20200530, 18:36  #1 
May 2016
161_{10} Posts 
f(n+6)= (n^21*(4*n+5))=N
Good evening,
The particular progression relation of this function .... generates successively prime and composite numbers which are connected to the next and previous function I tried with large numbers but ... does anyone have an opinion please? This Function : odd integer prime number or composite number Example: n = 6 f(n) = N = 7 factor = 7=1*7 NOTE : n = 12 f(n) = 91 factor = 91=1*7*13 NOTE : n = 18 f(n) = 247 factor = 247=1*13*19 NOTE : n = 24 f(n) = 475 factor = 475=1*5*5*19 NOTE : n = 30 f(n) = 775 factor = 775 =1*5*5*31 NOTE : n = 36 f(n) = 1147 factor = 1147=1*31*37 NOTE : n = 42 f(n) = 1591 factor = 1591=1*37*43 NOTE : n = 48 f(n) = 2107 factor = 2107=1*7*7*43 NOTE : n = 54 f(n) = 2695 factor= 2695=1*5*7*7*11 NOTE : n = 60 f(n) = 3355 factor = 3355=1*5*11*61 NOTE : n = 66 f(n) = 4087 factor = 4087=1*61*67 NOTE : n = 72 f(n) = 4891 factor = 4891=1*67*73 NOTE : n = 78 f(n) = 5767 factor = 5767=1*73*79 NOTE : n = 84 f(n) = 6715 factor = 6715=1*5*17*79 NOTE : n = 90 f(n) = 7735 factor = 7735=1*5*7*13*17 NOTE : n = 96 f(n) = 8827 factor = 8827=1*7*13*97 NOTE : 
20200530, 20:21  #2 
Mar 2006
Germany
5·569 Posts 

20200531, 06:55  #3 
"Curtis"
Feb 2005
Riverside, CA
4308_{10} Posts 

20200531, 07:16  #4  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}·2,281 Posts 
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