mersenneforum.org f(n+6)= (n^2-1*(4*n+5))=N
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 2020-05-30, 18:36 #1 Godzilla     May 2016 16110 Posts f(n+6)= (n^2-1*(4*n+5))=N Good evening, The particular progression relation of this function $n + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6$.... generates successively prime and composite numbers which are connected to the next and previous function $f(n+6)$ I tried with large numbers but ... does anyone have an opinion please? This Function $f(n+6)$: $f(n+6)= (n^2-1*(4*n+5))=N$ $\frac{N}{n+1}=$ odd integer prime number or composite number Example: n = 6 f(n) = N = 7 factor = 7=1*7 NOTE : $\frac{N}{n+1}= \frac{7}{6+1}=1$ n = 12 f(n) = 91 factor = 91=1*7*13 NOTE : $\frac{N}{n+1}= \frac{91}{12+1}=7$ n = 18 f(n) = 247 factor = 247=1*13*19 NOTE : $\frac{N}{n+1}= \frac{247}{18+1}=13$ n = 24 f(n) = 475 factor = 475=1*5*5*19 NOTE : $\frac{N}{n+1}= \frac{475}{24+1}=19$ n = 30 f(n) = 775 factor = 775 =1*5*5*31 NOTE : $\frac{N}{n+1}= \frac{775}{30+1}=25$ n = 36 f(n) = 1147 factor = 1147=1*31*37 NOTE : $\frac{N}{n+1}= \frac{1147}{36+1}=31$ n = 42 f(n) = 1591 factor = 1591=1*37*43 NOTE : $\frac{N}{n+1}= \frac{1591}{42+1}=37$ n = 48 f(n) = 2107 factor = 2107=1*7*7*43 NOTE : $\frac{N}{n+1}= \frac{2107}{48+1}=43$ n = 54 f(n) = 2695 factor= 2695=1*5*7*7*11 NOTE : $\frac{N}{n+1}= \frac{2695}{54+1}=49$ n = 60 f(n) = 3355 factor = 3355=1*5*11*61 NOTE : $\frac{N}{n+1}= \frac{3355}{60+1}=55$ n = 66 f(n) = 4087 factor = 4087=1*61*67 NOTE : $\frac{N}{n+1}= \frac{4087}{66+1}=61$ n = 72 f(n) = 4891 factor = 4891=1*67*73 NOTE : $\frac{N}{n+1}= \frac{4891}{72+1}=67$ n = 78 f(n) = 5767 factor = 5767=1*73*79 NOTE : $\frac{N}{n+1}= \frac{5767}{78+1}=73$ n = 84 f(n) = 6715 factor = 6715=1*5*17*79 NOTE : $\frac{N}{n+1}= \frac{6715}{84+1}=79$ n = 90 f(n) = 7735 factor = 7735=1*5*7*13*17 NOTE : $\frac{N}{n+1}= \frac{7735}{90+1}=85$ n = 96 f(n) = 8827 factor = 8827=1*7*13*97 NOTE : $\frac{N}{n+1}= \frac{8827}{96+1}=91$
2020-05-30, 20:21   #2
kar_bon

Mar 2006
Germany

3×947 Posts

Quote:
 Originally Posted by Godzilla [...] generates successively prime and composite numbers [..]
Why? Because...
Quote:
 Originally Posted by Godzilla $f(n+6)= (n^2-1*(4*n+5))=N$
N = n2 - 4n - 5 = (n-5)*(n+1), so your N/(n+1) = n-5.

So your composite/prime generator f(n) = n-5.

2020-05-31, 06:55   #3
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

22×1,061 Posts

Quote:
 Originally Posted by kar_bon Why? Because... N = n2 - 4n - 5 = (n-5)*(n+1), so your N/(n+1) = n-5. So your composite/prime generator f(n) = n-5.
Soooo.... f(n+6) would be.... n + 1?

Awesome. Godzilla, indeed.

2020-05-31, 07:16   #4
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

27×71 Posts

Quote:
 Wolfgang Mozart: I never knew that music like that was possible! ... Wolfgang Mozart: No, no! One hears such sounds, and what can one say but... "Salieri."
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