci sequence A305390.
%C A107793 Average value is 4.38095...
%C A107793 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.
%p A107793 # 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.
%p A107793 with(ListTools); S := Array(0..30);
%p A107793 psi:=proc(T) Flatten(subs( {1=[2], 2=[3], 3=[1,2,3]}, T)); end;
%p A107793 S[0]:=[1];
%p A107793 for n from 1 to 16 do S[n]:=psi(S[n-1]): od:
%p A107793 # Get differences between indices of 1's in S:
%p A107793 Bag:=proc(S) local i,a; global DIFF; a:=[];
%p A107793 for i from 1 to nops(S) do if S[i]=1 then a:=[op(a),i]; fi; od:
%p A107793 DIFF(a); end;
%p A107793 Bag(S[16]);
%t A107793 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]}]
%Y A107793 Cf. A000045, A000213, A000931.
%Y A107793 Cf. also A059832, A275925, A305389, A305390, A305391.
%K A107793 nonn
%O A107793 0,1
%A A107793 _Roger L. Bagula_, Jun 11 2005
%E A107793 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).