1248136

You start with $1$ , $128$ , $16$ , $2$ , $256$ , $32$ , $4096$ , $512$ , $64$ , $8192$ to obtain the numbers $124$ , $125$ , $136$ , $248$ , $251$ , $361$ , $481$ , $512$ , $612$ , $813$ respectively. This just leaves $137$ , $249$ , $371$ , $491$ , $712$ , $713$ and $913$ to find.

If we can obtain $137$ by starting with $2^n$ , we must obtain $371$ by starting with $2^{n+1}$ (and one of $712$ or $713$ next). If we can obtain $249$ by starting with $2^n$ , we must obtain $491$ by starting with $2^{n+1}$ and then obtain $913$ by starting with $2^{n+2}$ . Thus finding three of these numbers follows from finding the others, and so we are left with having to find $137$ , $249$ , $712$ and $713$ explicity.

To find the number $137$ , we need to find positive integers $a,n$ such that $1.75 \times 10^a \le 2^n < 2\times10^a$ in which case we obtain $137$ by starting with $2^n$ . These inequalities reduce to finding $a,n$ such that $\frac{\log1.75 + a}{\log 2} \le n < 1 + \frac{a}{\log 2}$ The smallest value of $a$ for which this is possible is $a=13$ , with $n=44$ . Starting with $2^{44}$ , we obtain the number $137$ . Then starting with $2^{45}$ we obtain the number $371$ , and starting with $2^{46}$ we obtain the number $712$ .

To find the number $249$ , we need to find positive integers $a,n$ such that $2.25 \times 10^a \le 2^n < 2.5 \times 10^a$ in which case we obtain the number $249$ by starting with $2^n$ . These inequalities reduce to requiring $\frac{\log2.25 + a}{\log 2} \le n < \frac{\log2.5 + a}{\log 2}$ The smallest value of $a$ for which this is true is $a=15$ , with $n=51$ . Then starting with $2^{51}$ yields the number $249$ .

Finally, to find the number $713$ , we need to find positive integers $a,n$ such that $7.5 \times 2^a \le 2^n < 8 \times 2^a$ in which case we obtain the number $713$ by starting with $2^n$ . These inequalities reduce to $\frac{\log7.5 + a}{\log 2} \le n < \frac{\log8 + a}{\log 2}$ The smallest value of $a$ for which this is true is $a=28$ , with $n=96$ . Thus starting with $2^{96}$ yields the number $713$ .

Thus we can obtain all the other numbers. Whether this argument is much more informative than plugging the following two lines into Mathematica:

a[n_] := IntegerDigits[2^n][[1]]*100 + 
  IntegerDigits[2^(n + 1)][[1]]*10 + IntegerDigits[2^(n + 2)][[1]]

DeleteDuplicates[Sort[Table[a[n], {n, 0, 99}]]]

is moot.

Yea, that's what I want to know too. I didn't post my solution (even though I wrote one) because I didn't have a justification for those 17 other than "I checked to see if they exist, and they do"

Yes, this works too.