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.
Are there any additional i.e. "numbers over 100" solutions?
(In reply to computer soln
by Steven Lord)
Looking over your logic on there being no further solutions, I notice a couple of discrepancies:
Due to the sparseness of larger primes, eventually it is the average, Sp(n)/n that will have fewer digits than p(n).
Also n=9085 is not the point at which the same number of digits becomes impossible. Just to take the point at which my own program is now at, at n=530522, p(n)=6276343 and the average, Sp(n)/n, is 3021610.23.... If an integral average were to occur in this region, it would still have 7 digits just like the primes in this neighborhood.
Posted by Charlie
on 2018-10-10 11:37:03