Ramanujan congruence mod 7
$begingroup$
Hello I am trying to prove this congruence:
$$P(7n+5)equiv 0 pmod{7}$$
In order to do that I have done the next thing:
We have that
$displaystylesum_{ngeq0};P(n)q^{n}=frac{1}{(q;q)_{infty}}$
we multiply by $q^{2}$ then
begin{eqnarray}
displaystylesum_{ngeq0};P(n)q^{n+2}&=&frac{q^{2}}{(q;q)_{infty}}\
&=& frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q;q)_{infty}^{7}}
end{eqnarray}
Therefore the coefficient of $q^{7n+7}$ in the LHS is $P(7n+5)$ therefore we have to check that the coefficient of $q^{7n+7}$ of $ frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q;q)_{infty}^{7}}$ is $equiv 0(mod ; 7)$.
Now we have that $(q;q)^{3}_{infty}=displaystylesum_{ngeq0} (-1)^{n}(2n+1)q^{frac{n(n+1)}{2}}$, therefore
begin{eqnarray}
q^{2}((q;q)^{3}_{infty})^{2}&=&(q(q;q)^{3}_{infty})^{2}\
&=& displaystyle sum_{n,m geq0} (-1)^{n}(2n+1)(2m+1)q^{frac{n(n+1)}{2}+frac{m(m+1)}{2}+2}
end{eqnarray}
Now we will check when $frac{n(n+1)}{2}+frac{m(m+1)}{2}+2$ is a mutiply of $7$
Note that $$(2n+1)^{2}+(2m+1)^{2}=8left(frac{n(n+1)}{2}+frac{m(m+1)}{2}+2right)-14$$
Then $frac{n(n+1)}{2}+frac{m(m+1)}{2}+2equiv 0(mod;7)$ if and only if $(2n+1)^{2}+(2m+1)^{2}equiv0(mod ;7)$ only if $(2n+1)^{2}equiv0(mod ;7)$ and $(2m+1)^{2}equiv0(mod ;7)$ Then $2n+1equiv0(mod ;7)$ and $2m+1equiv0(mod ;7)$.
Therefore the coefficient of $q^{7n+7}$ in $q^{2}((q;q)^{3}_{infty})^{2}$ is is multiple of 7.
I do not know if that is correct this idea and also I do not know how to do it for $frac{1}{(q;q)^{7}_{infty}}$. I think I can use $$frac{1}{(1-q)^{7}}equiv frac{1}{1-q^{7}}(mod;7)$$ so that I can have $$frac{1}{(q;q)^{7}_{infty}}equivfrac{1}{(q^{7};q^{7})_{infty}}(mod; 7 )$$
But I do not know how to procede with this. I woud appreciate any hint you can give me.
Thank you for your time!
integer-partitions q-series
$endgroup$
add a comment |
$begingroup$
Hello I am trying to prove this congruence:
$$P(7n+5)equiv 0 pmod{7}$$
In order to do that I have done the next thing:
We have that
$displaystylesum_{ngeq0};P(n)q^{n}=frac{1}{(q;q)_{infty}}$
we multiply by $q^{2}$ then
begin{eqnarray}
displaystylesum_{ngeq0};P(n)q^{n+2}&=&frac{q^{2}}{(q;q)_{infty}}\
&=& frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q;q)_{infty}^{7}}
end{eqnarray}
Therefore the coefficient of $q^{7n+7}$ in the LHS is $P(7n+5)$ therefore we have to check that the coefficient of $q^{7n+7}$ of $ frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q;q)_{infty}^{7}}$ is $equiv 0(mod ; 7)$.
Now we have that $(q;q)^{3}_{infty}=displaystylesum_{ngeq0} (-1)^{n}(2n+1)q^{frac{n(n+1)}{2}}$, therefore
begin{eqnarray}
q^{2}((q;q)^{3}_{infty})^{2}&=&(q(q;q)^{3}_{infty})^{2}\
&=& displaystyle sum_{n,m geq0} (-1)^{n}(2n+1)(2m+1)q^{frac{n(n+1)}{2}+frac{m(m+1)}{2}+2}
end{eqnarray}
Now we will check when $frac{n(n+1)}{2}+frac{m(m+1)}{2}+2$ is a mutiply of $7$
Note that $$(2n+1)^{2}+(2m+1)^{2}=8left(frac{n(n+1)}{2}+frac{m(m+1)}{2}+2right)-14$$
Then $frac{n(n+1)}{2}+frac{m(m+1)}{2}+2equiv 0(mod;7)$ if and only if $(2n+1)^{2}+(2m+1)^{2}equiv0(mod ;7)$ only if $(2n+1)^{2}equiv0(mod ;7)$ and $(2m+1)^{2}equiv0(mod ;7)$ Then $2n+1equiv0(mod ;7)$ and $2m+1equiv0(mod ;7)$.
Therefore the coefficient of $q^{7n+7}$ in $q^{2}((q;q)^{3}_{infty})^{2}$ is is multiple of 7.
I do not know if that is correct this idea and also I do not know how to do it for $frac{1}{(q;q)^{7}_{infty}}$. I think I can use $$frac{1}{(1-q)^{7}}equiv frac{1}{1-q^{7}}(mod;7)$$ so that I can have $$frac{1}{(q;q)^{7}_{infty}}equivfrac{1}{(q^{7};q^{7})_{infty}}(mod; 7 )$$
But I do not know how to procede with this. I woud appreciate any hint you can give me.
Thank you for your time!
integer-partitions q-series
$endgroup$
$begingroup$
To be clear: you're satisfied with your proof that $$[q^{7n+7}]q^{2}((q;q)^{3}_{infty})^{2} equiv 0 pmod 7$$ and you want to show that $$[q^{7n+7}]frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q^{7};q^{7})_{infty}} equiv 0 pmod 7$$?
$endgroup$
– Peter Taylor
Dec 5 '18 at 16:33
$begingroup$
Of course but I do not know how to finish it!
$endgroup$
– Liddo
Dec 5 '18 at 16:53
$begingroup$
It is of interest to know that Ramanujan proved this congruence using the same method. He provided another more complicated proof by giving a closed form for $sum_{n=0}^{infty} p(7n+5)q^n$. See this post for more details.
$endgroup$
– Paramanand Singh
Dec 6 '18 at 17:13
add a comment |
$begingroup$
Hello I am trying to prove this congruence:
$$P(7n+5)equiv 0 pmod{7}$$
In order to do that I have done the next thing:
We have that
$displaystylesum_{ngeq0};P(n)q^{n}=frac{1}{(q;q)_{infty}}$
we multiply by $q^{2}$ then
begin{eqnarray}
displaystylesum_{ngeq0};P(n)q^{n+2}&=&frac{q^{2}}{(q;q)_{infty}}\
&=& frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q;q)_{infty}^{7}}
end{eqnarray}
Therefore the coefficient of $q^{7n+7}$ in the LHS is $P(7n+5)$ therefore we have to check that the coefficient of $q^{7n+7}$ of $ frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q;q)_{infty}^{7}}$ is $equiv 0(mod ; 7)$.
Now we have that $(q;q)^{3}_{infty}=displaystylesum_{ngeq0} (-1)^{n}(2n+1)q^{frac{n(n+1)}{2}}$, therefore
begin{eqnarray}
q^{2}((q;q)^{3}_{infty})^{2}&=&(q(q;q)^{3}_{infty})^{2}\
&=& displaystyle sum_{n,m geq0} (-1)^{n}(2n+1)(2m+1)q^{frac{n(n+1)}{2}+frac{m(m+1)}{2}+2}
end{eqnarray}
Now we will check when $frac{n(n+1)}{2}+frac{m(m+1)}{2}+2$ is a mutiply of $7$
Note that $$(2n+1)^{2}+(2m+1)^{2}=8left(frac{n(n+1)}{2}+frac{m(m+1)}{2}+2right)-14$$
Then $frac{n(n+1)}{2}+frac{m(m+1)}{2}+2equiv 0(mod;7)$ if and only if $(2n+1)^{2}+(2m+1)^{2}equiv0(mod ;7)$ only if $(2n+1)^{2}equiv0(mod ;7)$ and $(2m+1)^{2}equiv0(mod ;7)$ Then $2n+1equiv0(mod ;7)$ and $2m+1equiv0(mod ;7)$.
Therefore the coefficient of $q^{7n+7}$ in $q^{2}((q;q)^{3}_{infty})^{2}$ is is multiple of 7.
I do not know if that is correct this idea and also I do not know how to do it for $frac{1}{(q;q)^{7}_{infty}}$. I think I can use $$frac{1}{(1-q)^{7}}equiv frac{1}{1-q^{7}}(mod;7)$$ so that I can have $$frac{1}{(q;q)^{7}_{infty}}equivfrac{1}{(q^{7};q^{7})_{infty}}(mod; 7 )$$
But I do not know how to procede with this. I woud appreciate any hint you can give me.
Thank you for your time!
integer-partitions q-series
$endgroup$
Hello I am trying to prove this congruence:
$$P(7n+5)equiv 0 pmod{7}$$
In order to do that I have done the next thing:
We have that
$displaystylesum_{ngeq0};P(n)q^{n}=frac{1}{(q;q)_{infty}}$
we multiply by $q^{2}$ then
begin{eqnarray}
displaystylesum_{ngeq0};P(n)q^{n+2}&=&frac{q^{2}}{(q;q)_{infty}}\
&=& frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q;q)_{infty}^{7}}
end{eqnarray}
Therefore the coefficient of $q^{7n+7}$ in the LHS is $P(7n+5)$ therefore we have to check that the coefficient of $q^{7n+7}$ of $ frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q;q)_{infty}^{7}}$ is $equiv 0(mod ; 7)$.
Now we have that $(q;q)^{3}_{infty}=displaystylesum_{ngeq0} (-1)^{n}(2n+1)q^{frac{n(n+1)}{2}}$, therefore
begin{eqnarray}
q^{2}((q;q)^{3}_{infty})^{2}&=&(q(q;q)^{3}_{infty})^{2}\
&=& displaystyle sum_{n,m geq0} (-1)^{n}(2n+1)(2m+1)q^{frac{n(n+1)}{2}+frac{m(m+1)}{2}+2}
end{eqnarray}
Now we will check when $frac{n(n+1)}{2}+frac{m(m+1)}{2}+2$ is a mutiply of $7$
Note that $$(2n+1)^{2}+(2m+1)^{2}=8left(frac{n(n+1)}{2}+frac{m(m+1)}{2}+2right)-14$$
Then $frac{n(n+1)}{2}+frac{m(m+1)}{2}+2equiv 0(mod;7)$ if and only if $(2n+1)^{2}+(2m+1)^{2}equiv0(mod ;7)$ only if $(2n+1)^{2}equiv0(mod ;7)$ and $(2m+1)^{2}equiv0(mod ;7)$ Then $2n+1equiv0(mod ;7)$ and $2m+1equiv0(mod ;7)$.
Therefore the coefficient of $q^{7n+7}$ in $q^{2}((q;q)^{3}_{infty})^{2}$ is is multiple of 7.
I do not know if that is correct this idea and also I do not know how to do it for $frac{1}{(q;q)^{7}_{infty}}$. I think I can use $$frac{1}{(1-q)^{7}}equiv frac{1}{1-q^{7}}(mod;7)$$ so that I can have $$frac{1}{(q;q)^{7}_{infty}}equivfrac{1}{(q^{7};q^{7})_{infty}}(mod; 7 )$$
But I do not know how to procede with this. I woud appreciate any hint you can give me.
Thank you for your time!
integer-partitions q-series
integer-partitions q-series
edited Dec 5 '18 at 15:36
gt6989b
33.9k22455
33.9k22455
asked Dec 5 '18 at 15:29
LiddoLiddo
20619
20619
$begingroup$
To be clear: you're satisfied with your proof that $$[q^{7n+7}]q^{2}((q;q)^{3}_{infty})^{2} equiv 0 pmod 7$$ and you want to show that $$[q^{7n+7}]frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q^{7};q^{7})_{infty}} equiv 0 pmod 7$$?
$endgroup$
– Peter Taylor
Dec 5 '18 at 16:33
$begingroup$
Of course but I do not know how to finish it!
$endgroup$
– Liddo
Dec 5 '18 at 16:53
$begingroup$
It is of interest to know that Ramanujan proved this congruence using the same method. He provided another more complicated proof by giving a closed form for $sum_{n=0}^{infty} p(7n+5)q^n$. See this post for more details.
$endgroup$
– Paramanand Singh
Dec 6 '18 at 17:13
add a comment |
$begingroup$
To be clear: you're satisfied with your proof that $$[q^{7n+7}]q^{2}((q;q)^{3}_{infty})^{2} equiv 0 pmod 7$$ and you want to show that $$[q^{7n+7}]frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q^{7};q^{7})_{infty}} equiv 0 pmod 7$$?
$endgroup$
– Peter Taylor
Dec 5 '18 at 16:33
$begingroup$
Of course but I do not know how to finish it!
$endgroup$
– Liddo
Dec 5 '18 at 16:53
$begingroup$
It is of interest to know that Ramanujan proved this congruence using the same method. He provided another more complicated proof by giving a closed form for $sum_{n=0}^{infty} p(7n+5)q^n$. See this post for more details.
$endgroup$
– Paramanand Singh
Dec 6 '18 at 17:13
$begingroup$
To be clear: you're satisfied with your proof that $$[q^{7n+7}]q^{2}((q;q)^{3}_{infty})^{2} equiv 0 pmod 7$$ and you want to show that $$[q^{7n+7}]frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q^{7};q^{7})_{infty}} equiv 0 pmod 7$$?
$endgroup$
– Peter Taylor
Dec 5 '18 at 16:33
$begingroup$
To be clear: you're satisfied with your proof that $$[q^{7n+7}]q^{2}((q;q)^{3}_{infty})^{2} equiv 0 pmod 7$$ and you want to show that $$[q^{7n+7}]frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q^{7};q^{7})_{infty}} equiv 0 pmod 7$$?
$endgroup$
– Peter Taylor
Dec 5 '18 at 16:33
$begingroup$
Of course but I do not know how to finish it!
$endgroup$
– Liddo
Dec 5 '18 at 16:53
$begingroup$
Of course but I do not know how to finish it!
$endgroup$
– Liddo
Dec 5 '18 at 16:53
$begingroup$
It is of interest to know that Ramanujan proved this congruence using the same method. He provided another more complicated proof by giving a closed form for $sum_{n=0}^{infty} p(7n+5)q^n$. See this post for more details.
$endgroup$
– Paramanand Singh
Dec 6 '18 at 17:13
$begingroup$
It is of interest to know that Ramanujan proved this congruence using the same method. He provided another more complicated proof by giving a closed form for $sum_{n=0}^{infty} p(7n+5)q^n$. See this post for more details.
$endgroup$
– Paramanand Singh
Dec 6 '18 at 17:13
add a comment |
1 Answer
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oldest
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$begingroup$
I think the bit you're missing is $$frac{1}{1 - z} = 1 + z + z^2 + cdots$$
Therefore $$frac{1}{(q^7;q^7)_infty} = prod_{k=1}^infty sum_{j=0}^infty q^{7jk}$$
which clearly only has non-zero coefficients for powers of seven.
Thus the coefficient $[q^{7n+7}]frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q^{7};q^{7})_{infty}}$ is some integer-weighted sum* of coefficients $[q^{7i}]q^{2}((q;q)^{3}_{infty})^{2}$.
* It is, of course, quite easy to be more precise than this, but unnecessary for the argument. On the other hand, maybe it's a nicer argument to say that $frac{1}{(q^7;q^7)_infty}$ is the generating function for the partition numbers inflated by a factor of seven, so that $$[q^{7n+7}]frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q^{7};q^{7})_{infty}} = [q^{7n+7}](q^{2}((q;q)^{3}_{infty})^{2})sum_{i=0}^infty P(i)q^{7i} = sum_{i=0}^{n+1} P(i)[q^{7(n-i+1)}](q^{2}((q;q)^{3}_{infty})^{2})$$
$endgroup$
add a comment |
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$begingroup$
I think the bit you're missing is $$frac{1}{1 - z} = 1 + z + z^2 + cdots$$
Therefore $$frac{1}{(q^7;q^7)_infty} = prod_{k=1}^infty sum_{j=0}^infty q^{7jk}$$
which clearly only has non-zero coefficients for powers of seven.
Thus the coefficient $[q^{7n+7}]frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q^{7};q^{7})_{infty}}$ is some integer-weighted sum* of coefficients $[q^{7i}]q^{2}((q;q)^{3}_{infty})^{2}$.
* It is, of course, quite easy to be more precise than this, but unnecessary for the argument. On the other hand, maybe it's a nicer argument to say that $frac{1}{(q^7;q^7)_infty}$ is the generating function for the partition numbers inflated by a factor of seven, so that $$[q^{7n+7}]frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q^{7};q^{7})_{infty}} = [q^{7n+7}](q^{2}((q;q)^{3}_{infty})^{2})sum_{i=0}^infty P(i)q^{7i} = sum_{i=0}^{n+1} P(i)[q^{7(n-i+1)}](q^{2}((q;q)^{3}_{infty})^{2})$$
$endgroup$
add a comment |
$begingroup$
I think the bit you're missing is $$frac{1}{1 - z} = 1 + z + z^2 + cdots$$
Therefore $$frac{1}{(q^7;q^7)_infty} = prod_{k=1}^infty sum_{j=0}^infty q^{7jk}$$
which clearly only has non-zero coefficients for powers of seven.
Thus the coefficient $[q^{7n+7}]frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q^{7};q^{7})_{infty}}$ is some integer-weighted sum* of coefficients $[q^{7i}]q^{2}((q;q)^{3}_{infty})^{2}$.
* It is, of course, quite easy to be more precise than this, but unnecessary for the argument. On the other hand, maybe it's a nicer argument to say that $frac{1}{(q^7;q^7)_infty}$ is the generating function for the partition numbers inflated by a factor of seven, so that $$[q^{7n+7}]frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q^{7};q^{7})_{infty}} = [q^{7n+7}](q^{2}((q;q)^{3}_{infty})^{2})sum_{i=0}^infty P(i)q^{7i} = sum_{i=0}^{n+1} P(i)[q^{7(n-i+1)}](q^{2}((q;q)^{3}_{infty})^{2})$$
$endgroup$
add a comment |
$begingroup$
I think the bit you're missing is $$frac{1}{1 - z} = 1 + z + z^2 + cdots$$
Therefore $$frac{1}{(q^7;q^7)_infty} = prod_{k=1}^infty sum_{j=0}^infty q^{7jk}$$
which clearly only has non-zero coefficients for powers of seven.
Thus the coefficient $[q^{7n+7}]frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q^{7};q^{7})_{infty}}$ is some integer-weighted sum* of coefficients $[q^{7i}]q^{2}((q;q)^{3}_{infty})^{2}$.
* It is, of course, quite easy to be more precise than this, but unnecessary for the argument. On the other hand, maybe it's a nicer argument to say that $frac{1}{(q^7;q^7)_infty}$ is the generating function for the partition numbers inflated by a factor of seven, so that $$[q^{7n+7}]frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q^{7};q^{7})_{infty}} = [q^{7n+7}](q^{2}((q;q)^{3}_{infty})^{2})sum_{i=0}^infty P(i)q^{7i} = sum_{i=0}^{n+1} P(i)[q^{7(n-i+1)}](q^{2}((q;q)^{3}_{infty})^{2})$$
$endgroup$
I think the bit you're missing is $$frac{1}{1 - z} = 1 + z + z^2 + cdots$$
Therefore $$frac{1}{(q^7;q^7)_infty} = prod_{k=1}^infty sum_{j=0}^infty q^{7jk}$$
which clearly only has non-zero coefficients for powers of seven.
Thus the coefficient $[q^{7n+7}]frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q^{7};q^{7})_{infty}}$ is some integer-weighted sum* of coefficients $[q^{7i}]q^{2}((q;q)^{3}_{infty})^{2}$.
* It is, of course, quite easy to be more precise than this, but unnecessary for the argument. On the other hand, maybe it's a nicer argument to say that $frac{1}{(q^7;q^7)_infty}$ is the generating function for the partition numbers inflated by a factor of seven, so that $$[q^{7n+7}]frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q^{7};q^{7})_{infty}} = [q^{7n+7}](q^{2}((q;q)^{3}_{infty})^{2})sum_{i=0}^infty P(i)q^{7i} = sum_{i=0}^{n+1} P(i)[q^{7(n-i+1)}](q^{2}((q;q)^{3}_{infty})^{2})$$
edited Dec 5 '18 at 17:20
answered Dec 5 '18 at 17:14
Peter TaylorPeter Taylor
8,84812341
8,84812341
add a comment |
add a comment |
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$begingroup$
To be clear: you're satisfied with your proof that $$[q^{7n+7}]q^{2}((q;q)^{3}_{infty})^{2} equiv 0 pmod 7$$ and you want to show that $$[q^{7n+7}]frac{q^{2}((q;q)^{3}_{infty})^{2}}{(q^{7};q^{7})_{infty}} equiv 0 pmod 7$$?
$endgroup$
– Peter Taylor
Dec 5 '18 at 16:33
$begingroup$
Of course but I do not know how to finish it!
$endgroup$
– Liddo
Dec 5 '18 at 16:53
$begingroup$
It is of interest to know that Ramanujan proved this congruence using the same method. He provided another more complicated proof by giving a closed form for $sum_{n=0}^{infty} p(7n+5)q^n$. See this post for more details.
$endgroup$
– Paramanand Singh
Dec 6 '18 at 17:13