Sunday, August 28

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?

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