A factorization balancing act

A couple of weeks ago, Claudio Meller presented 26487 and 65821 as examples of the property of having one each of the base-ten digits when combined with the digits of their respective factorizations. Surprisingly, he missed two:

    26487 = 3^5 * 109
    28651 = 7 * 4093
    61054 = 2 * 7^3 * 89
    65821 = 7 * 9403

I wondered how this might be turned into a sequence. Base-ten k-balanced factorization integers: The combined digits of an integer and its factorization primes and exponents contain exactly k copies of each of the ten digits. So,

 45849660 = 2^2 * 3 * 5 * 19 * 37 * 1087

 84568740 = 2^2 * 3 * 5 * 67 * 109 * 193

104086845 = 3^2 * 5 * 19 * 23 * 67 * 79

106978404 = 2^2 * 3 * 13 * 685759

and so on. For any given k, k-balanced integers are necessarily finite. For k=2, there are a little over 13000. Is the largest of these greater than the smallest 3-balanced integer?

It's not too difficult to generate very large terms:

345655692176023955231233047798293565182493067678373538829683596019374251932061880049408412541410264671864570007449201127

is an example of a 13-balanced integer. Can you come up with a larger one?

My 100th Leyland prime find

Last October I found my first previously unknown Leyland (probable) prime. Today I found my 100th. The graph shows (in order of discovery) the number of decimal digits of those 100 primes, ranging from 43633 to 61184. The finds have had the unintended (but certainly not unpleasant) consequence of pushing me up a list of probable prime contributors! My search for Leyland primes is of course in aid of my Leyland prime indexing effort which has reached #1137. There'll be a significant jump in about three weeks when I finish rounding up the few remaining unknowns with decimal digit lengths of between 54334 and 55390.