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).

Order k= Members of the cycle 3 4 5 6 8 9 28
Known distinct cycles 0 46 1 2 2 1 1

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 'quasi-amicable' 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 Munch-Pedersen wrote (April, 2004):

"the answer to problem 23c was found back in 1997. The following cycle: R8 Dickerman 1997

1215571544=2^3*11*13813313
1270824975=3^2*5^2*7*19*42467
1467511664=2^4*19*599*8059
1530808335=3^3*5*7*1619903
1579407344=2^4*31^2*59*1741
1638031815=3^4*5*7*521*1109
1727239544=2^3*2671*80833
1512587175=3*5^2*11*1833439

can be found on my page http://amicable.homepage.dk/tables.htm under reduced sociable numbers. In the same place you can see that there at the moment are 1946 known betrothed numbers (called reduced amicable pairs on my page).

***

On May 12, Andre Needham wrote:

Just a quick note to say I'm really enjoying the Prime Puzzles website.
Also, I've found another answer for Problem 23, part b. Here's a sociable 6-cycle I recently found:

2118923133656=2^3*7*863*43844627
2426887897384=2^3*59*5141711647
2200652585816=2^3*43*1433*4464233
2024477041144=2^3*253059630143
1771417411016=2^3*11*20129743307
1851936384424=2^3*7*1637*20201767

Also, you may want to update some of the other data for this problem. Check Jan Munch Pedersen's website for updated stats on how many sociable 4-cycles have been found, and how far the exhaustive search for sociable numbers has been carried to.

***


 

 


Records   |  Conjectures  |  Problems  |  Puzzles