cp's OEIS Frontend

For the definition of unitary divisors see A034448. This sequence is a permutation of A327157; the starting numbers of successive cycles are in increasing order; the numbers in a cycle are kept in the order of the iteration with the smallest number in the cycle as the starting number. In order to be consistent with A327157 the terminal 1-cycle consisting of 1 is not included in the sequence.

Sequence A336218 gives the cycle lengths, therefore the start of the k-th cycle in this sequence is at index 1 + Sum_{i=1..k-1} A336218(i). Sequence A336219 is the first column of the triangle.

From the formula of Vladeta Jovovic in A034448, it follows that all unitary aliquot sequences, and hence cycles, contain only odd numbers or only even numbers (except for the possible terminal 1). The table of Antti Karttunen in the link of A327157 includes just 2 odd cycles, the 2-cycles: 8619765, 9627915 and 17257695, 17578785.

This is the first column of the irregular triangle in A336216.

From the formula of Vladeta Jovovic in A034448 we get for an even number n not divisible by 4 and odd prime p: usigma(2^m * p * n) = (2^(m+1) + 1) * (p + 1) * usigma(n) / 3 so that usigma(2^m * p * n) = (2^m * p * n) * usigma(n) when 3* 2^m * p = (2^(m+1) + 1) * (p + 1), and consequently, p = (2^(m+1) + 1) / (2^m - 1), i.e. p = 5 for m = 1, and p = 3 for m = 2.

Therefore, if all members a_1, a_2, ... , a_k, a_1 of a cycle are even and not divisible by 4 and 5 then 10*a_1, 10*a_2, ... , 10*a_k, 10*a_1 form a cycle, and if all members a_1, a_2, ... , a_k, a_1 of a cycle are even and not divisible by 3 and 4 then 12*a_1, 12*a_2, ... , 12*a_k, 12*a_1 form a cycle.

If all members a_1, a_2, ... , a_k, a_1 of a cycle are odd, divisible by 3, but not divisible by 5 and 9 then 15*a_1, 15*a_2, ... , 15*a_k, 15*a_1 form a cycle. No such cycles exist in the current data up to 27287260.