- Paul Zimmermann
- Wolfgang Creyaufmueller
- Juan Luis Varona
- Wieb Bosma
- Sander Hoogendoorn
- Christophe Clavier
- Chris Caldwell

An aliquot sequence can end in one of four ways:

n Divisors Sum 121, 2, 3, 4, 6 16 161, 2, 4, 8 15 151, 3, 5 9 91, 3 4 41, 2 3 31 1 1end of the sequence .

- with a prime number, followed by 1, as above.
- with a perfect number, such as 6.
- with a cycle of two or more numbers.
- increases without limit.

An example of the third case is 220:

n Divisors Sum 251, 5 6 61, 2, 3 6

It is currently not known whether any number leads to a sequence that increases without limit (case 4). In an attempt to decide this question, many aliquot sequences have been computed. The ultimate fate of the sequences for all numbers up to 1000 has been found to be one of the first three cases, except for 276, 552, 564, 660, and 966. The ends of these five sequences (the “Lehmer five”) are currently unknown. Various people, including most of those listed at the top of this page, are working on sequences with larger starting values. I have done some work on the sequence that starts with 966. See below for the status of this sequence.

n Divisors Sum 2201, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110 284 2841, 2, 4, 71, 142 220

It is not actually necessary to list
all of the divisors of a number
to compute the sum of the divisors.
Instead, there is a formula that uses
the prime factorization of the number.
If the factorization of *N* is

whereN=p*^{a}q*^{b}r* ...^{c}

ThisSum= [ (p- 1) / (^{(a+1)}p- 1) ] * [ (q- 1) / (^{(b+1)}q- 1) ] * [ (r- 1) / (^{(c+1)}r- 1) ] * ...

As an example: 12 = 2^{2} * 3^{1}
(usually written: 2^{2} * 3).
The sum of its divisors (including 12 itself) is:

The sum of the aliquot divisors is then 28 - 12 = 16.Sum= [ (2^{3}- 1) / (2 - 1) ] * [ (3^{2}- 1) / (3 - 1) ]

= [ (8 - 1) / (2 - 1) ] * [ (9 - 1) / (3 - 1) ]

= [ 7/1 ] * [ 8/2 ] = 7 * 4 = 28

Starting value: 966 #0 = 966 = 2 * 3 * 7 * 23 #1 = 1338 = 2 * 3 * 223 #2 = 1350 = 2 * 3^3 * 5^2 #3 = 2370 = 2 * 3 * 5 * 79 #4 = 3390 = 2 * 3 * 5 * 113 #5 = 4818 = 2 * 3 * 11 * 73 #6 = 5838 = 2 * 3 * 7 * 139 #7 = 7602 = 2 * 3 * 7 * 181 #8 = 9870 = 2 * 3 * 5 * 7 * 47 #9 = 17778 = 2 * 3 * 2963 #10 = 17790 = 2 * 3 * 5 * 593 #11 = 24978 = 2 * 3 * 23 * 181 #12 = 27438 = 2 * 3 * 17 * 269 #13 = 30882 = 2 * 3 * 5147 #14 = 30894 = 2 * 3 * 19 * 271 #15 = 34386 = 2 * 3 * 11 * 521 #16 = 40782 = 2 * 3 * 7 * 971 #17 = 52530 = 2 * 3 * 5 * 17 * 103 #18 = 82254 = 2 * 3 * 13709 #19 = 82266 = 2 * 3 * 13711 #20 = 82278 = 2 * 3^2 * 7 * 653Click here to see all of terms (so far) in the sequence.

The sequence has been calculated
up to the 959^{th} term, which is:

#959 = 75521278771217953409751209776054073314771137357812063153906615114\ 56191440344410174143346010432586733455633829814351208029335820367\ 94114579306773775418042614407002005671675856948450923010096 = 2^4 * 3^3 * 23 * 29 * 262095614592765955250677472985917018278260652166319837143603944\ 953085659959756586086933825116351086035302967606972597313472979\ 4952503312855054326223756325359566751366232424957481158459The last factor above (2620956...1158459) is a 184-digit composite number, whose factors are currently unknown.

Usually, each term in the sequence is larger than the preceding term.
However, each of the terms #658, #659, and #660 is **smaller**
than the preceding term.
Then the sequence increases again, starting with term #661.
The sequence decreases again from #704 thru #708.

Send e-mail to me at jrhowell@ix.netcom.com

Go to my math index page.

Go to my home page.

Last updated on May 10, 2017.