mersenneforum.org In honor of my old friend, Gottfried Wilhelm Leibniz, or Monadic factorisation of 2^n+1
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 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 3410 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 428 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++))

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