20060916, 23:13  #1 
Feb 2004
France
929 Posts 
A property of Fermat numbers. Already known ?
I'd like to know if the following property of Fermat numbers (prime or not) is already known:
where: Do you know ? or do you think this can be easily built from known properties of Fermat numbers ? Examples: I don't have a proof ... It just appeared while playing with the number of cycles under x^2 modulo a Mersenne number Mq where q is a Fermat number ... I already seen: 3, 30, 4080 when playing with LLT, long time ago, but cannot remember when ... Ki numbers have Fermat numbers as factors, plus 2^k*3 , where k is quite strange ... Interestingly, we have a relation for the sum S(i) of the Ki mod Fn (from which one could find a formula for the Ki. But too late for me now ...) : 1+1+3 = 5 = 5 + 0 : A0 = 0 1+1+3+13 = 18 = 17 + 1 : A1 = 1 = A0 + 2^0 1+1+3+30+225 = 260 = 257 + 3 : A2 = 3 = A1 + 2^1 1+1+3+30+4080+61441 = 65537 + 19 : A3 = 19 = A2 + 2^4 1+1+3+30+4080+134215680+4160749569= 4294967297 + 2067 : A4 = 2067 = A3 + 2^11 So: A(i+1) = A(i) + 2^(2^i1) . Funny ! Useful ? Tony Last fiddled with by T.Rex on 20060916 at 23:15 
20060917, 08:32  #2 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
Hi,
what is the K_{n} sequence? How is it generated? Alex 
20060917, 10:43  #3 
Feb 2004
France
3A1_{16} Posts 
PARI code
Here is the PARI code that generates the values.
It comes from the formula computing the number of cycles of length L (L divides q1, and here q=2^(2^n)+1 ) under x^2 modulo a Mersenne number 2^q1 . I(ve only optimized the code since divisors of q1 here are powers of 2. Code:
HGFn(n,f)= { Fn=2^(2^n)+1; print("\n",Fn); print("L= 1 > N= 1"); SS_ = 1; SS_m= 1; SS_p= 1; for(j=1, 2^n, m=2^j; S = 0; for(i=0, j, S+ = moebius(2^(ji)) * 2^(2^i); ); SS_ += S; SS_m+= (S/m)%Fn; SS_p+= S%Fn; if(S!=0, print1("L= ",m," > N%Fn= ",(S/m)%Fn, " (", SS_p,") ")); if(f==1, print(factor(S/m)), print(" ")); ); print("S/mi % Fn: ", SS_%Fn,"\nS % Fn", SS_m, "\n", SS_p); } HGFn(1,0) HGFn(2,0) \\ No factorisation HGFn(3,1) \\ Factorisation 
20060917, 11:10  #4 
Einyen
Dec 2003
Denmark
2^{3}·3^{2}·47 Posts 
It seems that:

20060917, 11:10  #5 
Feb 2004
France
929 Posts 
Seems:
Example: 
20060917, 12:23  #6 
Einyen
Dec 2003
Denmark
110100111000_{2} Posts 
Actually:
for n>0 Inserting in your F4 formula: rewriting to: So: for n>0 So its trivial. Last fiddled with by ATH on 20060917 at 13:20 
20060917, 22:11  #7 
Feb 2004
France
929 Posts 
Yes, thanks to show it. I was 95% sure of that today (I wrote the first post of this thread after midnight. Not a good moment to use my mind ...).
What is not trivial is that I've found where I saw 4080: In the paper A LLTlike test for proving the primality of Fermat numbers. I wrote 2 years ago, Chapter 5 page 6, the residues before the last step of the LLTlike iteration, for n=2, 3 and 4, are: 6, 60 and 4080 . To be compared with K2=3, K3=30 and K4=4080 here. Remind that this paper proves how to build a LLTlike test for Fermat numbers. For n=5, the residue before the last one is: 3443904229, which is prime and has no link with K5, probably because F5 is not prime ... Also, in my other paper A primality test for Fermat numbers faster than Pépin test ?, in chapter 3.3 page 7, , and in chapter 3.4 page 8 . Vn here is a Pell number. E. Lucas thought that this could be used to speed up the primality proof for Fermat numbers (but he gave no hints !). That may be a coincidence, but I think there is a link. But I do not see what to do with it. Or, simply, all these numbers are made of the product of the first Fermat numbers. What is not trivial too is that one can build the Fermat numbers by using the moebius function: Code:
SS=1; for(j=1, 2^n, m=2^j; S = 0; for(i=0, j, S += moebius(2^(ji)) * 2^(2^i); ); SS += S % Fn; if(S!=0, print1("L= ",m," > N%Fn= ",(S/m)%Fn, " (", S%Fn,") ")); ); Just for fun. Tony Last fiddled with by T.Rex on 20060917 at 22:13 
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