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Twelve Divisor Trial (Posted on 2010-04-09) Difficulty: 3 of 5
Barbara writes a sequence of integers starting with the number 12. Each subsequent integer she writes is chosen randomly with equal probability from amongst the positive divisors of the previous integer (including the possibility of the integer itself). She keeps writing integers until she writes the integer 1 for the first time, and then she stops.

An example of one such sequence is 12, 6, 6, 3, 3, 3, 1.

What is the expected value of the number of terms in Barbara’s sequence?

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: BARBARA IS EXPECTING (tiny error) | Comment 4 of 7 |
(In reply to BARBARA IS EXPECTING by Ady TZIDON)

At the end:  30*L(12)=36+22+21+42=111

but actually 36+22+21+42 = 121

So L(12) = 121/30 as Daniel and I got.

All 3 of our solutions are basically the same.

  Posted by Jer on 2010-04-09 15:30:47

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