Problems & Puzzles: Problems Problem 23. Divisors (V) Aliquot sequences, Sociable Numbers. As we saw in the Problem 22, one possible end of an aliquot sequence is to reach to a cycle of numbers. If two numbers form the cycle (order 2), they are named "amicable numbers", If more than 2 numbers form the cycle (order k=>3) they are called "sociable numbers" How many cycles of sociable numbers are known? The answer is, 53 cycles are known. The table below summarizes the number of known sociable cycles as given in the compilation by D. Moews (1995).
D. Moews says: "P. Moews and I have exhaustively searched for all aliquot cycles of length greater than 2 such that the number preceding the largest number of the cycle is below 1.54*10^11. As of November 1997, Jan Otto Munch Pedersen has carried this search up to 4.27*10^11…" If you want to know one by one all these 53 cycles please see: "Moews, D. ``A List of Aliquot Cycles of Length Greater than 2.'' Rev. Dec. 18, 1995. http://xraysgi.ims.uconn.edu:8080/sociable.txt. I have verified the existence of some of these cycles, and found that the least number that generates  in some instant of the develop of the aliquot sequence  the cycle of order k = 5 is n = 9464. The least number that generates the cycle of order k = 28 is n = 2856. Outstanding is the fact that it hasn't been discovered any cycle of order 3. Questions: a) Can you find the least number that generate each of the other 51 sequences? b) Can you generate a different cycle (regarding the order or another specific cycle for some of the known orders) to the 53 already found? Jud McCranie proposes to define a similar sequences to the aliquot ones, using the c(n) = s(n)  n  1 function (this is the Chowla's function) instead of the already used s(n) = s(n)  n. And the new questions are: c) At the end of these new sequences, do exist cycles of 3 or more members? Can you find the first one of these? d) Hagis & Lord have discovered 46 pairs of cycles of two numbers (also named 'quasiamicable' or 'betrothed' numbers) below 10^7, all of them with opposite parity. But, apparently nothing inhibits pairs of the same and even parity. Would you like to find the first one of them, or to explain why they can not exist? See Ref 2, B5, p. 59 Jan MunchPedersen wrote (April, 2004):
*** On May 12, Andre Needham wrote:





