Comments on A230624, Numbers that are generated in every base
An email exchange between N. J. A. Sloane and David L. Applegate
Dec 31 2021 - 

NJAS:
Kaprekar says that n is "generated by k in base b" if n = k + S_b(k), where S_b(k) is the sum of the digits of k when k is written in base b. For example in base 10, 101 has two generators, 91 and 100. 101 is the smallest number with two generators in base 10. For more background see the paper that Max Alekseyev and I just finished: http://neilsloane.com/doc/colombian12302021.pdf (it is also on the arXiv but this version is better). A "self-number in base b" has no generator. All numbers here are nonnegative, by the way.

What I am writing about is A230624, the list of numbers that have a generator in every base b >= 2. There are 90 known terms (the b-file is from Lars Blomberg). It begins 0, 2, 10, 14, 22, 38, 62, 94, ... It seems it is not known whether this sequence is infinite. 
I created subsidiary sequences A349820 - A349823, hoping some structure would emerge, without much success.

It is easy to see that all terms must be even, and if the number is n = 2t, once b is greater than t, n is generated by the base-b single-digit number t. So we only need to find generators for bases 2 through n/2.  See A349223 for certificates for the first few terms of A230624.

DLA: (Jan 02 2022)
A couple of comments about A230624 (some probably already known, but not
explicitly mentioned in the email or sequence comments, and I didn't
notice them in a quick glance at the linked paper:

1. If b is odd, then n is generated by k in base b iff n is even.  Hence
we don't need to consider odd b or odd n.
2a. (as mentioned above) If n is even, then n is generated by k=n/2 in every base b > n/2.
2b. Claim: If n is even, let 2^i be the largest power of 2 dividing n+2.  If (n+2)/2^i - 1 > sqrt(n), then there is no k such that n is generated by k in base b=(n+2)/2^i - 1 > sqrt(n).  Otherwise n is generated by some k in every base b with sqrt(n) <= b <= n/2.

Proof sketch:
If n is even, b is even, and 2b <= n < b^2 + 1 and n is generated by k in base b, 
then k must have exactly 2 digits. 
Set x = 2*floor(n /(2b+2)), y=mod(n, 2b+2)/2. Then x < b, and if 2b+2 does not divide n+2, y < b.  
In that case, x is generated by k=x*b+y in base b
 (k + S_b(k) = (x*b + y) + (x + y) = x*(b+1) + 2y = (2b+2)*floor(n / (2b+2)) + mod(n, 2b+2) = n).

Thus, for sqrt(n) <= b <= n/2, the bases b without generators are
exactly the even b where b+1 divides (n+2)/2. Since b+1 is odd, we can
let 2^i be the largest power of 2 that divides (n+2),
and write (b+1) (j 2^i) = n+2, or b = (n+2) / (j 2^i) - 1, for j odd.  
But if j=1 gives a b >= sqrt(n), there is no generator.  
If j=1 gives a b < sqrt(n), then
all larger j will also give b < sqrt(n). So it suffices to consider j=1.

Possible gaps:
I might have gotten a < vs <= mixed up, or an off-by-one error to make
sure that for the examples cited, 1-digit and 3-digit k are excluded.
Also, it is possible I overlooked something in the possible 2-digit
solutions.  Or, there's something more fundamental wrong.

Given that claim, n is in A230624 iff:
1. n is even,
2. n has a generator for each base b <= sqrt(n), and
3. (n+2) / 2^i - 1 <= sqrt(n) (where 2^i is the largest power of 2
dividing n+2).

I'm generating a b-file containing entries <= 10^9 based on this claim,
and will upload it once it's done (it's currently up to 0.8 * 10^9).

NJAS (Jan 5 2022):
Dave, The first point in your argument is certainly well-known.
See for example Santanu Bandyopadhyay, <a href="https://www.ese.iitb.ac.in/~santanu/RM8.pdf">Self-Number</a>, Indian Institute of Technology Bombay (Mumbai, India, 2020). 
I added a link to it to A230624.  I also added a link to one of the main sequences related to this topic, A003052.

NJAS (Jan 8 2022):
Dave, I have now checked the arguments in your message of Jan 2 pretty carefully, and I believe your theorem 2b:

2b. Claim: If n is even, let 2^i be the largest power of 2 dividing
n+2.  If (n+2)/2^i - 1 > sqrt(n), then there is no k such that n is
generated by k in base b=(n+2)/2^i - 1 > sqrt(n).  Otherwise n is
generated by some k in every base b with sqrt(n) <= b <= n/2.

  It is easy to see that for bases b > n/2, an even n has the single-digit generator (n/2)_b.
So you looked at the range of bases where we must use two-digit generators, and basically showed that a two-digit generator exists iff b+1 does not divide (n+2)/2.

This cut down the search for "Universally generated numbers", A230624, to searching for generators to the range of bases b <= sqrt(n). Very nice!

I took a look at the small even bases. Sequences A228082, A349831, A349832, A349833 now list the even numbers that are generated in base 2, bases 2 and 4, bases 2,4,and 6, and bases 2,4,6, and 8, respectively.  It is a pity that even the simplest of these, A228082 (or its complement, A010061, Kaprekar's "self-numbers") doesn't have a simple formula. The sequence we are trying to prove is infinite, A230624, is of course the limiting intersection.

DLA (Jan 9 2022):
For me, the tricky part of "a two-digit generator exists iff b+1 does not divide (n+2)/2" is what relationships are required between b and n, and whether they're inside the "iff" or outside.
One version of a precise statement is: a two-digit generator exists iff:
(1) b is even, n is even, 2b+2 <= n <= b^2 + b - 2, and b+1 does not divide n+2, or
(2) b is even, n is odd, b+1 <= n <= b^2 + 2b - 2, and b+1 does not divide n+2, or
(3) b is odd, n is even, and b+1 <= n <= b^2 + 2b - 3.

The result for one-digit generators can be written as: a one-digit generator exists iff n is even, n <= 2b-2.

An interesting minor consequence is that no multiples of 4 are universally generated: n=4k does not have a generator for b=2k.

As for proving that A230624 is infinite: I suspect that this might just be a probabilistic result: for a given even b, and n > b^2, consider

generator(n,b)
   if n <= b^2 + 2b - 2, return the 2-digit generator for b, n   // fails if b+1 divides n+2.
   else:
     write n = x (b^k+1) + y, where 1 <= x < b, 0 <= y < b^k +1
     // Since y < b^k + 1, the recursive call will only use terms using b^(k-1),...
     return x b^k + generator(y,b)

For large n, this will only fail with probability 1/(b+1).  If the failures were independent, the probability that n survives bases 2..n is then at least 1/2 * 2/3 * 4/5 * 6/7 * 8/9 * 10/11 * 12/13 ... >= 1/(n+1).  Of course, the failures aren't completely independent, but intuitively failures are likely to be positively correlated rather than negatively correlated.

Experiments seem to bear this out: In the plot below, 'remaining' plots the number of even n between 0 and 10^6 that have generators for even bases 2..x, and 'prediction' = 10^6 * 1/2 * prod(b/(b+1), b=2..x by 2).

[Insert Figure https://oeis.org/A230624/a230624.pdf here]

Below the top graph, add the sentence
To see the relationship more clearly, here's a plot of remaining / prediction:
and below the second graph, add:
As far as I know, though, that doesn't help at all with proving that the sequence is infinite.

-Dave