Timeline for Sequence of least prime-multiples with smallest Hamming weight
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| Aug 15, 2020 at 15:26 | answer | added | Christian Elsholtz | timeline score: 2 | |
| Apr 1, 2020 at 6:46 | comment | added | Gerry Myerson | Certainly the inequality holds if $2^n-1$ is prime, even with $k=n-1$, and there's only one prime factor, so that gives about $40$ examples. | |
| Apr 1, 2020 at 3:26 | comment | added | Manfred Weis | @GerryMyerson checking the linked paper for examples I saw that Theorem 5 states that if the number $\Omega(2^n-1) \lt \frac{\log n}{\log k}$ then one of the prime factors of $2^n-1$ has the property that all of its multiples have Hamming weight of at least $k$; $\Omega()$ denotes the number of primefactors counted with multiples. Unfortunately no clue is given as to which of the factors has that property. The linked paper is real gem by the way. | |
| Mar 31, 2020 at 22:59 | comment | added | Gerry Myerson | Do you know any value of $n$ for which $H>6$? | |
| Mar 31, 2020 at 17:11 | history | edited | Manfred Weis | CC BY-SA 4.0 |
added 2 characters in body
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| Mar 31, 2020 at 16:43 | history | edited | Manfred Weis | CC BY-SA 4.0 |
improved the definition of $\mu(n)$
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| Mar 31, 2020 at 15:28 | history | edited | Manfred Weis | CC BY-SA 4.0 |
added value for $\mu(4)$ as provided by Gerry Myerson and fixed the second part of the question
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| Mar 31, 2020 at 15:23 | history | edited | Manfred Weis | CC BY-SA 4.0 |
added value for $\mu(4)$ as provided by Gerry Myerson
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| Mar 31, 2020 at 14:15 | comment | added | Manfred Weis | @Sylvain yes, that's the definition of Hamming weights at least in the context of this question. | |
| Mar 31, 2020 at 11:45 | comment | added | Sylvain JULIEN | If I understand correctly, the Hamming weight is the number of $1$ in the binary expansion, right? | |
| Mar 31, 2020 at 11:00 | comment | added | Gerry Myerson | $7m=2^k+1$ is impossible, since $2^k+1\equiv0\bmod7$ has no solution, so $\mu(4)=1$. | |
| Mar 31, 2020 at 10:50 | history | edited | Manfred Weis | CC BY-SA 4.0 |
deleted 11 characters in body
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| Mar 31, 2020 at 10:50 | comment | added | Manfred Weis | @GerryMyerson today is not my day... | |
| Mar 31, 2020 at 6:27 | comment | added | Gerry Myerson | $11\times3=33=2^5+1$, so $\mu(5)=3$, so I still don't follow you. $3^2\cdot331$? | |
| Mar 31, 2020 at 3:36 | comment | added | Manfred Weis | @GerryMyerson shame on me! I wanted to know the $p_n$ with a Hamming weight larger than 6; I edited the question accordingly. Thank you. | |
| Mar 31, 2020 at 3:31 | history | edited | Manfred Weis | CC BY-SA 4.0 |
clarified definition of the sequence in reply to Gerry Myerson's comment
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| Mar 31, 2020 at 3:03 | comment | added | Gerry Myerson | OK, then $\mu(8)=27$, since the eighth prime is $19$, and $19\times27=513=2^9+1$ has Hamming weight $2$, and $19m$ has Hamming weight greater than $2$ for all $m<27$. It seems to me there will be a lot of $n$ for which $\mu(n)$ will be considerably greater than $6$. | |
| Mar 31, 2020 at 2:51 | comment | added | Manfred Weis | @GerryMyerson $\mu(6)=5$. I chose that on purpose because knowing the smallest factor $m$ that yields the smallest Hamming weight allows for easy calculation of that weight, but knowing the minimal Haming weight gives no way of easily calculating the minimal $m$ that yields that minimal weight, especially if the minimal Haming weight were proven by contradiction, i.e. the assumption that a higher Hamming weight were minimal led to cotradiction. | |
| Mar 30, 2020 at 22:18 | comment | added | Gerry Myerson | Let me see if I understand the question. If you want to calculate $\mu(6)$, say, you first find $p_6=13$, the sixth prime, then you find $13\times5=65=(1000001)_2$ has Hamming weight $2$, and obviously no multiple of $13$ is a power of $2$, so $\mu(6)=5$ (or is it $\mu(6)=2$?). Are the values of $\mu(n)$ tabulated at the Online Encyclopedia of Integer Sequences? | |
| Mar 30, 2020 at 13:40 | history | asked | Manfred Weis | CC BY-SA 4.0 |