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It is easy to see that {sigma_k(N) mod N: k >= A051903(N)} is purely periodic.

All terms are nonsquarefree: if N is squarefree and N is here, then A328919(N) < A051903(N) = 1, so A328919(N) = 0. By the property mentioned in A328919, a necessary condition is that for every prime p dividing N, write N = p*s, we have p divides d(s), d = A000005. But d(s) is a power of 2, so N = 2, and 2 is not here.

Although it seems that for most N we have A328919(N) = A051903(N), this sequence is infinite. See A328934 for more information.

If N is squarefree then A328919(N) = 1, but the converse is not true. While it is conjectured that 12 is the only N such that 0 = A328919(N) < A051903(N), there are infinitely many N such that 1 = A328919(N) < A051903(N).

Let p_1, p_2, ..., p_(k-1) be k-1 distinct odd primes, k >= 3. Let N = 2^k*p_1*p_2*...*p_(k-1), then N is here. It is easy to see that {sigma_t(N) mod 2^k: t >= k} and {sigma_t(N) mod p_i: t >= 1} are both purely periodic.

To show this, it is sufficient to show that:

(a) sigma_t(N) == sigma_(t+2^(k-2))(N) (mod 2^k) for all 1 <= t <= k-1, so {sigma_t(N) mod 2^k: t >= 1} is purely periodic;

Proof. Write N = 2^k*M, then sigma_(t+2^(k-2))(N) - sigma_t(N) = Sum_{d|M} (d^(t+2^(k-2)) + (2d)^(t+2^(k-2)) + ... + (2^k*t)^(t+2^(k-2)) - d^t - (2d)^t - ... - (2^k*d)^t). This is divisible by 2^k if and only if 2^k | Sum_{d|M} (2d)^t, or 2^(k-t) | sigma_t(M). This is obvious because sigma_t(p_i) is even, as sigma_t() is multiplicative, 2^(k-1) | sigma_t(M).

(b) there exists some i such that sigma_0(N) !== sigma_(p_i-1)(N) (mod p_i).

Proof. Write N = M*p_i, then sigma_(p_i-1)(N) - sigma_0(N) = Sum_{d|M} (d^(p_i-1) + (p_i*d)^(p_i-1) - d - 1) == -(Sum_{d|M} 1) = (k+1)*2^(k-1) (mod p_i). As max{p_1, p_2, ..., p_(k-1)} >= A000040(k) > k+1, there exists some prime p_i that does not divide p_i.

This shows that A051903(N) - A328919(N) is unbounded, and can take every natural number as its value for infinitely many times. It is conjectured that the smallest N such that A328919(N) = 1 and A051903(N) = k is N = 2^(k-1)*A002110(k-1), k >= 3.