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Message boards : General discussion : Prime Score

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robish
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Joined: 7 Jan 12
Posts: 1964
ID: 126266
Credit: 5,722,827,831
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Message 122888 - Posted: 25 Nov 2018 | 14:03:13 UTC

The scoring of primes seems to be exponential. Can anyone explain how it's calculated?

For example if a gfn 21 or 22 was found tomorrow, what prime score would they have?

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My lucky numbers 10590941048576+1 and 224584605939537911+81292139*23#*n for n=0..26

recoil44

Joined: 20 Dec 15
Posts: 167
ID: 433037
Credit: 411,347,492
RAC: 0

Message 122890 - Posted: 25 Nov 2018 | 15:50:51 UTC - in response to Message 122888.

For those who are interested, the formula is:

score = X^3 * ln(X) / Q

X is the size of the number expressed in base e -- i.e., it's similar to the standard length calculation LOG10(k) + LOG10(b)*n, except using the natural logarithm instead of the base 10 logarithm. So X = ln(k) + ln(b)*n.

Q is a scaling factor equal to 150732545640984000.

Furthermore, if you're the double checker, the score is reduced by 50%.

Michael posted this a while back

robish
Volunteer moderator
Volunteer tester

Joined: 7 Jan 12
Posts: 1964
ID: 126266
Credit: 5,722,827,831
RAC: 2,075,174

Message 122891 - Posted: 25 Nov 2018 | 15:53:54 UTC - in response to Message 122890.

For those who are interested, the formula is:

score = X^3 * ln(X) / Q

X is the size of the number expressed in base e -- i.e., it's similar to the standard length calculation LOG10(k) + LOG10(b)*n, except using the natural logarithm instead of the base 10 logarithm. So X = ln(k) + ln(b)*n.

Q is a scaling factor equal to 150732545640984000.

Furthermore, if you're the double checker, the score is reduced by 50%.

Michael posted this a while back

Thanks recoil44, I must have missed it. Cheers.
____________
My lucky numbers 10590941048576+1 and 224584605939537911+81292139*23#*n for n=0..26

Message boards : General discussion : Prime Score