Divisor chains

[Richard Guy]

The following might be an acceptable sequence
for Neil Sloane's OEIS, but someone needs to
do some work

1  2  3  4  5  6  7  8  9 10 11 12 13 14 ...
1  1  1  1  1  1  1  5  4  1  2  5  5  4 ...

It arises from a problem I got recently from
Paul Vaderlind.

If the sequence of numbers from  1  to  37,
is arranged so that each term is a divisor
of the sum of preceding ones, starting
                 37, 1, ...
what is the next term?

The answer is either  2  or  19.  P'r'aps
I won't spoil your fun by pointing out
which.  But I will spoil it by asking:
How do you know that there is such a
`divisor chain' ?

The sequence is (my present state of knowledge
of) the number of divisor chains of length  n.
Here are the ones I found

1        2 1         3 1 2        4 2 3 1

5 1 2 4 3     6 2 4 3 5 1     7 1 2 5 3 6 4

8  2  5  3  6  4  7  1
8  4  2  7  3  1  5  6
8  4  2  7  3  6  5  1
8  4  3  5  1  7  2  6
8  4  6  3  7  2  5  1

9  1  2  4  8  6  5  7  3
9  1  2  6  3  7  4  8  5
9  3  4  8  6  5  7  2  1
9  3  6  2  1  7  4  8  5

10  2  4  8  3  9  6  7  1  5

11  1  2  7  3  8  4  9  5  10  6
11  1  4  8  6  10  5  9  2  7  3

12  2  1  5  10  3  11  4  8  7  9  6
12  2  7  3  8  4  9  5  10  6  11  1
12  3  5  10  2  8  4  11  1  7  9  6
12  4  8  3  9  6  2  11  5  10  7  1
12  4  8  6  10  5  9  2  7  3  11  1

13  1  2  8  3  9  4  10  5  11  6  12  7
13  1  2  8  6  10  4  11  5  12  9  3  7
13  1  2  8  6  10  4  11  5  3  9  12  7
13  1  2  8  12  4  10  5  11  3  9  3  7
13  1  2  8  12  4  10  5  11  6  9  3  7 

14  2  8  6  10  4  11  5  3  9  12  7  13  1
14  2  8  6  10  4  11  5  12  9  3  7  13  1
14  2  8  12  4  10  5  11  6  9  3  7  13  1
14  2  8  12  9  3  4  13  1  11  7  6  10  5
		 
Enough sins of omission, to say nothing of
commission, for today.  Please check and extend.
Any ideas for proving anything?      R.