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

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Message 142688 - Posted: 25 Aug 2020 | 17:49:21 UTC

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.

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Message 142690 - Posted: 25 Aug 2020 | 18:43:07 UTC - in response to Message 142688.

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).

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The "Shut up already!" badge:  This loud mouth has mansplained on the forums over 10 thousand times!  Sheesh!!!Discovered the World's First GFN-19 prime!!!Discovered 1 mega primeFound 1 prime in the 2018 Tour de PrimesFound 1 prime in the 2019 Tour de PrimesFound 1 prime in the 2020 Tour de PrimesFound 2 primes in the 2021 Tour de Primes321 LLR Turquoise: Earned 5,000,000 credits (5,132,712)Cullen LLR Turquoise: Earned 5,000,000 credits (5,038,114)ESP LLR Turquoise: Earned 5,000,000 credits (6,177,890)Generalized Cullen/Woodall LLR Ruby: Earned 2,000,000 credits (2,500,221)PPS LLR Sapphire: Earned 20,000,000 credits (20,751,038)PSP LLR Turquoise: Earned 5,000,000 credits (7,956,186)SoB LLR Sapphire: Earned 20,000,000 credits (36,067,618)SR5 LLR Jade: Earned 10,000,000 credits (10,007,110)SGS LLR Ruby: Earned 2,000,000 credits (3,718,606)TRP LLR Turquoise: Earned 5,000,000 credits (5,084,329)Woodall LLR Turquoise: Earned 5,000,000 credits (5,032,821)321 Sieve (suspended) Jade: Earned 10,000,000 credits (10,061,196)Cullen/Woodall Sieve (suspended) Ruby: Earned 2,000,000 credits (4,170,256)Generalized Cullen/Woodall Sieve (suspended) Turquoise: Earned 5,000,000 credits (5,059,304)PPS Sieve Sapphire: Earned 20,000,000 credits (22,885,121)Sierpinski (ESP/PSP/SoB) Sieve (suspended) Amethyst: Earned 1,000,000 credits (1,035,522)TRP Sieve (suspended) Ruby: Earned 2,000,000 credits (2,051,121)AP 26/27 Jade: Earned 10,000,000 credits (10,118,303)GFN Emerald: Earned 50,000,000 credits (78,031,938)WW Sapphire: Earned 20,000,000 credits (32,204,000)PSA Jade: Earned 10,000,000 credits (12,445,029)
Message 142691 - Posted: 25 Aug 2020 | 18:43:12 UTC - in response to Message 142688.
Last modified: 26 Aug 2020 | 8:21:05 UTC

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 * 211366286 + 1 = 32 * 211366286 + 1


2) Divide n by 2, and square 2n:

32 * 211366286 + 1 = 32 * (25683143)2 + 1


3) Combine the k and 2n parts:

32 * (25683143)2 + 1 = (3 * 25683143)2 + 1


That last line is now a GFN where:

b is (3 * 25683143)
n is 21 (or just 2)


It's a GFN-1, 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 75898524288+1

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Message 142693 - Posted: 25 Aug 2020 | 19:22:47 UTC
Last modified: 25 Aug 2020 | 19:26:42 UTC

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 n-m differences are high, 63686-63679=7 and 2141884-2141872=12. I wonder if this is a coincidence. The difference 12 is the absolutely highest one, among all known Fermat divisors.

/JeppeSN

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Message 142696 - Posted: 26 Aug 2020 | 6:10:16 UTC

Ah, thanks!

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Message 142711 - Posted: 26 Aug 2020 | 18:02:29 UTC - in response to Message 142696.

The n-m differences are high, 63686-63679=7 and 2141884-2141872=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?

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Message 142718 - Posted: 27 Aug 2020 | 8:17:55 UTC - in response to Message 142711.

The n-m differences are high, 63686-63679=7 and 2141884-2141872=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 n-m=11-8=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 n-m 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 n-m for the known Fermat divisors here: http://www.prothsearch.com/fermat.html#Count

/JeppeSN

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

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