Crowdmath

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EdwinNational

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EdwinNational

#1 h Jan 30, 2021, 4:51 AM

This topic is linked to equalsn.

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We give an explicit construction with $n$ letters $1, 2, \ldots n$ , and show that it has leading constant $\frac{1}{2}$ :
Note: $S_x$ is a sequence $S$ repeated $x$ times. We call each $S$ in the construction a $\textit{snippit}$ .
Construction

Proof that the construction works

This post has been edited 1 time. Last edited by EdwinNational, Jan 30, 2021, 5:02 AM
Reason: forgot to say what x was in the construction

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penrose43

26 posts

penrose43

#2 h Jan 31, 2021, 12:14 AM • 1 Y

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Quote:

The sequence itself has length ${n \choose 2}(2x)$

Isn't the length of the sequence half that since we have ${n \choose 2}$ pairs and each pair is repeated $x$ times?

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EdwinNational

36 posts

EdwinNational

#3 h Jan 31, 2021, 12:24 AM

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Quote:

Isn't the length of the sequence half that since we have ${n \choose 2}$ pairs and each pair is repeated $x$ times?

Each pair has length $2$ because it has two letters(so a pair $ij$ repeated $x$ times should have length $2x$ ).

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penrose43

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penrose43

#4 h Jan 31, 2021, 12:57 AM

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Ohh, right! My bad. That looks like it works then!

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JGeneson

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JGeneson

#5 h Jan 31, 2021, 6:31 PM

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EdwinNational wrote:

We give an explicit construction with $n$ letters $1, 2, \ldots n$ , and show that it has leading constant $\frac{1}{2}$ :
Note: $S_x$ is a sequence $S$ repeated $x$ times. We call each $S$ in the construction a $\textit{snippit}$ .
Construction

Proof that the construction works

penrose43 wrote:

Ohh, right! My bad. That looks like it works then!

This does not work. Look at the letters $n-1$ and $n$ . Their alternation has length $2x+2(n-2)$ in this construction.

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LoneShadow379

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LoneShadow379

#6 h Jul 13, 2022, 2:44 PM

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You haven't counted the (1,i), (2,i), (3,i)... terms - for instance, (2,i) would come between (1,j) and (2,j) and thus contribute to the alternating sequence

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