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copied!<p>I don't think it's possible in the general case. If <code>m</code> is a power of <code>p</code> (or vice-versa), or if they're both powers of a common base, you can do it, since each group of log<sub><code>m</code></sub>(<code>p</code>) is then independent. However, in the general case, suppose you're converting the number <code>a</code><sub><code>1</code></sub> <code>a</code><sub><code>2</code></sub> <code>a</code><sub><code>3</code></sub> <code>...</code> <code>a</code><sub><code>n</code></sub>. The equivalent number in base <code>p</code> is</p> <p><code>sum(a</code><sub><code>i</code></sub> <code>*</code> <code>m</code><sup><code>i-1</code></sup> for <code>i</code> in <code>1..n)</code></p> <p>If we've processed the first <code>i</code> digits, then we have the <code>i</code>th partial sum. To compute the <code>i+1</code>'th partial sum, we need to add <code>a</code><sub><code>i+1</code></sub> <code>* m</code><sup><code>i</code></sup>. In the general case, this number is going have non-zero digits in most places, so we'll need to modify all of the digits we've processed so far. In other words, we'll have to process <em>all</em> of the input digits before we'll know what the final output digits will be.</p> <p>In the special case where <code>m</code> are both powers of a common base, or equivalently if log<sub><code>m</code></sub>(<code>p</code>) is a rational number, then <code>m</code><sup><code>i</code></sup> will only have a few non-zero digits in base <code>p</code> near the front, so we can safely output most of the digits we've computed so far.</p>

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