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Lower density of numbers not summable by consecutive integers
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Let us call a positive integer $ninmathbbN$ consecutively summable if there are positive integers $m, k < n$ such that $$n=sum_i=0^k (m+i).$$For $Asubseteq mathbbN$ we set the lower density of $A$ to be $$textld(A)=textlim inf_ntoinftyfracAcap1,ldots,nn.$$
If $N$ is the set of positive integers that are not consecutively summable, do we have $textld(N)=0$?
nt.number-theory
asked 1 hour ago
Dominic van der Zypen
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Let us call a positive integer $ninmathbbN$ consecutively summable if there are positive integers $m, k < n$ such that $$n=sum_i=0^k (m+i).$$For $Asubseteq mathbbN$ we set the lower density of $A$ to be $$textld(A)=textlim inf_ntoinftyfracAcap1,ldots,nn.$$
If $N$ is the set of positive integers that are not consecutively summable, do we have $textld(N)=0$?
nt.number-theory
asked 1 hour ago
Dominic van der Zypen
13.4k43172
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let us call a positive integer $ninmathbbN$ consecutively summable if there are positive integers $m, k < n$ such that $$n=sum_i=0^k (m+i).$$For $Asubseteq mathbbN$ we set the lower density of $A$ to be $$textld(A)=textlim inf_ntoinftyfracAcap1,ldots,nn.$$
If $N$ is the set of positive integers that are not consecutively summable, do we have $textld(N)=0$?
nt.number-theory
asked 1 hour ago
Dominic van der Zypen
13.4k43172
Let us call a positive integer $ninmathbbN$ consecutively summable if there are positive integers $m, k < n$ such that $$n=sum_i=0^k (m+i).$$For $Asubseteq mathbbN$ we set the lower density of $A$ to be $$textld(A)=textlim inf_ntoinftyfracAcap1,ldots,nn.$$
If $N$ is the set of positive integers that are not consecutively summable, do we have $textld(N)=0$?
nt.number-theory
nt.number-theory
asked 1 hour ago
Dominic van der Zypen
13.4k43172
asked 1 hour ago
Dominic van der Zypen
13.4k43172
asked 1 hour ago
Dominic van der Zypen
13.4k43172
asked 1 hour ago
Dominic van der Zypen
13.4k43172
asked 1 hour ago
Dominic van der Zypen
13.4k43172
13.4k43172
add a comment |Â
add a comment |Â
2 Answers
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It is known that the only numbers not "consecutive summable" are the powers of $2$.
This is easy to prove: If $m+m+1+ldots m+k=2^a$ then $2^a=frac(k+1)(2m+k)2$.
This means that $2^a+1=(k+1)(2m+k)$ which is impossible since the differences form the two parentheses is an odd number. (And they should be both powers of two).
It is not that difficult to show that every number not of the form $2^a, ain mathbbN$ can be "consecutive summable".
I think the shortest way to prove both is here.
This shows that $ld(A)=0$.
answered 1 hour ago
Konstantinos Gaitanas
1,87411130
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up vote
0
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Let us denote $B=mathbb Nsetminus N$, i.e., the set of consecutive-summable numbers.
There already is an answer precisely characterizing the sets $N$ and $B$. (Which is much better result than just estimating the density.) If we are only interested in the density, another way to get that it is equal to zero is to notice the following.
If we fix a $kinmathbb N$, then we get the numbers of the form
$$n=km+sum_i=0^k k = km + T_k$$
where $T_k$ is the $k$-th triangular number. So we see that with finitely many exceptions $B$ contains the arithmetic progression $kmathbb N+a$, where $a=T_k bmod k$.
That means that $N$ is almost (i.e., with finitely many exceptions) contained in the union of the remaining $(k-1)$ congruence classes modulo $k$.
If we take any coprime $k_1,ldots,k_n$, then we have that $N$ is almost contained in the union of corresponding congruence classes modulo $k_1cdots k_n$. (Here we are using the Chinese remainder theorem which gives us the correspondence between the system of remainder modulo $k_1,dots,k_n$ and a single congruence class modulo $k_1cdots k_n$.) This gives us an estimate for the upper density
$$operatornameud(N) le frac(k_1-1)cdot (k_2-1) cdot (k_n-1)k_1cdot k_2 cdot k_n = prod_j=1^n left(1-frac1k_jright).$$
In particular, if we take $k_n=p_n$ to be the $n$-th prime number, we get that
$$operatornameud(N) le prod_j=1^n left(1-frac1k_jright)$$
and since this is true for every $n$ we have
$$operatornameud(N) le prod_j=1^infty left(1-frac1p_jright)=0.$$
You may notice that this argument would work for any set which can be related in a similar way to the union of distinct congruence classes if we have $sum frac1k_j=infty$.
answered 2 mins ago
Martin Sleziak
2,75432028
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
It is known that the only numbers not "consecutive summable" are the powers of $2$.
This is easy to prove: If $m+m+1+ldots m+k=2^a$ then $2^a=frac(k+1)(2m+k)2$.
This means that $2^a+1=(k+1)(2m+k)$ which is impossible since the differences form the two parentheses is an odd number. (And they should be both powers of two).
It is not that difficult to show that every number not of the form $2^a, ain mathbbN$ can be "consecutive summable".
I think the shortest way to prove both is here.
This shows that $ld(A)=0$.
answered 1 hour ago
Konstantinos Gaitanas
1,87411130
add a comment |Â
up vote
2
down vote
It is known that the only numbers not "consecutive summable" are the powers of $2$.
This is easy to prove: If $m+m+1+ldots m+k=2^a$ then $2^a=frac(k+1)(2m+k)2$.
This means that $2^a+1=(k+1)(2m+k)$ which is impossible since the differences form the two parentheses is an odd number. (And they should be both powers of two).
It is not that difficult to show that every number not of the form $2^a, ain mathbbN$ can be "consecutive summable".
I think the shortest way to prove both is here.
This shows that $ld(A)=0$.
answered 1 hour ago
Konstantinos Gaitanas
1,87411130
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It is known that the only numbers not "consecutive summable" are the powers of $2$.
This is easy to prove: If $m+m+1+ldots m+k=2^a$ then $2^a=frac(k+1)(2m+k)2$.
This means that $2^a+1=(k+1)(2m+k)$ which is impossible since the differences form the two parentheses is an odd number. (And they should be both powers of two).
It is not that difficult to show that every number not of the form $2^a, ain mathbbN$ can be "consecutive summable".
I think the shortest way to prove both is here.
This shows that $ld(A)=0$.
answered 1 hour ago
Konstantinos Gaitanas
1,87411130
It is known that the only numbers not "consecutive summable" are the powers of $2$.
This is easy to prove: If $m+m+1+ldots m+k=2^a$ then $2^a=frac(k+1)(2m+k)2$.
This means that $2^a+1=(k+1)(2m+k)$ which is impossible since the differences form the two parentheses is an odd number. (And they should be both powers of two).
It is not that difficult to show that every number not of the form $2^a, ain mathbbN$ can be "consecutive summable".
I think the shortest way to prove both is here.
This shows that $ld(A)=0$.
answered 1 hour ago
Konstantinos Gaitanas
1,87411130
edited 47 mins ago
answered 1 hour ago
Konstantinos Gaitanas
1,87411130
answered 1 hour ago
Konstantinos Gaitanas
1,87411130
answered 1 hour ago
Konstantinos Gaitanas
1,87411130
1,87411130
add a comment |Â
add a comment |Â
up vote
0
down vote
Let us denote $B=mathbb Nsetminus N$, i.e., the set of consecutive-summable numbers.
There already is an answer precisely characterizing the sets $N$ and $B$. (Which is much better result than just estimating the density.) If we are only interested in the density, another way to get that it is equal to zero is to notice the following.
If we fix a $kinmathbb N$, then we get the numbers of the form
$$n=km+sum_i=0^k k = km + T_k$$
where $T_k$ is the $k$-th triangular number. So we see that with finitely many exceptions $B$ contains the arithmetic progression $kmathbb N+a$, where $a=T_k bmod k$.
That means that $N$ is almost (i.e., with finitely many exceptions) contained in the union of the remaining $(k-1)$ congruence classes modulo $k$.
If we take any coprime $k_1,ldots,k_n$, then we have that $N$ is almost contained in the union of corresponding congruence classes modulo $k_1cdots k_n$. (Here we are using the Chinese remainder theorem which gives us the correspondence between the system of remainder modulo $k_1,dots,k_n$ and a single congruence class modulo $k_1cdots k_n$.) This gives us an estimate for the upper density
$$operatornameud(N) le frac(k_1-1)cdot (k_2-1) cdot (k_n-1)k_1cdot k_2 cdot k_n = prod_j=1^n left(1-frac1k_jright).$$
In particular, if we take $k_n=p_n$ to be the $n$-th prime number, we get that
$$operatornameud(N) le prod_j=1^n left(1-frac1k_jright)$$
and since this is true for every $n$ we have
$$operatornameud(N) le prod_j=1^infty left(1-frac1p_jright)=0.$$
You may notice that this argument would work for any set which can be related in a similar way to the union of distinct congruence classes if we have $sum frac1k_j=infty$.
answered 2 mins ago
Martin Sleziak
2,75432028
add a comment |Â
up vote
0
down vote
Let us denote $B=mathbb Nsetminus N$, i.e., the set of consecutive-summable numbers.
There already is an answer precisely characterizing the sets $N$ and $B$. (Which is much better result than just estimating the density.) If we are only interested in the density, another way to get that it is equal to zero is to notice the following.
If we fix a $kinmathbb N$, then we get the numbers of the form
$$n=km+sum_i=0^k k = km + T_k$$
where $T_k$ is the $k$-th triangular number. So we see that with finitely many exceptions $B$ contains the arithmetic progression $kmathbb N+a$, where $a=T_k bmod k$.
That means that $N$ is almost (i.e., with finitely many exceptions) contained in the union of the remaining $(k-1)$ congruence classes modulo $k$.
If we take any coprime $k_1,ldots,k_n$, then we have that $N$ is almost contained in the union of corresponding congruence classes modulo $k_1cdots k_n$. (Here we are using the Chinese remainder theorem which gives us the correspondence between the system of remainder modulo $k_1,dots,k_n$ and a single congruence class modulo $k_1cdots k_n$.) This gives us an estimate for the upper density
$$operatornameud(N) le frac(k_1-1)cdot (k_2-1) cdot (k_n-1)k_1cdot k_2 cdot k_n = prod_j=1^n left(1-frac1k_jright).$$
In particular, if we take $k_n=p_n$ to be the $n$-th prime number, we get that
$$operatornameud(N) le prod_j=1^n left(1-frac1k_jright)$$
and since this is true for every $n$ we have
$$operatornameud(N) le prod_j=1^infty left(1-frac1p_jright)=0.$$
You may notice that this argument would work for any set which can be related in a similar way to the union of distinct congruence classes if we have $sum frac1k_j=infty$.
answered 2 mins ago
Martin Sleziak
2,75432028
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let us denote $B=mathbb Nsetminus N$, i.e., the set of consecutive-summable numbers.
There already is an answer precisely characterizing the sets $N$ and $B$. (Which is much better result than just estimating the density.) If we are only interested in the density, another way to get that it is equal to zero is to notice the following.
If we fix a $kinmathbb N$, then we get the numbers of the form
$$n=km+sum_i=0^k k = km + T_k$$
where $T_k$ is the $k$-th triangular number. So we see that with finitely many exceptions $B$ contains the arithmetic progression $kmathbb N+a$, where $a=T_k bmod k$.
That means that $N$ is almost (i.e., with finitely many exceptions) contained in the union of the remaining $(k-1)$ congruence classes modulo $k$.
If we take any coprime $k_1,ldots,k_n$, then we have that $N$ is almost contained in the union of corresponding congruence classes modulo $k_1cdots k_n$. (Here we are using the Chinese remainder theorem which gives us the correspondence between the system of remainder modulo $k_1,dots,k_n$ and a single congruence class modulo $k_1cdots k_n$.) This gives us an estimate for the upper density
$$operatornameud(N) le frac(k_1-1)cdot (k_2-1) cdot (k_n-1)k_1cdot k_2 cdot k_n = prod_j=1^n left(1-frac1k_jright).$$
In particular, if we take $k_n=p_n$ to be the $n$-th prime number, we get that
$$operatornameud(N) le prod_j=1^n left(1-frac1k_jright)$$
and since this is true for every $n$ we have
$$operatornameud(N) le prod_j=1^infty left(1-frac1p_jright)=0.$$
You may notice that this argument would work for any set which can be related in a similar way to the union of distinct congruence classes if we have $sum frac1k_j=infty$.
answered 2 mins ago
Martin Sleziak
2,75432028
Let us denote $B=mathbb Nsetminus N$, i.e., the set of consecutive-summable numbers.
There already is an answer precisely characterizing the sets $N$ and $B$. (Which is much better result than just estimating the density.) If we are only interested in the density, another way to get that it is equal to zero is to notice the following.
If we fix a $kinmathbb N$, then we get the numbers of the form
$$n=km+sum_i=0^k k = km + T_k$$
where $T_k$ is the $k$-th triangular number. So we see that with finitely many exceptions $B$ contains the arithmetic progression $kmathbb N+a$, where $a=T_k bmod k$.
That means that $N$ is almost (i.e., with finitely many exceptions) contained in the union of the remaining $(k-1)$ congruence classes modulo $k$.
If we take any coprime $k_1,ldots,k_n$, then we have that $N$ is almost contained in the union of corresponding congruence classes modulo $k_1cdots k_n$. (Here we are using the Chinese remainder theorem which gives us the correspondence between the system of remainder modulo $k_1,dots,k_n$ and a single congruence class modulo $k_1cdots k_n$.) This gives us an estimate for the upper density
$$operatornameud(N) le frac(k_1-1)cdot (k_2-1) cdot (k_n-1)k_1cdot k_2 cdot k_n = prod_j=1^n left(1-frac1k_jright).$$
In particular, if we take $k_n=p_n$ to be the $n$-th prime number, we get that
$$operatornameud(N) le prod_j=1^n left(1-frac1k_jright)$$
and since this is true for every $n$ we have
$$operatornameud(N) le prod_j=1^infty left(1-frac1p_jright)=0.$$
You may notice that this argument would work for any set which can be related in a similar way to the union of distinct congruence classes if we have $sum frac1k_j=infty$.
answered 2 mins ago
Martin Sleziak
2,75432028
answered 2 mins ago
Martin Sleziak
2,75432028
answered 2 mins ago
Martin Sleziak
2,75432028
answered 2 mins ago
Martin Sleziak
2,75432028
2,75432028
add a comment |Â
add a comment |Â
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