The number of decompositions of $2n-1$ into a difference of two squares?
$begingroup$
Is there a exact formula for number of decompositions of $2n-1$ into a difference of two squares?
Examples:
3: 1 | 21: 1 | 39: 2 | 57: 1
4: 1 | 22: 1 | 40: 1 | 58: 2
5: 2 | 23: 3 | 41: 3 | 59: 3
6: 1 | 24: 1 | 42: 1 | 60: 2
7: 1 | 25: 2 | 43: 2 | 61: 2
8: 2 | 26: 2 | 44: 2 | 62: 2
9: 1 | 27: 1 | 45: 1 | 63: 2
10: 1 | 28: 2 | 46: 2 | 64: 1
11: 2 | 29: 2 | 47: 2 | 65: 2
12: 1 | 30: 1 | 48: 2 | 66: 1
13: 2 | 31: 1 | 49: 1 | 67: 2
14: 2 | 32: 3 | 50: 3 | 68: 4
15: 1 | 33: 2 | 51: 1 | 69: 1
16: 1 | 34: 1 | 52: 1 | 70: 1
17: 2 | 35: 2 | 53: 4 | 71: 2
18: 2 | 36: 1 | 54: 1 | 72: 2
19: 1 | 37: 1 | 55: 1 | 73: 2
20: 2 | 38: 3 | 56: 2 | 74: 3
number-theory square-numbers divisor-counting-function
$endgroup$
add a comment |
$begingroup$
Is there a exact formula for number of decompositions of $2n-1$ into a difference of two squares?
Examples:
3: 1 | 21: 1 | 39: 2 | 57: 1
4: 1 | 22: 1 | 40: 1 | 58: 2
5: 2 | 23: 3 | 41: 3 | 59: 3
6: 1 | 24: 1 | 42: 1 | 60: 2
7: 1 | 25: 2 | 43: 2 | 61: 2
8: 2 | 26: 2 | 44: 2 | 62: 2
9: 1 | 27: 1 | 45: 1 | 63: 2
10: 1 | 28: 2 | 46: 2 | 64: 1
11: 2 | 29: 2 | 47: 2 | 65: 2
12: 1 | 30: 1 | 48: 2 | 66: 1
13: 2 | 31: 1 | 49: 1 | 67: 2
14: 2 | 32: 3 | 50: 3 | 68: 4
15: 1 | 33: 2 | 51: 1 | 69: 1
16: 1 | 34: 1 | 52: 1 | 70: 1
17: 2 | 35: 2 | 53: 4 | 71: 2
18: 2 | 36: 1 | 54: 1 | 72: 2
19: 1 | 37: 1 | 55: 1 | 73: 2
20: 2 | 38: 3 | 56: 2 | 74: 3
number-theory square-numbers divisor-counting-function
$endgroup$
1
$begingroup$
This is OEIS sequence A193773.
$endgroup$
– Somos
Dec 24 '18 at 1:05
add a comment |
$begingroup$
Is there a exact formula for number of decompositions of $2n-1$ into a difference of two squares?
Examples:
3: 1 | 21: 1 | 39: 2 | 57: 1
4: 1 | 22: 1 | 40: 1 | 58: 2
5: 2 | 23: 3 | 41: 3 | 59: 3
6: 1 | 24: 1 | 42: 1 | 60: 2
7: 1 | 25: 2 | 43: 2 | 61: 2
8: 2 | 26: 2 | 44: 2 | 62: 2
9: 1 | 27: 1 | 45: 1 | 63: 2
10: 1 | 28: 2 | 46: 2 | 64: 1
11: 2 | 29: 2 | 47: 2 | 65: 2
12: 1 | 30: 1 | 48: 2 | 66: 1
13: 2 | 31: 1 | 49: 1 | 67: 2
14: 2 | 32: 3 | 50: 3 | 68: 4
15: 1 | 33: 2 | 51: 1 | 69: 1
16: 1 | 34: 1 | 52: 1 | 70: 1
17: 2 | 35: 2 | 53: 4 | 71: 2
18: 2 | 36: 1 | 54: 1 | 72: 2
19: 1 | 37: 1 | 55: 1 | 73: 2
20: 2 | 38: 3 | 56: 2 | 74: 3
number-theory square-numbers divisor-counting-function
$endgroup$
Is there a exact formula for number of decompositions of $2n-1$ into a difference of two squares?
Examples:
3: 1 | 21: 1 | 39: 2 | 57: 1
4: 1 | 22: 1 | 40: 1 | 58: 2
5: 2 | 23: 3 | 41: 3 | 59: 3
6: 1 | 24: 1 | 42: 1 | 60: 2
7: 1 | 25: 2 | 43: 2 | 61: 2
8: 2 | 26: 2 | 44: 2 | 62: 2
9: 1 | 27: 1 | 45: 1 | 63: 2
10: 1 | 28: 2 | 46: 2 | 64: 1
11: 2 | 29: 2 | 47: 2 | 65: 2
12: 1 | 30: 1 | 48: 2 | 66: 1
13: 2 | 31: 1 | 49: 1 | 67: 2
14: 2 | 32: 3 | 50: 3 | 68: 4
15: 1 | 33: 2 | 51: 1 | 69: 1
16: 1 | 34: 1 | 52: 1 | 70: 1
17: 2 | 35: 2 | 53: 4 | 71: 2
18: 2 | 36: 1 | 54: 1 | 72: 2
19: 1 | 37: 1 | 55: 1 | 73: 2
20: 2 | 38: 3 | 56: 2 | 74: 3
number-theory square-numbers divisor-counting-function
number-theory square-numbers divisor-counting-function
edited Dec 23 '18 at 22:39
François Huppé
asked Dec 23 '18 at 22:23
François HuppéFrançois Huppé
362111
362111
1
$begingroup$
This is OEIS sequence A193773.
$endgroup$
– Somos
Dec 24 '18 at 1:05
add a comment |
1
$begingroup$
This is OEIS sequence A193773.
$endgroup$
– Somos
Dec 24 '18 at 1:05
1
1
$begingroup$
This is OEIS sequence A193773.
$endgroup$
– Somos
Dec 24 '18 at 1:05
$begingroup$
This is OEIS sequence A193773.
$endgroup$
– Somos
Dec 24 '18 at 1:05
add a comment |
1 Answer
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oldest
votes
$begingroup$
Any odd composite integer $m$ can be written as $m=pq$. We can switch it to a difference of squares:
$$m=pq$$
$$4m=4pq$$
$$4m=2pq+2pq$$
$$4m=2pq+2pq+q^{2}-q^{2}+p^{2}-p^{2}$$
$$4m=left(p+qright)^{2}-left(q-pright)^{2}$$
$$m=frac{left(p+qright)^{2}}{4}-frac{left(q-pright)}{4}^{2}$$
$$m=left(frac{p+q}{2}right)^{2}-left(frac{q-p}{2}right)^{2}$$
Since $m$ is odd, $p$ and $q$ are also odd. Therefore, every possible pairs $p,q$ such that $m=pq$ will be solutions. We can now use the number of divisors function $sigma(m)$ to conclude that:
The number of decompositions of $2n-1$ into a difference of two squares is defined by: $$leftlceil frac{sigmaleft(2n-1right)}{2}rightrceil$$
This is OEIS sequence A193773
$endgroup$
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
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active
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votes
$begingroup$
Any odd composite integer $m$ can be written as $m=pq$. We can switch it to a difference of squares:
$$m=pq$$
$$4m=4pq$$
$$4m=2pq+2pq$$
$$4m=2pq+2pq+q^{2}-q^{2}+p^{2}-p^{2}$$
$$4m=left(p+qright)^{2}-left(q-pright)^{2}$$
$$m=frac{left(p+qright)^{2}}{4}-frac{left(q-pright)}{4}^{2}$$
$$m=left(frac{p+q}{2}right)^{2}-left(frac{q-p}{2}right)^{2}$$
Since $m$ is odd, $p$ and $q$ are also odd. Therefore, every possible pairs $p,q$ such that $m=pq$ will be solutions. We can now use the number of divisors function $sigma(m)$ to conclude that:
The number of decompositions of $2n-1$ into a difference of two squares is defined by: $$leftlceil frac{sigmaleft(2n-1right)}{2}rightrceil$$
This is OEIS sequence A193773
$endgroup$
add a comment |
$begingroup$
Any odd composite integer $m$ can be written as $m=pq$. We can switch it to a difference of squares:
$$m=pq$$
$$4m=4pq$$
$$4m=2pq+2pq$$
$$4m=2pq+2pq+q^{2}-q^{2}+p^{2}-p^{2}$$
$$4m=left(p+qright)^{2}-left(q-pright)^{2}$$
$$m=frac{left(p+qright)^{2}}{4}-frac{left(q-pright)}{4}^{2}$$
$$m=left(frac{p+q}{2}right)^{2}-left(frac{q-p}{2}right)^{2}$$
Since $m$ is odd, $p$ and $q$ are also odd. Therefore, every possible pairs $p,q$ such that $m=pq$ will be solutions. We can now use the number of divisors function $sigma(m)$ to conclude that:
The number of decompositions of $2n-1$ into a difference of two squares is defined by: $$leftlceil frac{sigmaleft(2n-1right)}{2}rightrceil$$
This is OEIS sequence A193773
$endgroup$
add a comment |
$begingroup$
Any odd composite integer $m$ can be written as $m=pq$. We can switch it to a difference of squares:
$$m=pq$$
$$4m=4pq$$
$$4m=2pq+2pq$$
$$4m=2pq+2pq+q^{2}-q^{2}+p^{2}-p^{2}$$
$$4m=left(p+qright)^{2}-left(q-pright)^{2}$$
$$m=frac{left(p+qright)^{2}}{4}-frac{left(q-pright)}{4}^{2}$$
$$m=left(frac{p+q}{2}right)^{2}-left(frac{q-p}{2}right)^{2}$$
Since $m$ is odd, $p$ and $q$ are also odd. Therefore, every possible pairs $p,q$ such that $m=pq$ will be solutions. We can now use the number of divisors function $sigma(m)$ to conclude that:
The number of decompositions of $2n-1$ into a difference of two squares is defined by: $$leftlceil frac{sigmaleft(2n-1right)}{2}rightrceil$$
This is OEIS sequence A193773
$endgroup$
Any odd composite integer $m$ can be written as $m=pq$. We can switch it to a difference of squares:
$$m=pq$$
$$4m=4pq$$
$$4m=2pq+2pq$$
$$4m=2pq+2pq+q^{2}-q^{2}+p^{2}-p^{2}$$
$$4m=left(p+qright)^{2}-left(q-pright)^{2}$$
$$m=frac{left(p+qright)^{2}}{4}-frac{left(q-pright)}{4}^{2}$$
$$m=left(frac{p+q}{2}right)^{2}-left(frac{q-p}{2}right)^{2}$$
Since $m$ is odd, $p$ and $q$ are also odd. Therefore, every possible pairs $p,q$ such that $m=pq$ will be solutions. We can now use the number of divisors function $sigma(m)$ to conclude that:
The number of decompositions of $2n-1$ into a difference of two squares is defined by: $$leftlceil frac{sigmaleft(2n-1right)}{2}rightrceil$$
This is OEIS sequence A193773
edited Dec 24 '18 at 1:11
answered Dec 23 '18 at 22:23
François HuppéFrançois Huppé
362111
362111
add a comment |
add a comment |
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$begingroup$
This is OEIS sequence A193773.
$endgroup$
– Somos
Dec 24 '18 at 1:05