

A328934


Nonsquarefree N such that A328919(N) = 1; nonsquarefree N such that {sigma_t(N) mod N: t >= 1} is purely periodic, while {sigma_t(N) mod N: t >= 0} is not.


1



18, 20, 24, 28, 44, 45, 52, 56, 60, 68, 76, 84, 88, 90, 92, 99, 116, 120, 124, 126, 132, 140, 148, 152, 153, 156, 164, 168, 172, 180, 184, 188, 198, 204, 207, 212, 220, 228, 234, 236, 240, 244, 248, 260, 261, 264, 268, 276, 280, 284, 292, 306, 308, 312, 315, 316, 332, 336
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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_(k1) be k1 distinct odd primes, k >= 3. Let N = 2^k*p_1*p_2*...*p_(k1), 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^(k2))(N) (mod 2^k) for all 1 <= t <= k1, so {sigma_t(N) mod 2^k: t >= 1} is purely periodic;
Proof. Write N = 2^k*M, then sigma_(t+2^(k2))(N)  sigma_t(N) = Sum_{dM} (d^(t+2^(k2)) + (2d)^(t+2^(k2)) + ... + (2^k*t)^(t+2^(k2))  d^t  (2d)^t  ...  (2^k*d)^t). This is divisible by 2^k if and only if 2^k  Sum_{dM} (2d)^t, or 2^(kt)  sigma_t(M). This is obvious because sigma_t(p_i) is even, as sigma_t() is multiplicative, 2^(k1)  sigma_t(M).
(b) there exists some i such that sigma_0(N) !== sigma_(p_i1)(N) (mod p_i).
Proof. Write N = M*p_i, then sigma_(p_i1)(N)  sigma_0(N) = Sum_{dM} (d^(p_i1) + (p_i*d)^(p_i1)  d  1) == (Sum_{dM} 1) = (k+1)*2^(k1) (mod p_i). As max{p_1, p_2, ..., p_(k1)} >= 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^(k1)*A002110(k1), k >= 3.


LINKS



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

(PARI) for(n=1, oo, if(A328919(n)==1 && !issquarefree(n), return(n))) \\ See A328919 for its program.


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



