Timeline for Simple question about 0,1-polynomials
Current License: CC BY-SA 4.0
10 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
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 |
added 161 characters in body
|
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 |
deleted 82 characters in body
|
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 |