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Generalized Fermat Prime Search :
Why is 9 * 2^11366286 + 1 a generalized Fermat number?
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BurVolunteer tester
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Joined: 25 Feb 20 Posts: 465 ID: 1241833 Credit: 284,583,935 RAC: 682,319

The number appears on PrimePage as 4th largest generalized Fermat prime: primes.utm.edu.
Why is it a generalized Fermat number? I tried forming it to b^2^n + 1 but couldn't make it.  

Ravi FernandoProject administrator Volunteer tester Project scientist Send message
Joined: 21 Mar 19 Posts: 176 ID: 1108183 Credit: 10,247,472 RAC: 6,176

Technically, any prime that's one more than a square is a generalized Fermat prime with n = 1. In this case, it's (3 * 2^5683143)^2 + 1. But of course the more "interesting" GFN primes are the ones where n is large (e.g. n >= 15).  

Michael GoetzVolunteer moderator Project administrator
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Joined: 21 Jan 10 Posts: 13650 ID: 53948 Credit: 285,528,436 RAC: 39,566

The number appears on PrimePage as 4th largest generalized Fermat prime: primes.utm.edu.
Why is it a generalized Fermat number? I tried forming it to b^2^n + 1 but couldn't make it.
1) change k into the square root of k, squared:
9 * 2^{11366286} + 1 = 3^{2} * 2^{11366286} + 1
2) Divide n by 2, and square 2^{n}:
3^{2} * 2^{11366286} + 1 = 3^{2} * (2^{5683143})^{2} + 1
3) Combine the k and 2^{n} parts:
3^{2} * (2^{5683143})^{2} + 1 = (3 * 2^{5683143})^{2} + 1
That last line is now a GFN where:
b is (3 * 2^{5683143})
n is 2^{1} (or just 2)
It's a GFN1, essentially.
(Hopefully I didn't screw up any of the equivalencies.)
EDIT: This will be true for any Proth prime where k is a perfect square, and n is even.
____________
My lucky number is 75898^{524288}+1  


Also noteworthy, there are two known generalized Fermat primes of this type, b^(2^1)+1, which divide Fermat numbers, namely:
June 1998: 169*2^63686 + 1 divides F(63679)
September 2011: 25*2^2141884 + 1 divides F(2141872) (PrimeGrid)
The nm differences are high, 6368663679=7 and 21418842141872=12. I wonder if this is a coincidence. The difference 12 is the absolutely highest one, among all known Fermat divisors.
/JeppeSN  

BurVolunteer tester
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Joined: 25 Feb 20 Posts: 465 ID: 1241833 Credit: 284,583,935 RAC: 682,319

Ah, thanks!  

BurVolunteer tester
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Joined: 25 Feb 20 Posts: 465 ID: 1241833 Credit: 284,583,935 RAC: 682,319

The nm differences are high, 6368663679=7 and 21418842141872=12. I wonder if this is a coincidence. The difference 12 is the absolutely highest one, among all known Fermat divisors. I was always wondering why the exponents of Fermat divisors are always so close to the exponents of the Fermat number.
Is there an obvious reason?  


The nm differences are high, 6368663679=7 and 21418842141872=12. I wonder if this is a coincidence. The difference 12 is the absolutely highest one, among all known Fermat divisors. I was always wondering why the exponents of Fermat divisors are always so close to the exponents of the Fermat number.
Is there an obvious reason?
Not obvious. It is a theorem that any divisor of Fermat number F(m), with m≥2, must be of the form:
c*2^(m+2) + 1
I write c instead of k here because this multiplier can be even or odd. You find k and n (in the form k*2^n + 1) by "moving" all the factors of 2 present inside c to the right, so that k becomes odd.
An example, F(8), so m=8. The first factor that was found had c=1209889024954, so:
1209889024954*2^(8+2) + 1
This c is divisible by 2 only once. We can put it to the form with k and n easily:
604944512477*2^11 + 1
Note that k must be odd. So nm=118=3 here.
Now, if we assume that the c in
c*2^(m+2) + 1
is "random" in some sense (this is not a precise term), then the chance c is already odd, is 50%. The chance c is singly even, i.e. divisible by two but not by four, is 25%. The chance that c is divisible exactly twice by two, i.e. c is divisible by four but not by eight, should be 12.5%. And so on.
This "model" (which is no proof of anything!) says that nm should be:
2: 50% of the times
3: 25% of the times
4: 12.5% of the times
and so on.
We do not know if this (if it can be made precise) is "true". You can see the actual distribution of nm for the known Fermat divisors here: http://www.prothsearch.com/fermat.html#Count
/JeppeSN  

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Generalized Fermat Prime Search :
Why is 9 * 2^11366286 + 1 a generalized Fermat number? 