40030624861domain: NAppears in sequencesAs p runs through the primes >= 5, sequence gives { numerator of Sum_{k=1..p-1} 1/k } / p^2.at n=9A061002Numerator((p-1)*H(p-1))/p^2 for p = prime(n) > 3, where H(k) is k-th harmonic number A001008(k)/A002805(k).at n=9A120308a(n) = squarefree part of A145609(n).at n=17A145738