[seqfan] Re: Cumulative multiplication

[David Wilson]

I guess these would be "base 10 partial cyclical digital product numbers" but for this discussion let's call them "good numbers".

Let p(n) be the product of the base 10 digits of n.

Let n be a single-digit number.
Then n = p(n), and n is a good number.

Now suppose n is a good number with two or more digits.
Since n has more than one digit, p(n) < n (prove).
This means that when forming the partial products of the cyclical digits of n, we first multiply all of n's digits to reach p(n) < n, then multiply by a few more digits to reach n (because n is good).
This implies that n = k*p(n), where k is the product of the additional digits.

So if n is good, we have n = k*p(n).
That is, positive good numbers are a subset of A007602.

Now suppose n is a good number that includes the digit 0.
We then have p(n) = 0, hence n = k*p(n) = 0.
So the only good number that includes the digit 0 is 0 itself.

By definition, a good number n is a product of base-10 digits.
This means that n is either 0, or of the form 2^i*3^j*5^k*7^l (with nonnegative exponents).
By the argument above, if n > 0, n cannot include the digit 0, hence cannot be a multiple of 10.
This means that a good number n is either 0, of the form 2^i*3^j*7^k, or of the form 3^i*5^j*7^k.

By testing numbers of these forms, I confirmed that the only good numbers < 2^64 are 0,1,2,3,4,5,6,7,8,9,735,18432,442368.
It seems that Harvey found them all up to 2^64.

Empirical observations indicate that the base-10 digits of large powers of 2 are uniformly distributed.
If so, then with probability 1, all sufficiently large powers of two include the digit 0, so that only a finite number of powers of 2 omit digit 0.
Extensive search corroborates that all powers of 2 beyond 2^86 = 77371252455336267181195264 include the digit 0 (see A007377).

A similar argument applies to numbers of the forms  2^i*3^j*7^k and 3^i*5^j*7^k.
Their base 10 digits should likewise be uniformly distributed, and sufficiently large numbers of this form should include the digit 0, hence are not good numbers.
This strong indicates that the set of good numbers is finite.

Note also that a very large number of the correct form that omits the digit 0, is still unlikely to be a good number on statistical grounds (the long sequence of digital products is much more likely to miss n than to hit it).
I would be very surprised if we found any more good numbers.