[math-fun] True So Far
Chuck Seggelin
seqfan at plastereddragon.com
Tue Feb 22 15:03:19 CET 2005
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 at comcast.net>
To: "math-fun" <math-fun at mailman.xmission.com>; <seqfan at 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 at skynet.be>
To: <ham>; "math-fun" <math-fun at mailman.xmission.com>;
<seqfan at 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,
É.
--
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
More information about the SeqFan
mailing list