The number of decompositions of $2n-1$ into a difference of two squares?












2












$begingroup$


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









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  • 1




    $begingroup$
    This is OEIS sequence A193773.
    $endgroup$
    – Somos
    Dec 24 '18 at 1:05


















2












$begingroup$


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









share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is OEIS sequence A193773.
    $endgroup$
    – Somos
    Dec 24 '18 at 1:05
















2












2








2


1



$begingroup$


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









share|cite|improve this question











$endgroup$




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






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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
















  • 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












1 Answer
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3












$begingroup$

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






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    3












    $begingroup$

    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






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      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






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        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






        share|cite|improve this answer











        $endgroup$



        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







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 24 '18 at 1:11

























        answered Dec 23 '18 at 22:23









        François HuppéFrançois Huppé

        362111




        362111






























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