As p runs through primes >= 5, sequence gives { numerator of Sum_{k=1..p-1} 1/k^2 } / p.
A125551
As p runs through primes >= 5, sequence gives { numerator of Sum_{k=1..p-1} 1/k^2 } / p.
Terms
- a(0) =41a(1) =767a(2) =178939a(3) =18500393a(4) =48409924397
External references
- oeis: A125551