Let Sp(n) be a sum of all primes from p(1) to p(n) inclusive.

Let m(n) be the average value of all those primes, i.e. m(n)= Sp(n)/n.

Find the n-th prime N such that m(n) equals the reversal of N.

Example for n=6: Sp(6)=2+3+5+7+11+13= 41; m(6)=41/6; not the reversal of 13,
that is 31 .

So 13 is not our prime.

Another 2-digit number is.

Find it.

Are there any additional i.e. "numbers over 100" solutions?

Thanks - I corrected my post accordingly.

Also I note, studying this page:

https://math.stackexchange.com/questions/623872/what-is-the-sum-of-the-prime-numbers-up-to-a-prime-number-n

Sp(n) ~

k

pk∼12cn2lnn.

*and so m(n) ~ *

12cn

lnn

while p(n) ~

nlnnWhile these are rough approximations, to the extent

that they are true, this would suggest that P(n) and m(n)

can persist with the same number of digits ad infinitum,

and if so, that would make the search for further examples,

ah, hard.

*Edited on ***October 11, 2018, 12:50 am**