Lengths Of Runs In The #-Of-Divisors Sequence

[Joshua Zucker]

On 7/17/07, Leroy Quet <qq-quet at mindspring.com> wrote:
> A harder question to answer:
> We can define sequence S(m) = {s(m,n)}, where s(m,n) = the length of the
> nth run of similar consecutive values in the sequence S(m-1), where S(0)
> = sequence A000005.
> (And S(1) = A131789, S(2) = A131790, of course.)
>
> Is every term of S(m) finite for every m = positive integer?
>
> It seems intuitive obvious that, yes, all terms are finite. But a proof
> would be harder to produce.

There seems to be a strongly alternating pattern:
Divisor sequence: 1 2 2 3 2 4 2 4 3 4 2 6 2 4 4 5 2 6 2 6 4 4 2 8 3 4
4 6 2 8 2 6 4 4 4 9 2 4 4 8 2 8 2 6 6 4 2 10 3 6 4 6 2 8 4 8 4 4 2 12
2 4 6 7 4 8 2 6 4 8 2 12 2 4 6 6 4 8 2 10 5 4 2 12 4 4 4 8 2 12 4 6 4
4 4 12 2 6 6 9

Run lengths in divisor sequence: 1 2 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 2
1 1 1 2 1 1 1 1 1 3 1 1 2 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 3 1 1 1 1 1 3 1 1 2 1 1 1 1 2
1 1 1 1 1 1 1 1 1 1 2

Run lengths in that: 1 1 10 1 5 1 3 1 5 1 2 1 4 1 11 1 16 1 8 1 5 1 2
1 4 1 10 2 2 1 9 2 4 1 1 2 9 1 11 1 4 1 10 1 10 1 1 1 6 1 1 1 10 1 9 1
7 1 30 1 9 2 1 1 22 1 4 2 8 1 28 1 4 1 4 1 4 1 33 1 3 1 9 1 5 1 26 1
18 1 4 1 5 1 10 1 9 1 3 1

Run lengths in that: 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 2 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 3 1 3 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1

And then: 1 25 1 4 1 10 1 1 1 10 1 55 1 15 1 4 1 63 1 3 1 12 1 33 1 5
1 32 1 9 1 7 1 3 1 13 1 56 1 61 1 5 1 103 1 47 1 17 1 13 1 25 1 5 1 5
1 47 1 3 1 21 1 7 1 11 1 1 1 17 1 3 1 1 1 8 1 1 1 5 1 7 1 9 1 2 1 15 1
36 1 5 1 11 1 7 1 1 1 15

And then: 1 1 1 1 1 1 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3
1 1 1 3 1 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 3 1 1 1 1 1 1 1 1 1 1 1

And then: 6 1 57 1 3 1 1 1 17 1 63 1 7 1 9 1 101 1 17 1 43 1 13 1 27 1
129 1 37 1 17 1 9 1 39 1 15 1 15 1 45 1 27 1 11 1 3 1 19 1 41 1 5 1 51
1 9 1 47 1 33 1 15 1 35 1 7 1 7 1 15 1 13 1 29 1 53 1 25 1 9 1 23 1 69
1 9 1 5 1 35 1 13 1 9 1 9 1 37 1

And then: 1 1 1 1 1 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

And then: 5 1 105 1 3 1 49 1 29 1 5 1 31 1 15 1 9 1 15 1 17 1 5 1 61 1
9 1 1 1 1 1 1 1 13 1 11 1 11 1 41 1 51 1 19 1 73 1 5 1 55 1 31 1 41 1
67 1 133 1 5 1 5 1 85 1 9 1 7 1 61 1 11 1 5 1 1 1 15 1 29 1

And so on, where alternately you get (long strings of 1s with the
occasional other number) and (alternately 1s and other numbers, with
occasionally 2 or 3 1s in a row).

By the way, using the first million terms of A000005, I get only the
first 82 terms of S(8), so it'll take some work to explore this any
further numerically.

I don't think it's at all obvious that this thing won't eventually
yield a sequence of all 1s, even if it is intuitively obvious to Leroy
Quet, it's sure not obvious to me!

--Joshua Zucker
PS: I also submitted extensions of A131789 and A131790 as part of this work.



Joshua Zucker wrote:
>I don't think it's at all obvious that this thing won't eventually
>yield a sequence of all 1s, even if it is intuitively obvious to Leroy
>Quet, it's sure not obvious to me!

I think my use of the phrase "intuitively obvious" is quite an 
exaggeration.
It just seems *likely* to me that the terms of each S(m) are finite. 
Nothing is really obvious.

:)

Thanks,
Leroy Quet