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How about 138? (Posted on 2018-05-10) Difficulty: 4 of 5
Let us define a sequence S(N): a1= N,
an= sum of all factors of an-1, excluding an-1 itself. Most sequences either converge or begin to repeat.

S(12) → 12,16,15,9,4,3,1,1,1...
S(6) → 6,6,6,6,...
S(220) → 220,284,220,284,220,284,...
Try to evaluate S(50), S(80), S(99).

How about S(138)?
Please comment!

No Solution Yet Submitted by Ady TZIDON    
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Comments: ( Back to comment list | You must be logged in to post comments.)
re: computer solution | Comment 3 of 4 |
(In reply to computer solution by Charlie)

Each of these is the Aliquot sequence of a number.  Many of the interesting things you noticed are here:

Some highlights:

6 and 28 are perfect numbers.  You table ends before the next which is 496.

220 and 284 are the first pair of amicable numbers.

276 is the first of a few numbers with unknown end behavior, but it is conjectured they do become periodic at some point.

  Posted by Jer on 2018-05-10 14:46:46
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