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.