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Jan 16 at 6:42 vote accept Denis Ivanov
Jan 15 at 16:36 answer added Peter Taylor timeline score: 2
Jan 15 at 16:35 comment added Denis Ivanov @DenisShatrov, could you explain more detailed please?
Jan 15 at 16:33 history edited Denis Ivanov CC BY-SA 4.0
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Jan 15 at 16:32 comment added Peter Taylor Yes, precisely.
Jan 15 at 16:31 comment added Denis Shatrov In other words, $P_n = Q_n$ if no carries occur when calculating the product $p_1^{m_1} \cdot \ldots \cdot p_k^{m_k}$ in binary.
Jan 15 at 16:24 history edited Denis Ivanov CC BY-SA 4.0
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Jan 15 at 16:24 comment added Denis Ivanov @PeterTaylor, yes, but there are other numbers (not $2^ap$) with this properties: $15, 30, 51, 60, 85\dots$
Jan 15 at 15:49 comment added Peter Taylor The examples you give can easily be explained by observing that the property always holds for $n$ of the form $2^a p$ when $p$ is prime. Perhaps the easiest characterisation, although maybe not the most useful, is that $Q_n = P_n$ whenever $Q_n$ is a $0,1$-polynomial.
Jan 15 at 15:19 history asked Denis Ivanov CC BY-SA 4.0