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All k-term sums of the first n natural numbers
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is there a formula that generates k-term sums of the first $n$ natural numbers? The numbers are not allowed to repeat. For example, for $n=10$ and $k=4$, some possible sums would be $1+2+3+4, 2+6+7+8, 5+4+8+9$, etc. Some impossible sums would be $1+2+5+8+9, 4+6, 2+2+2+2, 6+2+3+2$, etc. Pardon the lack of formatting, I am writing from my phone in the car.
sequences-and-series number-theory
edited Jul 18 at 22:52
Key Flex
4,346425
asked Jul 18 at 22:39
user575235
161
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up vote
2
down vote
favorite
is there a formula that generates k-term sums of the first $n$ natural numbers? The numbers are not allowed to repeat. For example, for $n=10$ and $k=4$, some possible sums would be $1+2+3+4, 2+6+7+8, 5+4+8+9$, etc. Some impossible sums would be $1+2+5+8+9, 4+6, 2+2+2+2, 6+2+3+2$, etc. Pardon the lack of formatting, I am writing from my phone in the car.
sequences-and-series number-theory
edited Jul 18 at 22:52
Key Flex
4,346425
asked Jul 18 at 22:39
user575235
161
1
It is not clear to me what exactly you are after. Are you looking for the number of possible summations? Do you treat $1+4$ differently than $2+3$ because the addends are different or do you treat them as the same since their total is the same? Or... are you looking for a way to generate the list of possible summations (or results of summations)?
– JMoravitz
Jul 18 at 22:49
In general, the sum $sum _k=n+1^mk$ equals $fracm(m+1)-n(n+1)2$. You may write down each of your sum as combinations of sums of this type (that is, of sums of a certain number of consecutive integers).
– Suzet
Jul 18 at 22:51
One way to interpret your Question is whether there is an expression which yields all values (and only those values) that are sums of $k$ summands drawn (without replacement) from the set $1,ldots,n$. However without seeing what use would be made, I'm doubtful that I've understood your request clearly enough to give a useful response.
– hardmath
Jul 18 at 22:52
2
Don't write from your phone in the car. That's dangerous to you and others.
– Cameron Buie
Jul 18 at 23:19
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
is there a formula that generates k-term sums of the first $n$ natural numbers? The numbers are not allowed to repeat. For example, for $n=10$ and $k=4$, some possible sums would be $1+2+3+4, 2+6+7+8, 5+4+8+9$, etc. Some impossible sums would be $1+2+5+8+9, 4+6, 2+2+2+2, 6+2+3+2$, etc. Pardon the lack of formatting, I am writing from my phone in the car.
sequences-and-series number-theory
edited Jul 18 at 22:52
Key Flex
4,346425
asked Jul 18 at 22:39
user575235
161
is there a formula that generates k-term sums of the first $n$ natural numbers? The numbers are not allowed to repeat. For example, for $n=10$ and $k=4$, some possible sums would be $1+2+3+4, 2+6+7+8, 5+4+8+9$, etc. Some impossible sums would be $1+2+5+8+9, 4+6, 2+2+2+2, 6+2+3+2$, etc. Pardon the lack of formatting, I am writing from my phone in the car.
sequences-and-series number-theory
edited Jul 18 at 22:52
Key Flex
4,346425
asked Jul 18 at 22:39
user575235
161
edited Jul 18 at 22:52
Key Flex
4,346425
edited Jul 18 at 22:52
Key Flex
4,346425
edited Jul 18 at 22:52
Key Flex
4,346425
4,346425
asked Jul 18 at 22:39
user575235
161
asked Jul 18 at 22:39
user575235
161
asked Jul 18 at 22:39
user575235
161
161
1
It is not clear to me what exactly you are after. Are you looking for the number of possible summations? Do you treat $1+4$ differently than $2+3$ because the addends are different or do you treat them as the same since their total is the same? Or... are you looking for a way to generate the list of possible summations (or results of summations)?
– JMoravitz
Jul 18 at 22:49
In general, the sum $sum _k=n+1^mk$ equals $fracm(m+1)-n(n+1)2$. You may write down each of your sum as combinations of sums of this type (that is, of sums of a certain number of consecutive integers).
– Suzet
Jul 18 at 22:51
One way to interpret your Question is whether there is an expression which yields all values (and only those values) that are sums of $k$ summands drawn (without replacement) from the set $1,ldots,n$. However without seeing what use would be made, I'm doubtful that I've understood your request clearly enough to give a useful response.
– hardmath
Jul 18 at 22:52
2
Don't write from your phone in the car. That's dangerous to you and others.
– Cameron Buie
Jul 18 at 23:19
add a comment |Â
1
It is not clear to me what exactly you are after. Are you looking for the number of possible summations? Do you treat $1+4$ differently than $2+3$ because the addends are different or do you treat them as the same since their total is the same? Or... are you looking for a way to generate the list of possible summations (or results of summations)?
– JMoravitz
Jul 18 at 22:49
In general, the sum $sum _k=n+1^mk$ equals $fracm(m+1)-n(n+1)2$. You may write down each of your sum as combinations of sums of this type (that is, of sums of a certain number of consecutive integers).
– Suzet
Jul 18 at 22:51
One way to interpret your Question is whether there is an expression which yields all values (and only those values) that are sums of $k$ summands drawn (without replacement) from the set $1,ldots,n$. However without seeing what use would be made, I'm doubtful that I've understood your request clearly enough to give a useful response.
– hardmath
Jul 18 at 22:52
2
Don't write from your phone in the car. That's dangerous to you and others.
– Cameron Buie
Jul 18 at 23:19
1
1
It is not clear to me what exactly you are after. Are you looking for the number of possible summations? Do you treat $1+4$ differently than $2+3$ because the addends are different or do you treat them as the same since their total is the same? Or... are you looking for a way to generate the list of possible summations (or results of summations)?
– JMoravitz
Jul 18 at 22:49
It is not clear to me what exactly you are after. Are you looking for the number of possible summations? Do you treat $1+4$ differently than $2+3$ because the addends are different or do you treat them as the same since their total is the same? Or... are you looking for a way to generate the list of possible summations (or results of summations)?
– JMoravitz
Jul 18 at 22:49
In general, the sum $sum _k=n+1^mk$ equals $fracm(m+1)-n(n+1)2$. You may write down each of your sum as combinations of sums of this type (that is, of sums of a certain number of consecutive integers).
– Suzet
Jul 18 at 22:51
In general, the sum $sum _k=n+1^mk$ equals $fracm(m+1)-n(n+1)2$. You may write down each of your sum as combinations of sums of this type (that is, of sums of a certain number of consecutive integers).
– Suzet
Jul 18 at 22:51
One way to interpret your Question is whether there is an expression which yields all values (and only those values) that are sums of $k$ summands drawn (without replacement) from the set $1,ldots,n$. However without seeing what use would be made, I'm doubtful that I've understood your request clearly enough to give a useful response.
– hardmath
Jul 18 at 22:52
One way to interpret your Question is whether there is an expression which yields all values (and only those values) that are sums of $k$ summands drawn (without replacement) from the set $1,ldots,n$. However without seeing what use would be made, I'm doubtful that I've understood your request clearly enough to give a useful response.
– hardmath
Jul 18 at 22:52
2
2
Don't write from your phone in the car. That's dangerous to you and others.
– Cameron Buie
Jul 18 at 23:19
Don't write from your phone in the car. That's dangerous to you and others.
– Cameron Buie
Jul 18 at 23:19
add a comment |Â
1 Answer
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They are, in fact, all positive integers between the minimum possible sum
$$frac(k)(k+1)2=1+2+3+...+k$$
and the maximum possible sum
$$kn-frack(k-1)2=(n-k+1)+(n-k+2)+...+n.$$
Can you show, by perturbing previous sums, why all of these sums can be reached?
answered Jul 18 at 22:52
Carl Schildkraut
8,26211238
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
They are, in fact, all positive integers between the minimum possible sum
$$frac(k)(k+1)2=1+2+3+...+k$$
and the maximum possible sum
$$kn-frack(k-1)2=(n-k+1)+(n-k+2)+...+n.$$
Can you show, by perturbing previous sums, why all of these sums can be reached?
answered Jul 18 at 22:52
Carl Schildkraut
8,26211238
add a comment |Â
up vote
5
down vote
They are, in fact, all positive integers between the minimum possible sum
$$frac(k)(k+1)2=1+2+3+...+k$$
and the maximum possible sum
$$kn-frack(k-1)2=(n-k+1)+(n-k+2)+...+n.$$
Can you show, by perturbing previous sums, why all of these sums can be reached?
answered Jul 18 at 22:52
Carl Schildkraut
8,26211238
add a comment |Â
up vote
5
down vote
up vote
5
down vote
They are, in fact, all positive integers between the minimum possible sum
$$frac(k)(k+1)2=1+2+3+...+k$$
and the maximum possible sum
$$kn-frack(k-1)2=(n-k+1)+(n-k+2)+...+n.$$
Can you show, by perturbing previous sums, why all of these sums can be reached?
answered Jul 18 at 22:52
Carl Schildkraut
8,26211238
They are, in fact, all positive integers between the minimum possible sum
$$frac(k)(k+1)2=1+2+3+...+k$$
and the maximum possible sum
$$kn-frack(k-1)2=(n-k+1)+(n-k+2)+...+n.$$
Can you show, by perturbing previous sums, why all of these sums can be reached?
answered Jul 18 at 22:52
Carl Schildkraut
8,26211238
answered Jul 18 at 22:52
Carl Schildkraut
8,26211238
answered Jul 18 at 22:52
Carl Schildkraut
8,26211238
answered Jul 18 at 22:52
Carl Schildkraut
8,26211238
8,26211238
add a comment |Â
add a comment |Â
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1
It is not clear to me what exactly you are after. Are you looking for the number of possible summations? Do you treat $1+4$ differently than $2+3$ because the addends are different or do you treat them as the same since their total is the same? Or... are you looking for a way to generate the list of possible summations (or results of summations)?
– JMoravitz
Jul 18 at 22:49
In general, the sum $sum _k=n+1^mk$ equals $fracm(m+1)-n(n+1)2$. You may write down each of your sum as combinations of sums of this type (that is, of sums of a certain number of consecutive integers).
– Suzet
Jul 18 at 22:51
One way to interpret your Question is whether there is an expression which yields all values (and only those values) that are sums of $k$ summands drawn (without replacement) from the set $1,ldots,n$. However without seeing what use would be made, I'm doubtful that I've understood your request clearly enough to give a useful response.
– hardmath
Jul 18 at 22:52
2
Don't write from your phone in the car. That's dangerous to you and others.
– Cameron Buie
Jul 18 at 23:19