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From a(n) is the smallest number m such that the sum of the digits of m^3 is equal to n^3. we can find the first 6 items are: {1, 2, 27, 1192, 341075, 3848163483}

I have found and verified that a(7)=2064403725539899.

My conjectures:

a(8) is a 23-digit integer.

a(9) is a 33-digit integer.

a(10) is a 45-digit integer.

a(n) has about $\lceil\frac{n^3 + 5.625}{23.2}\rceil$ digits.

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    $\begingroup$ How did you verify $a(7)$ ? $\endgroup$
    – Peter
    Commented Jun 19 at 9:30
  • $\begingroup$ I've found that the largest sum of cubes of 15-digit integers is 342. So a(7) needs at least 16 digits. As 343 is only a bit larger than 342, a 46-digit cube seems enough,we can narrow the candidate numbers between $10^{16}, and \ [10^{46/3}]$,and it's a luck to get 2064403725539899 as the solution. $\endgroup$
    – Mrexcel
    Commented Jun 20 at 3:40
  • $\begingroup$ And as $343\equiv 1 \bmod 9 $, so $a(7) \equiv \{1,4,7\} \bmod 9 $,only 1/3 candidate numbers need to be verified. $\endgroup$
    – Mrexcel
    Commented Jun 20 at 3:45
  • $\begingroup$ $a(8)<= 99995999799995999999999.$ $a(9)<= 999699989999999949999999999999999.$ $a(10)<=199999999929999999999949999999999999999999999.$ $\endgroup$
    – Mrexcel
    Commented Jun 20 at 13:43
  • $\begingroup$ $a(11)<=5999999999999599999999999999799999999999999999999999999999.$$a(12)<=999499998999999999999999999999999999996999999999999999999999999999999999999999.$ $\endgroup$
    – Mrexcel
    Commented Jun 20 at 14:13

1 Answer 1

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K^3 must have a digit sum >= 1000. The first x with digitsum(x) >= 1000 is 1 followed by 111 9s, and we can find the smallest k with k^3 >= 1 followed by 111 nines, some k >= 10^31. Actually a big bigger. We check this k.

If the digit sum of k^3 is not 1000, we find the first x > k^3 with a digit sum of 1000. That’s probably 28 followed by 110 nines, or 37 followed by 110 nines, and we go on like this. Since digitsums >= 1000 won’t be much larger than 1000, it should not take long to find a k where the digitsum of k^3 equals 1000.

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  • $\begingroup$ I think that your arithmetic is a bit off: If $k^3 \ge 2 \times 10^{111} - 1$, then the minimum $k$ should be on the order of $10^{37}$, not $10^{31}$. $\endgroup$
    – Dan
    Commented Jun 19 at 17:00
  • $\begingroup$ To be precise, $k \ge 12599210498948731647672106072782283506$. $\endgroup$
    – Dan
    Commented Jun 19 at 17:03
  • $\begingroup$ $a(5)^3 = 341075^3 = 39677989979796875$ So you can't just go on finding next $x>k^3$ with digit sum $=n^3$ because its too slow, and you won't find one in time $\endgroup$
    – EnEm
    Commented Jun 19 at 21:23

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