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#221 |
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"Richard B. Woods"
Aug 2002
Wisconsin USA
22·3·641 Posts |
Factoris (http://wims.unice.fr/wims/wims.cgi) can factor not only integers (278914005382139703576000 = 26 × 35 × 53 × 72 × 11 × 13 × 17 × 19 × 23 × 29 × 31 × 37 × 41 × 43 × 47), but also rational numbers, polynomials, and rational functions. It uses
"GP/PARI CALCULATOR Version 2.2.10 (alpha) for factorizations other than multivariate polynomials, and Maxima version: 5.9.1 for factorizations of multivariate polynomials." Factoris is part of the WWW Interactive Multipurpose Server (WIMS) at University of Nice. See http://wims.unice.fr/ |
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#222 | ||
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Sep 2002
Vienna, Austria
DB16 Posts |
196883 is the minimal dimension of the faithful irreducible representations of "The Monster" , the largest sparodic finite simple group (its order is roughly 8E53).
Also, the so-called "Laurent series" of the j-function begins with: Quote:
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#223 | |
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Sep 2002
Vienna, Austria
3×73 Posts |
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#224 | |
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Bronze Medalist
Jan 2004
Mumbai,India
22·33·19 Posts |
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Thank you Cheesehead you are right on target. Thats the information I need: 'To build, not to destroy' I have used factoris, the basic factorisation ,but I didnt explore any further Mally
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#225 | |
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Bronze Medalist
Jan 2004
Mumbai,India
1000000001002 Posts |
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rank / prime / digits / who / when comment 1 / 2^ [(22478785)^(3)] +1 / 746190 / g245 / Oct 2003 / Divides Fermat F(2478782), GF(2478782,3), GF(2478776,6), GF(2478782,12) Mally
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#226 | |
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Aug 2005
Brazil
2·181 Posts |
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#227 | |
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Bronze Medalist
Jan 2004
Mumbai,India
1000000001002 Posts |
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:surprised A million distinct factors? fetofs: you are a master indeed and someone to reckon with in dealing with numbers. .Since this number can be represented by 15 different prime factors then the other factors must be combinations of these 15. So starting with nC1 + nc2 +.... how far can one go to arrive at 1032192 to be exact ? Mally
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#228 |
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Aug 2002
Termonfeckin, IE
22×691 Posts |
The number of factors of any number that has a prime factor representation of:
p_1^n_1 * p_2^n_2......p_r^n_r where p_1,p_2,....p_r are it's prime factors and n_1,n_2,...n_r are their respective exponents, is: (n_1+1)*(n_2+1)*.....(n_r+1). This includes 1 and the number itself, so you can remove those two if you do not wish to count them. So for 2^6×3^5×5^3×7^2×11×13×17×19×23×29×31×37×41×43×47 the total number of factors would be: 7*6*4*3*2^11 = 1,032,192 It is easy to see why the number of factors follows the relation given above. Note that each factor can have p_i in it from 0 to n_i times for a total of (n_i +1). See C&P ch. 1 for more details. Last fiddled with by garo on 2005-12-27 at 18:31 Reason: typos |
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#229 |
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Bronze Medalist
Jan 2004
Mumbai,India
40048 Posts |
Thanks a million Garo. You explained it so brilliantly and lucidly that I had no problem to follow through. I don't have C&P but Im sure my other number theory books will have it. Your last para where the exponent is increased by 1 is a bit tricky but I'll fathom it out. The notation you used is also self explanatory. After reading so many posts in the forum and studying them I am able to not only understand it (the notation) but perhaps use them in my posts should the need arise. I wish the other expert pollsters could be as simple. I know there are posts explaining the notation used but I for one would welcome a comprehensive revival of such a topic for newbies like myself and others following this complex language in Number theory. Thanks once again. Mally
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#230 |
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Aug 2002
Termonfeckin, IE
22×691 Posts |
Glad I could be of help. The notation I am using is the standard LateX math notation where _ is used for subscripts and ^ for superscripts. Let me give you an example to make things clearer.
30 = 2*3*5 so should have 2*2*2 = 8 factors. The factors of 30 can be formed by taking 0 or 1 occurence of 2,3 and 5 and hence there are 2*2*2 possible factors since there are two possibilities for each of 2,3 and 5. In the list below the first three numebrs refer to the number of occurences of 2,3,5 in the factor. So here we go: 0,0,0 -> 1 1,0,0 -> 2 0,1,0 -> 3 0,0,1 -> 5 1,1,0 -> 6 1,0,1 ->10 0,1,1 ->15 1,1,1 -> 30 As an exercise try this on 24. Not that 24 = 2^3 * 3^1 so we expect (3+1)*(1+1) = 8 factors. |
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#231 |
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Bronze Medalist
Jan 2004
Mumbai,India
22·33·19 Posts |
Excellent Garo! Crystal clear and thank you. You are as dedicated as your beloved father, my fellow contemporary. God Bless you Mally
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