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Search: id:A103685
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| A103685 |
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Consider the morphism 1->{1,2}, 2->{1,2}, 3->{1}; sequence gives number of each after n substitutions. |
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2
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| 1, 0, 0, 3, 1, 0, 8, 4, 1, 22, 12, 5, 62, 34, 17, 176, 96, 51, 500, 272, 147, 1420, 772, 419, 4032, 2192, 1191, 11448, 6224, 3383, 32504, 17672, 9607
(list; graph; listen)
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OFFSET
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0,4
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COMMENT
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It seems that this is really three sequences? If so it would be better to split it into 3 separate entries. - njas, Apr 03 2005
Counting "crows" using improved Rauzy substitution starting at a[0]={1}: counting "1"'s, "2"'s and "3"'s ( crows).
This substitution group software in Mathematica was originally developed to find probability of transforms for IFS tile programs. Starting at{1} instead of {1,2,3} gives a different result. I've added a Markov like functioning to the method as well. Here the for every third n counts: probability of transforms= {a(n),a(n+1) and a(n+2)}/Sum[a[n+i],{i,0,2}]
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EXAMPLE
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Probability of transforms after 10th iteration is: p(10)={32504, 17672, 9607}/(32504+ 17672+ 9607)
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MATHEMATICA
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s[1] = {1, 2}; s[2] = {1, 3}; s[3] = {1}; t[a_] := Join[a, Flatten[s /@ a]]; (* Markov recussion*) p[0] = {1}; p[1] = t[{1}]; p[n_] := t[p[n - 1]] a = Flatten[Table[Table[ Count[Flatten[Table[p[n], {n, 0, i}]], m], {m, 1, 3}], {i, 0, 10}]]
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CROSSREFS
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Cf. A073058, A103684.
Sequence in context: A143395 A090536 A052420 this_sequence A078521 A137432 A135871
Adjacent sequences: A103682 A103683 A103684 this_sequence A103686 A103687 A103688
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KEYWORD
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nonn,uned
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AUTHOR
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Roger Bagula (rlbagulatftn(AT)yahoo.com), Mar 26 2005
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