swotting the old CCNA books?

Fun with number representation - the factorial basis.

Consider this series:

0,1,10,11,20,21,100,101,110,111,120,121,200,201,210,211,220,221,300,301,310,311,320,321,1000,1001...

The kth digit can take on values 0...k. The last number before a turnover of digits is k.(k-1).(k-2)...321. Thus 7654321 + 1 = 10000000.

Now, it so happens that

100....00 (k zeros) is the representation of (k+1)! = (k+1)k(k-1)(k-2)..1 and by definition 0! = 1.

Thus a number in this representation can be written

A = sum(0..k) An * n!

where 0 <= An <= n.

What are the primes?

2 = 10
3 = 11
5 = 21
7 = 101
11 = 121
13 = 201
17 = 221
19 = 301
23 = 321
29 = 1021
31 = 1101
37 = 1201
41 = 1221

Here are all the primes up to 719 = 6!-1 ;

0 0 0 1 0
0 0 0 1 1
0 0 0 2 1
0 0 1 0 1
0 0 1 2 1
0 0 2 0 1
0 0 2 2 1
0 0 3 0 1
0 0 3 2 1
0 1 0 2 1
0 1 1 0 1
0 1 2 0 1
0 1 2 2 1
0 1 3 0 1
0 1 3 2 1
0 2 0 2 1
0 2 1 2 1
0 2 2 0 1
0 2 3 0 1
0 2 3 2 1
0 3 0 0 1
0 3 1 0 1
0 3 1 2 1
0 3 2 2 1
0 4 0 0 1
0 4 0 2 1
0 4 1 0 1
0 4 1 2 1
0 4 2 0 1
0 4 2 2 1
1 0 1 0 1
1 0 1 2 1
1 0 2 2 1
1 0 3 0 1
1 1 0 2 1
1 1 1 0 1
1 1 2 0 1
1 1 3 0 1
1 1 3 2 1
1 2 0 2 1
1 2 1 2 1
1 2 2 0 1
1 2 3 2 1
1 3 0 0 1
1 3 0 2 1
1 3 1 0 1
1 3 3 0 1
1 4 1 0 1
1 4 1 2 1
1 4 2 0 1
1 4 2 2 1
1 4 3 2 1
2 0 0 0 1
2 0 1 2 1
2 0 2 2 1
2 0 3 2 1
2 1 0 2 1
2 1 1 0 1
2 1 2 0 1
2 1 2 2 1
2 1 3 0 1
2 2 0 2 1
2 2 3 0 1
2 2 3 2 1
2 3 0 0 1
2 3 0 2 1
2 3 3 0 1
2 4 0 0 1
2 4 1 2 1
2 4 2 0 1
2 4 2 2 1
2 4 3 2 1
3 0 1 0 1
3 0 2 0 1
3 0 3 0 1
3 0 3 2 1
3 1 0 2 1
3 1 2 0 1
3 1 2 2 1
3 2 0 0 1
3 2 1 2 1
3 2 2 0 1
3 2 3 2 1
3 3 0 0 1
3 3 1 0 1
3 3 1 2 1
3 3 2 2 1
3 4 0 0 1
3 4 0 2 1
3 4 1 0 1
3 4 1 2 1
3 4 3 2 1
4 0 1 0 1
4 0 1 2 1
4 0 3 0 1
4 0 3 2 1
4 1 0 2 1
4 1 2 2 1
4 1 3 0 1
4 2 2 0 1
4 2 3 0 1
4 3 0 2 1
4 3 1 2 1
4 3 2 2 1
4 3 3 0 1
4 4 0 0 1
4 4 1 2 1
4 4 2 2 1
4 4 3 2 1
5 0 0 0 1
5 0 1 0 1
5 0 2 0 1
5 0 2 2 1
5 0 3 0 1
5 1 1 0 1
5 1 2 2 1
5 1 3 0 1
5 1 3 2 1
5 2 0 2 1
5 2 1 2 1
5 2 2 0 1
5 3 0 0 1
5 3 0 2 1
5 3 1 2 1
5 3 3 0 1
5 4 0 2 1
5 4 2 0 1
5 4 3 2 1

Can you see a pattern?

Post #67,688 by boxley 12/9/02 9:27:33 PM Reply	swotting the old CCNA books? subneting in binary gives me a headache unless i is the easy stuff, would rather decipher ss7 ain messages. thanx, bill will work for cash and other incentives [link\|http://home.tampabay.rr.com/boxley/resume/Resume.html\|skill set] Opera was the television of the nineteenth century:loud, vulgar and garish with plots that could only be called infantile. "Pendergast"
Post #67,702 by folkert 12/9/02 10:34:39 PM Reply	No... actually Bill... Was pulling those from memory... If I were to pull out those old CCNA or Cisco IOS manuals or the Bay Network manuals for SiteMangler, then I'd be spouting off the OOB SNM and Variable SNM networking scenarios... Head hurts... must stop thinking in BINARY..... Must convert to 0x(HEX)... then Octal THEN Decimal... ahhh better. As teh saying goes: There are 10 people that understand Binary, those that do, and those that don't! Personally I like "Safe Hex" better. `[link\|mailto:curley95@attbi.com\|greg] - Grand-Master Artist in IT [link\|http://www.iwethey.org/ed_curry/\|REMEMBER ED CURRY!!!]` Your friendly Geheime Staatspolizei reminds: [link\|http://www.wired.com/news/wireless/0,1382,56742,00.html\|Wi-Fi enabled device use] comes with an all inclusive free trip to the (county)Photographer! Overbooking, is a problem, please be prepared for "room-ies". Why You ask? Here is the answer to your query: `SELECT * FROM politicians WHERE iq > 40 OR \\ WHERE ego < 1048575; 0 rows found`
Post #67,707 by deSitter 12/9/02 11:00:54 PM 12/9/02 11:05:55 PM Reply	Re: swotting the old CCNA books? Fun with number representation - the factorial basis. Consider this series: 0,1,10,11,20,21,100,101,110,111,120,121,200,201,210,211,220,221,300,301,310,311,320,321,1000,1001... The kth digit can take on values 0...k. The last number before a turnover of digits is k.(k-1).(k-2)...321. Thus 7654321 + 1 = 10000000. Now, it so happens that 100....00 (k zeros) is the representation of (k+1)! = (k+1)k(k-1)(k-2)..1 and by definition 0! = 1. Thus a number in this representation can be written A = sum(0..k) An * n! where 0 <= An <= n. What are the primes? 2 = 10 3 = 11 5 = 21 7 = 101 11 = 121 13 = 201 17 = 221 19 = 301 23 = 321 29 = 1021 31 = 1101 37 = 1201 41 = 1221 Here are all the primes up to 719 = 6!-1 ; 0 0 0 1 0 0 0 0 1 1 0 0 0 2 1 0 0 1 0 1 0 0 1 2 1 0 0 2 0 1 0 0 2 2 1 0 0 3 0 1 0 0 3 2 1 0 1 0 2 1 0 1 1 0 1 0 1 2 0 1 0 1 2 2 1 0 1 3 0 1 0 1 3 2 1 0 2 0 2 1 0 2 1 2 1 0 2 2 0 1 0 2 3 0 1 0 2 3 2 1 0 3 0 0 1 0 3 1 0 1 0 3 1 2 1 0 3 2 2 1 0 4 0 0 1 0 4 0 2 1 0 4 1 0 1 0 4 1 2 1 0 4 2 0 1 0 4 2 2 1 1 0 1 0 1 1 0 1 2 1 1 0 2 2 1 1 0 3 0 1 1 1 0 2 1 1 1 1 0 1 1 1 2 0 1 1 1 3 0 1 1 1 3 2 1 1 2 0 2 1 1 2 1 2 1 1 2 2 0 1 1 2 3 2 1 1 3 0 0 1 1 3 0 2 1 1 3 1 0 1 1 3 3 0 1 1 4 1 0 1 1 4 1 2 1 1 4 2 0 1 1 4 2 2 1 1 4 3 2 1 2 0 0 0 1 2 0 1 2 1 2 0 2 2 1 2 0 3 2 1 2 1 0 2 1 2 1 1 0 1 2 1 2 0 1 2 1 2 2 1 2 1 3 0 1 2 2 0 2 1 2 2 3 0 1 2 2 3 2 1 2 3 0 0 1 2 3 0 2 1 2 3 3 0 1 2 4 0 0 1 2 4 1 2 1 2 4 2 0 1 2 4 2 2 1 2 4 3 2 1 3 0 1 0 1 3 0 2 0 1 3 0 3 0 1 3 0 3 2 1 3 1 0 2 1 3 1 2 0 1 3 1 2 2 1 3 2 0 0 1 3 2 1 2 1 3 2 2 0 1 3 2 3 2 1 3 3 0 0 1 3 3 1 0 1 3 3 1 2 1 3 3 2 2 1 3 4 0 0 1 3 4 0 2 1 3 4 1 0 1 3 4 1 2 1 3 4 3 2 1 4 0 1 0 1 4 0 1 2 1 4 0 3 0 1 4 0 3 2 1 4 1 0 2 1 4 1 2 2 1 4 1 3 0 1 4 2 2 0 1 4 2 3 0 1 4 3 0 2 1 4 3 1 2 1 4 3 2 2 1 4 3 3 0 1 4 4 0 0 1 4 4 1 2 1 4 4 2 2 1 4 4 3 2 1 5 0 0 0 1 5 0 1 0 1 5 0 2 0 1 5 0 2 2 1 5 0 3 0 1 5 1 1 0 1 5 1 2 2 1 5 1 3 0 1 5 1 3 2 1 5 2 0 2 1 5 2 1 2 1 5 2 2 0 1 5 3 0 0 1 5 3 0 2 1 5 3 1 2 1 5 3 3 0 1 5 4 0 2 1 5 4 2 0 1 5 4 3 2 1 Can you see a pattern? -drl Edited by deSitter Dec. 9, 2002, 11:05:55 PM EST Expand All History

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