My Math Forum Mystery of prime numbers solved

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December 3rd, 2018, 05:56 PM   #11
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 Originally Posted by Micrm@ss Now that Riemann, Goldbach and the like are all solved. The world's most famous problem is this: is there going to be anybody at all dumb enough to pay him the \$3? They're presumably worth \$1m a pop, so \\$3 would be a bargain, right?

 December 3rd, 2018, 06:02 PM #12 Senior Member   Joined: Aug 2012 Posts: 2,101 Thanks: 605 I bet if he set up a GoFundMe page he'd make a bundle. "Unknown amateur has made world-class mathematical discovery, needs funds to bring it to the attention of the hidebound academics. Will revolutionize the world." I don't think there's any money on Amazon. Could be wrong. OP if you do this I hope you'll remember where you got the idea! Last edited by Maschke; December 3rd, 2018 at 06:04 PM.
 December 5th, 2018, 09:31 AM #13 Newbie   Joined: Dec 2018 From: US Posts: 2 Thanks: 0 Well, here is the free solution for finding prime numbers. Just calculate the sum of the following series: 0.111111111111111111111111... = 1/9 0.010101010101010101010101... = 1/99 0.001001001001001001001001... = 1/999 0.000100010001000100010001... = 1/9999 ... Just look for the decimal places where number 2 appears. These places will indicate prime numbers. However this method does not work for numbers greater than 60. No matter, just use bigger base numbers, i.e.: 1/9999 +1/99999999 + 1/999999999999 + ... Now if anyone can tell me how to calculate the sum of such series: 1/(a^1-1) + 1/(a^2-1) + 1/(a^3-1) + 1/(a^4-1) + ... where a>1 ... then I would be really grateful.
December 5th, 2018, 12:02 PM   #14
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Quote:
 Originally Posted by mig19farmer Well, here is the free solution for finding prime numbers. Just calculate the sum of the following series: 0.111111111111111111111111... = 1/9 0.010101010101010101010101... = 1/99 0.001001001001001001001001... = 1/999 0.000100010001000100010001... = 1/9999 ... Just look for the decimal places where number 2 appears.
I gotta tell ya, my eyes ain't that good. What? There are no 2's.

 December 5th, 2018, 12:44 PM #15 Newbie   Joined: Dec 2018 From: US Posts: 2 Thanks: 0 You have to add them all silly (to infinity... don't worry the sum converges). You get something like 0.122324243426244... So, 2s are located on the following decimal places above: 2,3,5,7. These also happen to be prime numbers. If you wonder why 10th place has number 4, that is because 10 is divisible by 1,2,5 and 10 (4 numbers). 9th decimal place has value of 3, meaning 9 is divisible by 1, 3 and 9 (3 numbers). However 60 is divisible by 12 numbers, meaning that +1 will carry over into 59th place, so this method is not that good for finding larger prime numbers. In fact, this method is not good at all, but hey, it's free advice

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