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Let $A:=\{n \in \mathbb N : 2^n=\sum_{j=0}^k 10^j2^{n_j} , $ for some integers $ 0 \le n_j \le 3 , j=0,...,k $ with $ k>1\}$ ; then is the set $A$ finite or infinite ? I have been unable to find any number other than $7$ in $A$ . Please help . Thanks in advance .

user
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    Is your sum written as you intended? If you don't bound the $n_j$ then the sum doesn't match the description in the header. For example , taking $k=2$ and $n_j=4,\forall j$ we get $10^0\times 16+10^1\times 16+10^2\times 16=111\times 16=1776$. – lulu Oct 19 '17 at 18:54
  • @lulu : Yes, thank you for pointing out ... I forgot to write $n_j \le 3$ – user Oct 19 '17 at 19:07
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    No problem, I figured that's what you meant. To your question...I certainly doubt that there are any others. To prove it I'd look at the last block. There are certainly exponents that end in a block of the form you want. For example, $2^{91}= 2475880078570760549798248448$ which ends in $8248448$ . I don't immediately see why arbitrarily long strings like that are possible among powers of $2$ (though I might be wrong). Maybe you can rule that out. – lulu Oct 19 '17 at 19:13
  • @lulu Yeah it's interesting. The set of allowable digits is too large for modular conditions to rule out the possibility (almost surely)! Since $2$ is a primitive root mod $5^2$, each time we widen the block we get 5 new choices of the leftmost digit. Assuming these are essentially random, this gives a fanout of $2\times$ since each choice has a 40% chance of being valid. So chances are there will always be valid blocks of arbitrary length. If the problem had been "only digits $1$ and $2$", then there would be a decent chance of proving finiteness this way. – Erick Wong Oct 19 '17 at 19:21
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    @ErickWong : Do you have any idea regarding what happens if I ask about say powers of $m$ , $9 \ge m \ge 4$, each of whose decimal digits consist of either $1$ or $m$ ? Is it possible to say whether such a set is finite or not ? – user Oct 19 '17 at 20:01
  • @users That sounds like a reasonable question. I suggest you try asking it separately. I suspect the set is provably finite for at least some of these values of $m$ (for instance $6$ and $4$ are not primitive mod $5$ so maybe there's more room for the argument to work). Certainly $m=5$ should be trivial. – Erick Wong Oct 19 '17 at 20:12
  • @ErickWong : I have asked ... https://math.stackexchange.com/questions/2481141/perfect-powers-of-a-fixed-integer-whose-representation-in-a-given-base-also-cons , can you please provide your arguments as an answer to that question ? – user Oct 20 '17 at 08:08

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