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Old 2020-03-27, 20:31   #32
R. Gerbicz
 
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"Robert Gerbicz"
Oct 2005
Hungary

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Quote:
Originally Posted by gd_barnes View Post
Put more simply for R1024, for k=81 all n's have a numeric covering set of factors [5, 41]. For k=9 and 36 that is not the case. They require algebraic factors for the even n to eliminate them.
You need less, for example for k=9 you need the algebraic factor only for n==0 mod 6, because the 9*1024^n-1 sequence for
n==1 mod 2 is divisible by 5
n==2 mod 6 is divisible by 7
n==4 mod 6 is divisible by 13

Ofcourse you can go further so where you need the algebraic factor for less than 1/6 part of the cases. But you will need in at least one remainder class the algebraic factor (without a proof).
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