Possibilities for changing one $1024$ banknote to banknotes from $\{2^i: i = 0,\ldots,9 \} $ How many possibilities do there exist to change one banknote $1024$ to denominations from the set set $\left\{2^i: i = 0,...,9 \right\} $.

I think that there really are a lot, but I don't have any idea on how to calculate it. I will be grateful for any help and hints.
I present a modest contribution. Suppose we treat the problem of decomposing $2^n$ into sums of powers of two excluding the trivial solution $2^n=2^n.$ This can be solved by dynamic programming, here is the algorithm.


#! /usr/bin/perl -w
#

my %memo;

sub change {
    my (&#36;val, &#36;mxpow) = @_;

    my &#36;key = "&#36;val-&#36;mxpow";
    return &#36;memo{&#36;key} if exists &#36;memo{&#36;key};

    my &#36;res = 0;
    my &#36;inc = 2 ** &#36;mxpow;

    if(&#36;mxpow&gt;0){
      my &#36;s = 0;
      while(&#36;s &lt;= &#36;val){
          &#36;res += change(&#36;val-&#36;s, &#36;mxpow-1);
          &#36;s += &#36;inc;
      }
    }
    else{
      &#36;res = 1;
    }

    &#36;memo{&#36;key} = &#36;res;
    return &#36;res;
}


MAIN: {
    my &#36;max = shift || 10;


    my @res;
    for(my &#36;n=1; &#36;n&lt;=&#36;max; &#36;n++){
      push @res, change(2**&#36;n, &#36;n-1);
    }

    print join(', ', @res) . "\n";
}


This algorithm will produce the following sequence:
$$1, 3, 9, 35, 201, 1827, 27337, 692003, 30251721, \\2320518947, 316359580361, 77477180493603,\ldots $$
which is A125792 from the OEIS. You may consult this link for additional references about this problem, including a recurrence for the terms of the sequence.

In particular we have that the count for $2^{10}$ ($n=10$) is
$$2320518947.$$
Let $N(x)$ return the number of ways to write $2^x$ as an ordered sum of powers of two less than or equal to $2^{x-1}$. 

If there are $n$ ways to write $2^k$ as the ordered sum of powers of two less than or equal to  $2^{k-1}$, then how many ways are there to write $2^{k+1}$ as the ordered sum of powers of two less than or equal to $2^k$? In other words, if $N(k)=n$, what is $N(k+1)$?

Since $2^{k+1} = 2^k + 2^k$, each of the $2^k$-sized halves can be written as ordered sums in $n + 1$ ways ($n$ from powers less than or equal to $2^{k-1}$, plus the $1$ way that is just $2^k$). Thus, $N(k+1) = (n+1)^2$.

$$N(k+1) = \left(N(k)+1\right)^2$$

Find $N(1)$. Then you should be able to find $N(10)$ by recursively applying the identity above.