cp's OEIS Frontend

There are two cases where a(n) = 0: (a) n divides A002805(k) for all k, which only happens for n = 1; (b) there are infinitely many k such that n does not divide A002805(k), which may happen for some primes p and their multiples.

For k > a(n) > 0, A002805(k) is always divisible by n.

For prime p and k >= p, A002805(k) = (the denominator of s + (Sum_{i=1..floor(k/p)} 1/i)/p) is not divisible by p if and only if p divides A001008(floor(k/p)) = (the numerator of Sum_{i=1..floor(k/p)} 1/i), because the denominator of s = Sum_{1 <= i <= k, i is not divisible by p} 1/i can never be divisible by p.

If k == -1 or 0 (mod p), then p divides A001008(k) iff p^2 divides A001008(floor(k/p)), otherwise p divides A001008(k) iff p divides the numerator of (Sum_{i=floor(k/p)*p+1..k} 1/i) + (Sum_{i=1..floor(k/p)} 1/i)/p, where p is an odd prime and k >= p. (Since Sum_{i=1..p-1} (p-1)!/i = (-1)^((p-1)/2)*((p-1)/2)!*(Sum_{i=1..(p-1)/2} ((p-1)/2)!/i) + ((p-1)/2)!*(Sum_{i=1..(p-1)/2} (-1)^((p-1)/2)*((p-1)/2)!/(-i)) == 0 (mod p), odd prime p divides the numerator of Sum_{1 <= i <= floor(k/p)*p, i is not divisible by p} 1/i.)