Is this bias on the differences of consecutive even abundant numbers easily explainable?
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Peter and I were chatting here a little about even abundant numbers and Peter did some serious computer check on the differences of successive even abundant numbers and found that:
In the range $[2, 10^7]$ the difference $2$ occured $ 804622 $ times, difference $4$ occured $758987$ times and difference $6$ occured $892466$ times.
The percentage of occurence of difference $4$ is $dfrac 758987804622+758987+892466 approx 0.309024$
Then he investigated the range $[2,10^8]$ and found that difference $2$ occured $8040361$ times, difference $4$ occured $7585187$ times and difference $6$ occured $ 8929753$ times.
So, percentage of occurence of difference $4$ is $dfrac 75851878040361+7585187+8929753 approx 0.308902$
It could be that situation even more "worsens" in some bigger ranges but even this seems to be some serious bias, is there any simple proof or heuristics of why difference of successive even abundant numbers more frequently has value $2$ 0r $6$ than that of $4$?
I am sensing that this can be easily explained but I do not see a way at this moment.
elementary-number-theory
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Peter and I were chatting here a little about even abundant numbers and Peter did some serious computer check on the differences of successive even abundant numbers and found that:
In the range $[2, 10^7]$ the difference $2$ occured $ 804622 $ times, difference $4$ occured $758987$ times and difference $6$ occured $892466$ times.
The percentage of occurence of difference $4$ is $dfrac 758987804622+758987+892466 approx 0.309024$
Then he investigated the range $[2,10^8]$ and found that difference $2$ occured $8040361$ times, difference $4$ occured $7585187$ times and difference $6$ occured $ 8929753$ times.
So, percentage of occurence of difference $4$ is $dfrac 75851878040361+7585187+8929753 approx 0.308902$
It could be that situation even more "worsens" in some bigger ranges but even this seems to be some serious bias, is there any simple proof or heuristics of why difference of successive even abundant numbers more frequently has value $2$ 0r $6$ than that of $4$?
I am sensing that this can be easily explained but I do not see a way at this moment.
elementary-number-theory
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Peter and I were chatting here a little about even abundant numbers and Peter did some serious computer check on the differences of successive even abundant numbers and found that:
In the range $[2, 10^7]$ the difference $2$ occured $ 804622 $ times, difference $4$ occured $758987$ times and difference $6$ occured $892466$ times.
The percentage of occurence of difference $4$ is $dfrac 758987804622+758987+892466 approx 0.309024$
Then he investigated the range $[2,10^8]$ and found that difference $2$ occured $8040361$ times, difference $4$ occured $7585187$ times and difference $6$ occured $ 8929753$ times.
So, percentage of occurence of difference $4$ is $dfrac 75851878040361+7585187+8929753 approx 0.308902$
It could be that situation even more "worsens" in some bigger ranges but even this seems to be some serious bias, is there any simple proof or heuristics of why difference of successive even abundant numbers more frequently has value $2$ 0r $6$ than that of $4$?
I am sensing that this can be easily explained but I do not see a way at this moment.
elementary-number-theory
Peter and I were chatting here a little about even abundant numbers and Peter did some serious computer check on the differences of successive even abundant numbers and found that:
In the range $[2, 10^7]$ the difference $2$ occured $ 804622 $ times, difference $4$ occured $758987$ times and difference $6$ occured $892466$ times.
The percentage of occurence of difference $4$ is $dfrac 758987804622+758987+892466 approx 0.309024$
Then he investigated the range $[2,10^8]$ and found that difference $2$ occured $8040361$ times, difference $4$ occured $7585187$ times and difference $6$ occured $ 8929753$ times.
So, percentage of occurence of difference $4$ is $dfrac 75851878040361+7585187+8929753 approx 0.308902$
It could be that situation even more "worsens" in some bigger ranges but even this seems to be some serious bias, is there any simple proof or heuristics of why difference of successive even abundant numbers more frequently has value $2$ 0r $6$ than that of $4$?
I am sensing that this can be easily explained but I do not see a way at this moment.
elementary-number-theory
edited Aug 26 at 21:21
asked Aug 26 at 21:16
Right
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1 Answer
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Ignoring the cases of odd abundant numbers (which are comparatively very rare), given two successive abundant numbers where the difference between them is $4$, you are guaranteed to have that the either the next or the previous abundant number appears with a difference of $2$ (given that every multiple of six is abundant). Therefore, for every occurence of a difference of $4$, there exists an occurence of difference $2$.
However, given two successive abundant numbers where the difference between them is $2$, there is no reason to think that the next/previous abundant number appears with difference $4$ - even if that is often the case. This would indicate the the number of occurences of difference $2$ would be higher than the occurences of difference $4$ by some amount. I think this answers one of your questions.
I can't immediately think why it should be the case that a successive difference of $6$ is more common than both of those. It is known that the natural density of abundant numbers is between $0.2474$ and $0.2480$, but I don't get any further.
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Ignoring the cases of odd abundant numbers (which are comparatively very rare), given two successive abundant numbers where the difference between them is $4$, you are guaranteed to have that the either the next or the previous abundant number appears with a difference of $2$ (given that every multiple of six is abundant). Therefore, for every occurence of a difference of $4$, there exists an occurence of difference $2$.
However, given two successive abundant numbers where the difference between them is $2$, there is no reason to think that the next/previous abundant number appears with difference $4$ - even if that is often the case. This would indicate the the number of occurences of difference $2$ would be higher than the occurences of difference $4$ by some amount. I think this answers one of your questions.
I can't immediately think why it should be the case that a successive difference of $6$ is more common than both of those. It is known that the natural density of abundant numbers is between $0.2474$ and $0.2480$, but I don't get any further.
add a comment |Â
up vote
0
down vote
Ignoring the cases of odd abundant numbers (which are comparatively very rare), given two successive abundant numbers where the difference between them is $4$, you are guaranteed to have that the either the next or the previous abundant number appears with a difference of $2$ (given that every multiple of six is abundant). Therefore, for every occurence of a difference of $4$, there exists an occurence of difference $2$.
However, given two successive abundant numbers where the difference between them is $2$, there is no reason to think that the next/previous abundant number appears with difference $4$ - even if that is often the case. This would indicate the the number of occurences of difference $2$ would be higher than the occurences of difference $4$ by some amount. I think this answers one of your questions.
I can't immediately think why it should be the case that a successive difference of $6$ is more common than both of those. It is known that the natural density of abundant numbers is between $0.2474$ and $0.2480$, but I don't get any further.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Ignoring the cases of odd abundant numbers (which are comparatively very rare), given two successive abundant numbers where the difference between them is $4$, you are guaranteed to have that the either the next or the previous abundant number appears with a difference of $2$ (given that every multiple of six is abundant). Therefore, for every occurence of a difference of $4$, there exists an occurence of difference $2$.
However, given two successive abundant numbers where the difference between them is $2$, there is no reason to think that the next/previous abundant number appears with difference $4$ - even if that is often the case. This would indicate the the number of occurences of difference $2$ would be higher than the occurences of difference $4$ by some amount. I think this answers one of your questions.
I can't immediately think why it should be the case that a successive difference of $6$ is more common than both of those. It is known that the natural density of abundant numbers is between $0.2474$ and $0.2480$, but I don't get any further.
Ignoring the cases of odd abundant numbers (which are comparatively very rare), given two successive abundant numbers where the difference between them is $4$, you are guaranteed to have that the either the next or the previous abundant number appears with a difference of $2$ (given that every multiple of six is abundant). Therefore, for every occurence of a difference of $4$, there exists an occurence of difference $2$.
However, given two successive abundant numbers where the difference between them is $2$, there is no reason to think that the next/previous abundant number appears with difference $4$ - even if that is often the case. This would indicate the the number of occurences of difference $2$ would be higher than the occurences of difference $4$ by some amount. I think this answers one of your questions.
I can't immediately think why it should be the case that a successive difference of $6$ is more common than both of those. It is known that the natural density of abundant numbers is between $0.2474$ and $0.2480$, but I don't get any further.
answered Aug 26 at 21:44
Matt
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