About Sequence Number that Count

9/09/2015

SEQUENCE NUMBERS THAT COUNT:

We already saw the sequence number 998,001.  It produced a sequence of numbers that counted (without any errors) from “000” to “997”, in three digit strings.  To understand this better it helps to know that 998,001 = 999².  So let’s first look at its predecessors.

92 = 81.  81 is a sequence number that produces a counting sequence. 1/81 = 0.012345679… It counts from 0 to 7 (accurately), using one digit strings.

992 = 9,801.  9,801 is a sequence number that produces a counting sequence. 1/9801 = 0. 00  01  02  03  04  05  06  07  08  09  10  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  30  31  32  33  34  35  36  37  38  39  40  41  42  43  44  45  46  47  48  49  50  51  52  53  54  55  56  57   58 59  60  61  62  63  64  65  66  67  68  69  70  71  72  73  74  75  76  77  78  79  80  81  82  83  84  85  86  87  88  89  90  91  92  93  94  95  96  97  99  00  01  02  ... You will notice that it counts (in two digit strings) from “00” to “97” (without error), then it skips 98, does 99, and begins to repeat the whole sequence again.

9992 = 998,001.  We have done this sequence number previously.  In summary it counted in three digit strings, from “000” to “997” (accurately).

So, if I were a mathematician and recognized this pattern, I might try 9,999² next.

9,9992 = 99,980,001.  I will check to see if it produces a known number sequence. 1/99980001 = 0. 0000  0001  0002  0003  0004  0005  0006  0007  0008  0009  0010  0011  0012  0013  0014  0015  0016  0017  0018  0019  0020  0021  0022  0023  0024  0025  0026  0027  0028  0029  …

Well, I am not able to show this whole sequence using Wolfram Alpha.  However, after seeing the behavior of the previous sequence numbers I would guess that this sequence counts from “0000” to “9997”.  And that further sequences can be produced to count to 99,997, 999,997, 9,999,997 and beyond.  This could also be the basis of a proof to show that we could find a sequence number that could count as high as we needed it to count because there are an infinite number of sequences that fit this pattern.

Comment: 81 is a sequence number that produces a digital sequence that counts from 0 to 7 accurately.  If I want a sequence that counts to 10, I can use the sequence number 9801 (which produces a sequence that can count to 97 accurately).  If I want to count higher I can always find a sequence number that can count from 0 to whatever number I specify.  But we may also be able to find a sequence number that counts from zero to any desired number. Suppose I wanted a sequence that counts from zero to 10, that did not skip any number, and did not keep on counting past 10? I can simple create the string of digits I desire: “012345678910” and divide by a string of nines of the same digit length – in this case 12 nines.  But then I have some work to do because 012345678910/999999999999 does not simplify to a unit fraction.  But this fraction can be re-written as the sum of two or more unit fractions whose sum will produce the string I want.  In this case I will need nine such fractions.  (If you are fraction-phobic please keep reading – we have calculator and computers to do the arithmetic.) It looks like a lot of work for such a simple problem – but it does show that we can produce the “right” solution, not just the “almost right” solution. 012345678910/999999999999 = 1/82 + 1/6643 + 1/44322935 + 1/3342126505735204 + 1/13777776364571245538106741530564 + 1/2837373133928592270994420052649404800004803647 26211938769978064 + 1/1211639332670151470619177485404809821093062603 140049804835240136696283172322539452816499055701 65476145628085154221859579459680 + 1/2291096678667782109419243152987347237995996443 884534813699339825439529970174908326181631920130 250818574590841574846920318160284278301627446520 767380402960694182076308529613867661634657040511 751782409282818590382632925014130579181000298831 3936928383139 + 1/5611132542106166045308830073608736935506155483 654607537208862712418700286711683287171939572047 317508097262422710923170314803350692509395180693 759783824248926790350783993785695711223184987137 890142972078853021019188233737784908200416410664 880546014671839411952610368072183227765982502400 291692137776402613562963719539071697400813720507 839577544288810854625242673675137216021748514095 434197759317668645621187460262243450067595807113 572261544833521442175314847918225229985218510678 92475647151900456592480 = 0. 0  1  2  3  4  5  6  7  8  9  10  0  1  2  3  4  5  6  7  8  9  10  0  1  2  3  4  5  6  7  8  9 10  … This fraction is a repeating decimal with a period of 12. It counts from 0 to 10, then does it again, and again – just to show you it can.  AND it never skips a number. It gives you something to think about.

So we have two ways to create sequence numbers that count as high as we want or need them to..

First we can use 81, 9801, 998001, 99980001, 9999800001, … (just add nines an zeros in equal numbers until you get to one that will count high enough for you).

Second, we can create a custom Sequence Number by using the method shown above.  This is not easy but it does work - theoretically.

David