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A360190
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Starting from 1, successively take the smallest "Choix de Bruxelles" with factor 13 which is not already in the sequence.
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2
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1, 13, 133, 1333, 13333, 133333, 1333333, 125641, 1256413, 12564133, 1197241, 117481, 9037, 90391, 9031, 90313, 903133, 90241, 902413, 9024133, 90241333, 6941641, 693241, 6932413, 69324133, 6717241, 671557, 65557, 5557, 55591, 5431, 54313, 543133, 54241
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OFFSET
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0,2
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COMMENTS
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At a given term t, the Choix de Bruxelles with factor 13 can choose to multiply any decimal digit substring (not starting 0) of t by 13, or divide by 13 if that substring is divisible by 13.
These choices on substrings give various possible next values and here take the smallest not yet in the sequence.
The sequence is finite and ends at a(6851) = 7, since the sole next Choix there is multiplication by 13 to 91, but 91 is already in the sequence at the preceding a(6850) = 91.
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LINKS
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EXAMPLE
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Below, square brackets [] represent multiplication by 13 (e.g., [4] = 52); curly brackets {} represent division by 13 (e.g., {26} = 2); digits outside the brackets are not affected by the multiplication or division (e.g., 1[3] = 139 and 1{169} = 113).
We begin with 1 and, at each step, we go to the smallest number possible that hasn't yet appeared in the sequence:
1 --> [1] = 13
13 --> [1]3 = 133
133 --> [1]33 = 1333
1333 --> [1]333 = 13333
13333 --> [1]3333 = 133333
133333 --> [1]33333 = 1333333
1333333 --> 1{333333} = 125641
... and so on.
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PROG
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(C#) // See Links
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CROSSREFS
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KEYWORD
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nonn,base,fini,full
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AUTHOR
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STATUS
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approved
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