How many numbers (positive integers) smaller than $n$ can be written as a sum of two or more consecutive power of 2 integers? [on hold]In how many ways can a number be expressed as a sum of consecutive numbers?How Many Natural Numbers Can Get Expressed as the Sum of Consecutive Natural Numbers in only One Way?If the difference of cubes of two consecutive integers is a square, then the square can be written as the sum of squares of two different integers.Total possible ways of representing n! as a sum of two or more consecutive positive integers.Determine all ways the integer $2015$ can be written as a sum of consecutive positive integers.Largest number of consecutive positive integers whose sum is exactly $2014$.writing numbers as sum of at least two consecutive odd positive integersSum of 3 Positive consecutive integersNumber of ways in which $1000$ can be written as a sum of two or more consecutive natural numbersWhich positive integers can NOT be written as a sum of consecutive positive integers
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How many numbers (positive integers) smaller than $n$ can be written as a sum of two or more consecutive power of 2 integers? [on hold]
In how many ways can a number be expressed as a sum of consecutive numbers?How Many Natural Numbers Can Get Expressed as the Sum of Consecutive Natural Numbers in only One Way?If the difference of cubes of two consecutive integers is a square, then the square can be written as the sum of squares of two different integers.Total possible ways of representing n! as a sum of two or more consecutive positive integers.Determine all ways the integer $2015$ can be written as a sum of consecutive positive integers.Largest number of consecutive positive integers whose sum is exactly $2014$.writing numbers as sum of at least two consecutive odd positive integersSum of 3 Positive consecutive integersNumber of ways in which $1000$ can be written as a sum of two or more consecutive natural numbersWhich positive integers can NOT be written as a sum of consecutive positive integers
$begingroup$
How many numbers (positive integers) smaller than $n$ can be written as a sum of two or more consecutive power of 2 integers?
My attempt so far: So basically we're searching how many $x$ and $y$ we have such that: $2^x+2^x+1+ldots+2^x+y leq n$, where $x,y,ninmathbbN$. By calculating the sum we arrive at:
$2^x(2^y+1-1)leq n$. And here I am stuck... Any ideas?
number-theory inequality
asked Mar 10 at 20:07
DemanedDemaned
63
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Demaned is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by Carl Mummert, Vinyl_cape_jawa, Riccardo.Alestra, Cesareo, Parcly Taxel 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, Vinyl_cape_jawa, Riccardo.Alestra, Cesareo, Parcly Taxel
add a comment |
$begingroup$
How many numbers (positive integers) smaller than $n$ can be written as a sum of two or more consecutive power of 2 integers?
My attempt so far: So basically we're searching how many $x$ and $y$ we have such that: $2^x+2^x+1+ldots+2^x+y leq n$, where $x,y,ninmathbbN$. By calculating the sum we arrive at:
$2^x(2^y+1-1)leq n$. And here I am stuck... Any ideas?
number-theory inequality
asked Mar 10 at 20:07
DemanedDemaned
63
New contributor
Demaned is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by Carl Mummert, Vinyl_cape_jawa, Riccardo.Alestra, Cesareo, Parcly Taxel 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, Vinyl_cape_jawa, Riccardo.Alestra, Cesareo, Parcly Taxel
$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. In particular, what is the source and background of the question?
$endgroup$
– Carl Mummert
Mar 10 at 20:43
add a comment |
$begingroup$
How many numbers (positive integers) smaller than $n$ can be written as a sum of two or more consecutive power of 2 integers?
My attempt so far: So basically we're searching how many $x$ and $y$ we have such that: $2^x+2^x+1+ldots+2^x+y leq n$, where $x,y,ninmathbbN$. By calculating the sum we arrive at:
$2^x(2^y+1-1)leq n$. And here I am stuck... Any ideas?
number-theory inequality
asked Mar 10 at 20:07
DemanedDemaned
63
New contributor
Demaned is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
How many numbers (positive integers) smaller than $n$ can be written as a sum of two or more consecutive power of 2 integers?
My attempt so far: So basically we're searching how many $x$ and $y$ we have such that: $2^x+2^x+1+ldots+2^x+y leq n$, where $x,y,ninmathbbN$. By calculating the sum we arrive at:
$2^x(2^y+1-1)leq n$. And here I am stuck... Any ideas?
number-theory inequality
number-theory inequality
asked Mar 10 at 20:07
DemanedDemaned
63
New contributor
Demaned is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 10 at 20:07
DemanedDemaned
63
New contributor
Demaned is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 10 at 20:33
Demaned
asked Mar 10 at 20:07
DemanedDemaned
63
New contributor
Demaned is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 10 at 20:07
DemanedDemaned
63
asked Mar 10 at 20:07
DemanedDemaned
63
63
New contributor
Demaned is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Demaned is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Demaned is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Carl Mummert, Vinyl_cape_jawa, Riccardo.Alestra, Cesareo, Parcly Taxel 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, Vinyl_cape_jawa, Riccardo.Alestra, Cesareo, Parcly Taxel
put on hold as off-topic by Carl Mummert, Vinyl_cape_jawa, Riccardo.Alestra, Cesareo, Parcly Taxel 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Carl Mummert, Vinyl_cape_jawa, Riccardo.Alestra, Cesareo, Parcly Taxel
$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. In particular, what is the source and background of the question?
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– Carl Mummert
Mar 10 at 20:43
add a comment |
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This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. In particular, what is the source and background of the question?
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– Carl Mummert
Mar 10 at 20:43
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This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. In particular, what is the source and background of the question?
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– Carl Mummert
Mar 10 at 20:43
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This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. In particular, what is the source and background of the question?
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– Carl Mummert
Mar 10 at 20:43
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1 Answer
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$begingroup$
If we only consider the case $n = 2^k$, and represent in binary the numbers matching the criteria, it seems simple enough to list them. E.g. for $n = 2^4 = 16$ we get the following four-bit numbers:
0011b = 2+1 = 3
0110b = 4+2 = 6
0111b = 4+2+1 = 7
1100b = 8+4 = 12
1110b = 8+4+2 = 14
1111b = 8+4+2+1 = 15
That is, there are three numbers with two consecutive ones, two with three ones, and one with four ones. That's $1+2+3 = 6$. The one-bits represent the terms of your sum as shown.
For $n = 32$, we have five bits and $1+2+3+4 = 10$ different numbers. Adding bits (or increasing $k$ by one; or doubling $n$) adds one possible position for each of the run lengths, so the amount of such numbers smaller than $n = 2^k$ is the triangular number $T_k-1$:
$$
T_k-1 = frac(k-1)k2
$$
The case where $n$ is not a power of two seems less simple.
answered Mar 10 at 21:18
ilkkachuilkkachu
542310
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If we only consider the case $n = 2^k$, and represent in binary the numbers matching the criteria, it seems simple enough to list them. E.g. for $n = 2^4 = 16$ we get the following four-bit numbers:
0011b = 2+1 = 3
0110b = 4+2 = 6
0111b = 4+2+1 = 7
1100b = 8+4 = 12
1110b = 8+4+2 = 14
1111b = 8+4+2+1 = 15
That is, there are three numbers with two consecutive ones, two with three ones, and one with four ones. That's $1+2+3 = 6$. The one-bits represent the terms of your sum as shown.
For $n = 32$, we have five bits and $1+2+3+4 = 10$ different numbers. Adding bits (or increasing $k$ by one; or doubling $n$) adds one possible position for each of the run lengths, so the amount of such numbers smaller than $n = 2^k$ is the triangular number $T_k-1$:
$$
T_k-1 = frac(k-1)k2
$$
The case where $n$ is not a power of two seems less simple.
answered Mar 10 at 21:18
ilkkachuilkkachu
542310
$endgroup$
add a comment |
$begingroup$
If we only consider the case $n = 2^k$, and represent in binary the numbers matching the criteria, it seems simple enough to list them. E.g. for $n = 2^4 = 16$ we get the following four-bit numbers:
0011b = 2+1 = 3
0110b = 4+2 = 6
0111b = 4+2+1 = 7
1100b = 8+4 = 12
1110b = 8+4+2 = 14
1111b = 8+4+2+1 = 15
That is, there are three numbers with two consecutive ones, two with three ones, and one with four ones. That's $1+2+3 = 6$. The one-bits represent the terms of your sum as shown.
For $n = 32$, we have five bits and $1+2+3+4 = 10$ different numbers. Adding bits (or increasing $k$ by one; or doubling $n$) adds one possible position for each of the run lengths, so the amount of such numbers smaller than $n = 2^k$ is the triangular number $T_k-1$:
$$
T_k-1 = frac(k-1)k2
$$
The case where $n$ is not a power of two seems less simple.
answered Mar 10 at 21:18
ilkkachuilkkachu
542310
$endgroup$
add a comment |
$begingroup$
If we only consider the case $n = 2^k$, and represent in binary the numbers matching the criteria, it seems simple enough to list them. E.g. for $n = 2^4 = 16$ we get the following four-bit numbers:
0011b = 2+1 = 3
0110b = 4+2 = 6
0111b = 4+2+1 = 7
1100b = 8+4 = 12
1110b = 8+4+2 = 14
1111b = 8+4+2+1 = 15
That is, there are three numbers with two consecutive ones, two with three ones, and one with four ones. That's $1+2+3 = 6$. The one-bits represent the terms of your sum as shown.
For $n = 32$, we have five bits and $1+2+3+4 = 10$ different numbers. Adding bits (or increasing $k$ by one; or doubling $n$) adds one possible position for each of the run lengths, so the amount of such numbers smaller than $n = 2^k$ is the triangular number $T_k-1$:
$$
T_k-1 = frac(k-1)k2
$$
The case where $n$ is not a power of two seems less simple.
answered Mar 10 at 21:18
ilkkachuilkkachu
542310
$endgroup$
If we only consider the case $n = 2^k$, and represent in binary the numbers matching the criteria, it seems simple enough to list them. E.g. for $n = 2^4 = 16$ we get the following four-bit numbers:
0011b = 2+1 = 3
0110b = 4+2 = 6
0111b = 4+2+1 = 7
1100b = 8+4 = 12
1110b = 8+4+2 = 14
1111b = 8+4+2+1 = 15
That is, there are three numbers with two consecutive ones, two with three ones, and one with four ones. That's $1+2+3 = 6$. The one-bits represent the terms of your sum as shown.
For $n = 32$, we have five bits and $1+2+3+4 = 10$ different numbers. Adding bits (or increasing $k$ by one; or doubling $n$) adds one possible position for each of the run lengths, so the amount of such numbers smaller than $n = 2^k$ is the triangular number $T_k-1$:
$$
T_k-1 = frac(k-1)k2
$$
The case where $n$ is not a power of two seems less simple.
answered Mar 10 at 21:18
ilkkachuilkkachu
542310
answered Mar 10 at 21:18
ilkkachuilkkachu
542310
answered Mar 10 at 21:18
ilkkachuilkkachu
542310
answered Mar 10 at 21:18
ilkkachuilkkachu
542310
542310
add a comment |
add a comment |
$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. In particular, what is the source and background of the question?
$endgroup$
– Carl Mummert
Mar 10 at 20:43