Sums of Powers of Two

Define $f(0)=1$ and $f(n)$ to be the number of different ways $n$ can be expressed as a sum of integer powers of $2$ using each power no more than twice.
For example, $f(10)=5$ since there are five different ways to express $10$:
\begin{align}
& 1 + 1 + 8\\
& 1 + 1 + 4 + 4\\
& 1 + 1 + 2 + 2 + 4\\
& 2 + 4 + 4\\
& 2 + 8
\end{align}
What is $f(10^{25})$?

Solving for a large number like $10^{25}$ directly can be computationally intensive. Hence, we would approach it in a smarter way. Notice that each integer can be expressed as a sum of powers of 2 (binary representation). Since each power of 2 can be used twice in this problem, we can think of it as binary representation but each digit can be 0, 1, or 2.

For example,
$f(2) = 2$ because we have two ways to express 2:
– 2
– 1 + 1

Similarly,
$f(3) = 2$ because we have two ways to express 3:
– 2 + 1
– 1 + 1 + 1

and
$f(4) = 4$ because we have four ways to express 4:
– 4
– 2 + 2
– 2 + 1 + 1
– 1 + 1 + 1 + 1

To find $f(10^{25})$, we need to find how many powers of 2 are less than $10^{25}$ (since we cannot use a power of 2 greater than this number). From this we can calculate $10^{25}=2^{x}$. Solving the equation we find $x \approx 83.38$.

Since 83 is the largest integer less than 83.38, we can say that there are 83 powers of 2 less than $10^{25}$.

The total number of ways $n$ can be expressed as a sum of integer powers of 2 using each power no more than twice is then equal to the number of ways 83 digits can be selected to be 0, 1 or 2. This is because any given power of 2 can either be selected 0, 1 or 2 times. Therefore, each of these 83 powers of 2 has three possible states.

So, the number of ways $10^{25}$ can be expressed as a sum of integer powers of 2 using each power no more than twice is equal to $3^{83}$.

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