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Let's start guessing what the next prime will be like with the Fermat prime pool. |
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In the Fermat prime pool, people express their bids as b values (the b in b^(2^n)+1).
Here we want to guess for the next prime that eliminates any of the five remaining multipliers in the Sierpiński problem, I presume.
Should we state our guess as the value of the exponent n in the expression
k·2^n + 1,
or should we give the number of decimal digits?
/JeppeSN |
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 Michael Goetz Volunteer moderator
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Date is another possibility, e.g., "We'll find the next prime in 2023."
Also, bear in mind that we're currently in the middle of a long double check of prior work, with much lower chances of finding a prime. While the numbers themselves are smaller (which increases chances), since the numbers have already been checked once, our chance of finding a prime is approximately 30 times lower than it normally would be.
Even without the double check, there were nine years between the discoveries of the last 2 primes. The gap between the 2 largest primes is pretty large: n=13M and n=31M.
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My lucky number is 75898524288+1 |
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MonkeydeeVolunteer tester
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The next K to be eliminated could be another possibility, or the date AND the next K.
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My Primes
Badge Score: 4*2 + 6*2 + 7*4 + 8*9 + 11*3 + 12*1 = 165
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I thought about it while I was participating in original SoB and my opinion was:
1. Next k would be 24737 or 55459 (but not 10223 because I did not know that it was PrimeGrid's part of search)
2. I expected it in the short period of time (n<2^25)
Now my expectation is:
1. One of these 2 k's also
2. n<2^26 (Maybe in 2024?) |
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It may be a matter of taste, but I find it uninteresting to bet on which k will be eliminated next (I was about to write "that is pure lottery", but gambling on the size of the next SoB prime is a lottery as well, I guess).
My guess for the exponent (of 2) of the next SoB prime shall be:
n = 17.8 M
that is k · 2^{17,800,000} + 1 for some k in { 21181; 22699; 24737; 55459; 67607 }.
I know I will win, and then Michael Goetz can predict when the leading edge will be near n = 17.8 M.
/JeppeSN
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2. n<2^2^26 (Maybe in 2024?)
You mean the prime will be (just) below 2^(2^26), then what we usually call n (the n in k·2^n + 1) will be close to 2^26 = 67 million?
/JeppeSN |
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n Just below 67000000. So I think. And 17.8 million range already passed or I mistake something?
I've corrected it. |
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n Just below 67000000. So I think. And 17.8 million range already passed or I mistake something?
You are right! My guess was meaningless :-(
The double-check is currently near n = 25 M according to stats_sob_llr, and the double check will end near n = 31.6 M.
I will come up with a meaningful guess at another time (I guess I committed hubris with the style in my post where I made the silly guess).
/JeppeSN |
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Michael GoetzVolunteer moderator
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...the double check will end near n = 31.6 M.
Nothing above 31M needs to be double checked. The 31M range was always PrimeGrid's, so the part that's done is fully double checked already.
The double check ends at n=30,999,908.
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My lucky number is 75898524288+1 |
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The guess should be stated as the exponent that will eliminate the next k. I am personally betting on the range from 3700097 to 37911100. |
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OK, I will bring a valid bid, then: n = 48,400,000.
/JeppeSN
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