OFFSET
1,1
COMMENTS
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.
EXAMPLE
For t > 0, sigma_t(24) == 12 (mod 24) if t is odd, sigma_t(24) == 10 (mod 24) if t is even. Note that sigma_0(24) = 8, so A328919(24) = 1, so 24 is a term.
PROG
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Oct 31 2019
STATUS
approved