Can the difference between consecutive even abundant numbers exceed 6?
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I came across an astonishing observation :
An abundant number is a positive integer $n$ with the property $S(n)>n$ , where $S(n)$ is the sum of the divisors of $n$ except $n$ itself.
The difference of consecutive even abundant numbers seems to be at most $6$. Can anyone prove/disprove this statement ?
Difference $6$ is not exceeded upto at least $10^7$.
elementary-number-theory divisor-sum
asked Aug 26 at 18:38
Peter
45.7k1039124
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up vote
3
down vote
favorite
I came across an astonishing observation :
An abundant number is a positive integer $n$ with the property $S(n)>n$ , where $S(n)$ is the sum of the divisors of $n$ except $n$ itself.
The difference of consecutive even abundant numbers seems to be at most $6$. Can anyone prove/disprove this statement ?
Difference $6$ is not exceeded upto at least $10^7$.
elementary-number-theory divisor-sum
asked Aug 26 at 18:38
Peter
45.7k1039124
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I came across an astonishing observation :
An abundant number is a positive integer $n$ with the property $S(n)>n$ , where $S(n)$ is the sum of the divisors of $n$ except $n$ itself.
The difference of consecutive even abundant numbers seems to be at most $6$. Can anyone prove/disprove this statement ?
Difference $6$ is not exceeded upto at least $10^7$.
elementary-number-theory divisor-sum
asked Aug 26 at 18:38
Peter
45.7k1039124
I came across an astonishing observation :
An abundant number is a positive integer $n$ with the property $S(n)>n$ , where $S(n)$ is the sum of the divisors of $n$ except $n$ itself.
The difference of consecutive even abundant numbers seems to be at most $6$. Can anyone prove/disprove this statement ?
Difference $6$ is not exceeded upto at least $10^7$.
elementary-number-theory divisor-sum
elementary-number-theory divisor-sum
asked Aug 26 at 18:38
Peter
45.7k1039124
asked Aug 26 at 18:38
Peter
45.7k1039124
edited Aug 26 at 18:44
asked Aug 26 at 18:38
Peter
45.7k1039124
asked Aug 26 at 18:38
Peter
45.7k1039124
asked Aug 26 at 18:38
Peter
45.7k1039124
45.7k1039124
add a comment |Â
add a comment |Â
1 Answer
1
active
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up vote
12
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accepted
This can be proven, observing that any multiple of $6$ greater than $6$ itself is abundant: the divisors include at least $1, fracn2,fracn3,fracn6$, which together sum to $n+1$.
edited Aug 26 at 19:12
Oscar Lanzi
10.6k11734
answered Aug 26 at 18:43
Matt
2,396719
Well done! (+1)
– Peter
Aug 26 at 18:47
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
accepted
This can be proven, observing that any multiple of $6$ greater than $6$ itself is abundant: the divisors include at least $1, fracn2,fracn3,fracn6$, which together sum to $n+1$.
edited Aug 26 at 19:12
Oscar Lanzi
10.6k11734
answered Aug 26 at 18:43
Matt
2,396719
Well done! (+1)
– Peter
Aug 26 at 18:47
add a comment |Â
up vote
12
down vote
accepted
This can be proven, observing that any multiple of $6$ greater than $6$ itself is abundant: the divisors include at least $1, fracn2,fracn3,fracn6$, which together sum to $n+1$.
edited Aug 26 at 19:12
Oscar Lanzi
10.6k11734
answered Aug 26 at 18:43
Matt
2,396719
Well done! (+1)
– Peter
Aug 26 at 18:47
add a comment |Â
up vote
12
down vote
accepted
up vote
12
down vote
accepted
This can be proven, observing that any multiple of $6$ greater than $6$ itself is abundant: the divisors include at least $1, fracn2,fracn3,fracn6$, which together sum to $n+1$.
edited Aug 26 at 19:12
Oscar Lanzi
10.6k11734
answered Aug 26 at 18:43
Matt
2,396719
This can be proven, observing that any multiple of $6$ greater than $6$ itself is abundant: the divisors include at least $1, fracn2,fracn3,fracn6$, which together sum to $n+1$.
edited Aug 26 at 19:12
Oscar Lanzi
10.6k11734
answered Aug 26 at 18:43
Matt
2,396719
edited Aug 26 at 19:12
Oscar Lanzi
10.6k11734
edited Aug 26 at 19:12
Oscar Lanzi
10.6k11734
edited Aug 26 at 19:12
Oscar Lanzi
10.6k11734
10.6k11734
answered Aug 26 at 18:43
Matt
2,396719
answered Aug 26 at 18:43
Matt
2,396719
answered Aug 26 at 18:43
Matt
2,396719
2,396719
Well done! (+1)
– Peter
Aug 26 at 18:47
add a comment |Â
Well done! (+1)
– Peter
Aug 26 at 18:47
Well done! (+1)
– Peter
Aug 26 at 18:47
Well done! (+1)
– Peter
Aug 26 at 18:47
add a comment |Â
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