 mersenneforum.org In honor of my old friend, Gottfried Wilhelm Leibniz, or Monadic factorisation of 2^n+1
 Register FAQ Search Today's Posts Mark Forums Read 2020-10-10, 09:23 #1 RMLabrador   "Roman V. Makarchuk" Aug 2020 Ukraine 1000102 Posts In honor of my old friend, Gottfried Wilhelm Leibniz, or Monadic factorisation of 2^n+1 2^n+1 rewrite as x^n+1 (1), let k*a^n+1/0/-1, a=3 so-called Monada for all n in(1), and for example if x=2, n=32, numeric value of (1) factored as 2^32+1=(a^6-a^4-a^2+a-1)*(a^14+a^13+a^12-a^11-a^10+a^9+a^8+a^7-a^6+a^5-a^4+a^3-a^2-a+1)   2020-10-10, 09:30 #2 RMLabrador   "Roman V. Makarchuk" Aug 2020 Ukraine 2216 Posts a^6-a^4-a^2+a-1=641 use Monade))) (641+1)/3=214 (214-1)/3=71 (71+1)/3=24 24/3=8 (8+1)/3=3 3=a so (((a^2-1)*a*a-1)*a+1)*a-1 - monadic polinomial, and if we do the same for the second part of 2^32+1=641*6700417 we obtain second polynomial As you can understand, everyone of factor 2^n+1 is divisible by k*3+1/0/-1 (2) till the very end and this is why i named (2) Monade. Good name, i like it! Last fiddled with by RMLabrador on 2020-10-10 at 09:34 Reason: grammatic error   2020-10-10, 09:38 #3 RMLabrador   "Roman V. Makarchuk" Aug 2020 Ukraine 2216 Posts And of course, in the Mother Nature exist an algorithm for get this polynomial not from factors, but direct from value of 2^n+1. I have the wireld one, and now going to learn C++))  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post diep Riesel Prime Search 7 2015-05-01 10:38 petrw1 Math 3 2008-03-30 14:20 fivemack Math 7 2007-11-17 01:27 ewmayer Math 10 2007-03-02 12:47 ewmayer Lounge 5 2006-07-14 00:17

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