 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 2·17 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 2·17 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 2×17 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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