Perhaps there is an error in my code then. I include 919 in the sequence, which leads to a few other divergences and a total of 2024 terms ending at 8945. -- Chuck ----- Original Message ----- From: "David Wilson" <davidwwilson@comcast.net> To: "math-fun" <math-fun@mailman.xmission.com>; <seqfan@ext.jussieu.fr> Sent: Tuesday, February 22, 2005 8:55 AM Subject: Re: [math-fun] True So Far Your values are correct as far as you gave them. At a(2021) = 8734, the largest digit count is 891 5's. None of the numbers 8735 through 8919 is in the sequence. None of the numbers 8920 or beyond are in the sequence, since their "count" parts exceed 891. So a(2021) = 8734 is the last element of the sequence. ----- Original Message ----- From: "Eric Angelini" <keynews.tv@skynet.be> To: <ham>; "math-fun" <math-fun@mailman.xmission.com>; <seqfan@ext.jussieu.fr> Sent: Tuesday, February 22, 2005 6:26 AM Subject: [math-fun] True So Far 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, Ã. -- No virus found in this outgoing message. Checked by AVG Anti-Virus. Version: 7.0.300 / Virus Database: 266.2.0 - Release Date: 2/21/2005