Product of the factorials of the digits of some primes

enzocreti · 2019-04-17, 20:13

30241 is an example of prime such that 3!*0!*2!*4!*1!+1=17^2.

up to primes 10^9 i found that if a prime with this property starts with digit 6, then the square of always 161^2 . why?

chalsall · 2019-04-17, 20:17

Quote:

Originally Posted by enzocreti

30241 is an example of prime such that 3!*0!*2!*4!*1!+1=17^2.

up to primes 10^9 i found that if a prime with this property starts with digit 6, then the square of always 161^2 . why?

Are you familiar with the concept of signal to noise?

Sorry, that was a rhetorical question. I just needed to yell at someone for being stupid. Nothing personal.

enzocreti · 2019-04-17, 20:19

Quote:

Originally Posted by chalsall

Are you familiar with the concept of signal to noise?

Sorry, that was a rhetorical question. I just needed to yell at someone for being stupid. Nothing personal.

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enzocreti · 2019-04-17, 20:22

as you can see all the primes which start with digit 6 ...The product of the factorials of their digits plus one is always 161^2 why is it not possible another square?

chalsall · 2019-04-17, 20:25

Quote:

Originally Posted by enzocreti

...why is it not possible another square?

You have presented your theorem. Run the empirical to see if it stands.

2019-04-17, 20:13	#1
enzocreti Mar 2018 2×5×53 Posts	Product of the factorials of the digits of some primes 30241 is an example of prime such that 3!0!2!4!1!+1=17^2. up to primes 10^9 i found that if a prime with this property starts with digit 6, then the square of always 161^2 . why?

2019-04-17, 20:22	#4
enzocreti Mar 2018 2×5×53 Posts	primes as you can see all the primes which start with digit 6 ...The product of the factorials of their digits plus one is always 161^2 why is it not possible another square? Last fiddled with by enzocreti on 2019-04-17 at 20:23

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