[seqfan] Re: Numbers that are divisible by the product of their digits

[hv at crypt.org]

Earlier I wrote:
:Charles Greathouse <charles.greathouse at case.edu> wrote:
::A007602 is the sequence of numbers that are divisible by the product of
::their digits. Does anyone know how this sequence grows?
:
:Taking the simpler case of base 3, from 1 .. 3^n-1 the number of values
:consisting of 1s and precisely k 2s is sum_k^n{ C(k, i) } = C(n+1, i+1).
:Assuming that asymptotically such a number is divisible by its digit
:product 1 / 2^k of the time, that gives me an expected count up to 3^n of:
:  sum_{i=0}^n { 2^(-i) C(n+1, i+1) }
[...]
:I do not know if it is true or provable that this will asymptotically
:approach the actual counts; but if it is, extending it to base 6 should
:give an example usefully similar in structure to base 10 but of lesser
:complexity.

For base 10 (or base 6), restrictions on the presence of 2 and 5 (or
respectively 2 and 3) introduce some complications. For example, if any
5s are present, the number must consist solely of odd digits; if there
are n 5s, that fixes the last n digits to be (the reverse of) the first
n digits of this sequence:

5 7 3 9 5 3 3 9 1 3 7 7 3 7 7 5 1 5 7 9 3 7 3 3 9 9 9 5 3 7 1 7 9 5 7 9 3 1 3
9 5 7 5 5 7 7 1 1 3 1 9 5 9 1 9 5 5 7 7 7 7 7 9 7 9 9 7 7 3 3 5 9 7 7 1 1 5 9
7 1 5 1 7 5 9 1 7 9 3 5 3 9 1 7 9 9 1 9 9 1 5 3 7 5 9 3 7 1 3 9 7 9 3 9 7 3 3

So for example, the only 3-odd-digit ending for a multiple of 5^3 is 375.

Before calculating it I was expecting a simple and obvious closed form
for this sequence, but if there is one it isn't obvious to me. Can someone
offer analysis?

Hugo