[seqfan] Re: D(n)

[David Seal]

> There are still only 8 terms up to now, and I can’t find any more,
> so perhaps this sequence may be finite after all; would you agree?

No, it's provably infinite. Define a sequence b(n) by:

b(0) = 2

For n > 0, b(n) = (10^b(n-1) - 1) / 9 = string of b(n-1) 1s

So b(1) = 11, b(2) = 11111111111, b(3) = string of 11111111111 1s, etc.

Then for n > 0, D(b(n)) = b(n-1) and b(n) does not equal b(n-1). Combining that with D(b(0)) = D(2) = 2 = b(0), b(n) requires n applications of D() to reduce it to an all-digits-different value for all n >= 0. So for all n >= 0, integers requiring n applications of D() to reduce them to having all different digits exist, and since a(n) is defined to be the smallest such integer, a(n) exists.

This is just an existence proof, of course, not an aid to determining a(n): b(n) is clearly hugely bigger than a(n) for all n > 1.

David