True So Far

[David Wasserman]

The sequence description doesn't make it clear that the terms are increasing.  I see that Eric's hand-calculated terms are increasing, and that Chuck's program computes an increasing sequence.  I started computing another version of this sequence, always using the smallest term that works, and I found the first decrease is at a(350) = 1002, a(351) = 920.  

The increasing sequence ends because after 8945, there isn't a next term ending with 5: 8955 would yield 896 fives, while 8965 or 8975 would yield 895 fives.  The non-monotonic version doesn't stop until all ten digits are simultaneously dead-ended.  The probability of this is small, but it doesn't seem to get smaller as the numbers get larger, so I guess this sequence does end after several billion terms.

>Hello math-fun and seqfan,
>
>I've just sent this to the OEIS :
>
>    10 12 13 14 15 16 17 18 19 20 23 24 25 26 27 28 29 30
>    34 35 36 37 38 39 40 45 46 47 48 49 50 56 57 58 59 60
>    67 68 69 70 78 79 80 89 90 90 102 103 104 105 106 107
>    108 109 112 113 114 115 116 117 118 119 123 124 125
>    126 127 128 129 134 135 136 137 138 139 145 146 147
>    148 149 156 157 158 159 167 168 169 178 179 180 189...
>
>[more hand calculated terms here (hope no errors)]:
>
>http://www.cetteadressecomportecinquantesignes.com/TrueSoFar.htm
>
>Description :
>
>    "True so far"-sequence. Last digit of a(n) must be seen
>    as a glyph and preceding digits as a quantity. So "10"
>    reads [one "0"] and "12" [one "2"] -- which are both true
>    statements: there is only one "0" glyph so far in the
>    sequence when [10] is read, and there is only one "2"
>    glyph when [12] is read. The sequence is built with
>    [a(n+1)-a(n)] being minimal and a(n+1) always "true so
>    far". This explains why integers [11], [21], [22], [31],
>    etc. are not in: their statements are false.
>
>    The nice substring ...1112,1113,1114,1115,1116,1117 1118...
>    appears in the sequence -- which means that so far the
>    whole sequence has used 111 "2", 111 "3", 111 "4", 111 "5",
>    111 "6", 111 "7" and 111 "8"...
>
>Question which ruined my sleep tonight:
>
>« Will the sequence ever stop? »
>
>... my intuition says yes...
>... could someone compute this and check for some more integers?
>
>Thanks,
>É.
>
>Attachment converted: Macintosh HD:Eric Angelini.vcf 5 (TEXT/ttxt) (0004B659)