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Let's define a function $p_2(n)$ that it is total distinct way to write $n$ positive integer as sum of square of positive integers. For example, $9$ can be partitioned as sum of squares in 4 distinct ways:

$$3^2$$ $$2^2 + 2^2+1^2$$ $$2^2 + 1^2+1^2+1^2+1^2+1^2$$ $$1^2+1^2+1^2+1^2+1^2+1^2+1^2+1^2+1^2$$

Thus $p_2(9)=4$

I used the similar method of partition function to find generating function for $p_2(n)$ and I have gotten

$$\sum\limits_{n=0}^{ \infty }p_2(n)q^n=(1+q^{1^2}+q^{2.1^2}+q^{3.1^2}+......)(1+q^{2^2}+q^{2.2^2}+q^{3.2^2}+......)(1+q^{3^2}+q^{2.3^2}+q^{3.3^2}+......)..$$

$$\sum\limits_{n=0}^{ \infty }p_2(n)q^n=\frac{1}{\prod\limits_{n=1}^{ \infty }(1-q^{n^2})} $$

I expanded first terms of that product function

$$\prod\limits_{n=1}^{ \infty }(1-q^{n^2})=1-q-q^4+q^5-q^9+q^{10}+q^{13}-q^{14}-q^{16}+q^{17}+q^{20}-q^{21}+q^{34}-q^{35}-q^{36}+q^{37}-q^{38}+q^{39}+q^{40}-q^{42}-2q^{49}+q^{51}+q^{52}+........$$

I have not found any references for sum formula for $\prod\limits_{n=1}^{ \infty }(1-q^{n^2})$.

Is there any sum formula of $\prod\limits_{n=1}^{ \infty }(1-q^{n^2})$ as we can express $\prod\limits_{n=1}^{ \infty }(1-q^{n})=\sum\limits_{n=-\infty}^{ \infty }(-1)^n q^{\frac{n(3n-1)}{2}}$?

Thanks for references and links to related subject.

UPDATE: 9th December,2014

I have found a relation for that square sum partition function I would like to share it.

$$\Phi_2(q)=\prod\limits_{n=1}^{ \infty }(1-q^{n^2})$$ $$\Phi_2(q)=\prod\limits_{n=1}^{ \infty }(1-q^{(2n-1)^2})\prod\limits_{n=1}^{ \infty }(1-q^{(2n)^2})$$

$$\Phi_2(q)=\prod\limits_{n=1}^{ \infty }(1-q^{(2n-1)^2})\prod\limits_{n=1}^{ \infty }(1-(q^4)^{n^2})$$

$$\Phi_2(q)=\Phi_2(q^4)\prod\limits_{n=1}^{ \infty }(1-q^{(2n-1)^2}) \tag 1$$ $$\Phi_2(-q)=\Phi_2(q^4)\prod\limits_{n=1}^{ \infty }(1+q^{(2n-1)^2})$$

$$\Phi_2(-q)\Phi_2(q)=\Phi^2_2(q^4)\prod\limits_{n=1}^{ \infty }(1+q^{(2n-1)^2})\prod\limits_{n=1}^{ \infty }(1-q^{(2n-1)^2})$$

$$\Phi_2(-q)\Phi_2(q)=\Phi^2_2(q^4)\prod\limits_{n=1}^{ \infty }(1-(q^2)^{(2n-1)^2}) \tag 2$$ If we use Equation 1 and put $q -->q^2$ $$\Phi_2(q^2)=\Phi_2(q^8)\prod\limits_{n=1}^{ \infty }(1-(q^2)^{(2n-1)^2}) \tag 3$$

Now we can find and function relation for $\Phi_2(q)$ if we divide Equation 2 and 3

$$\frac{\Phi_2(-q)\Phi_2(q)}{\Phi_2(q^2)}=\frac{\Phi^2_2(q^4)}{\Phi_2(q^8)}$$

$$\Phi_2(-q)\Phi_2(q)\Phi_2(q^8)=\Phi^2_2(q^4)\Phi_2(q^2) \tag 4$$

$$\sum\limits_{n=0}^{ \infty }p_2(n)q^n=\frac{1}{\prod\limits_{n=1}^{ \infty }(1-q^{n^2})} =\frac{1}{\Phi_2(q)}$$

If we define $$P_2(q)=\sum\limits_{n=0}^{ \infty }p_2(n)q^n$$

$$P_2(q)=\frac{1}{\Phi_2(q)}$$

If we put in Equation 4,

$$P_2(-q)P_2(q)P_2(q^8)=P^2_2(q^4)P_2(q^2) \tag 5$$

Finally we can write square sum partition as function relation,

$$\left( \sum\limits_{n=0}^{ \infty }p_2(n)(-1)^nq^n \right) \left( \sum\limits_{n=0}^{ \infty }p_2(n)q^n \right) \left( \sum\limits_{n=0}^{ \infty }p_2(n)q^{8n}\right)=\left( \sum\limits_{n=0}^{ \infty }p_2(n)q^{4n} \right)^2 \left( \sum\limits_{n=0}^{ \infty }p_2(n)q^{2n} \right) $$

n terms can be sum both side and a relation can be found for square sum partition function but It is quite complex relationship. I could not get further. Can someone please help me or advice how to simplify it and get further from that point

Thanks a lot

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  • $\begingroup$ I tried computing the first ten thousand coefficients, and didn't see a simple pattern. The coefficient on $q^n$ is occasionally greater than $n$: for instance, the coefficient on $q^{9911}$ is $19304$. $\endgroup$
    – zeb
    Nov 17, 2014 at 11:13
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    $\begingroup$ Have you checked the OEIS references: oeis.org/A001156 $\endgroup$
    – joro
    Nov 17, 2014 at 13:43
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    $\begingroup$ @Mathlover I meant that the coefficient of $q^{9911}$ in $\prod_n (1-q^{n^2})$ is $19304$. $\endgroup$
    – zeb
    Nov 17, 2014 at 20:54
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    $\begingroup$ @Mathlover I just used pari to compute the coefficients. The command I used was something like "polcoeff(prod(k=1, sqrt(10000), 1-x^k^2+O(x^10001)), 9911)" and it ran in under a second. (I also didn't expect the coefficients to get so large, which is why I felt the need to comment on it.) $\endgroup$
    – zeb
    Nov 18, 2014 at 8:26
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    $\begingroup$ @Mathlover: you might also want to read about proofs of Euler's pentagonal theorem and other results of that sort (Gauss, Macdonald) that use Lie algebra cohomology. In these results it is important that exponents correspond to degrees of basis elements of certain infinite-dimensional Lie algebras. (Converting products like that into sums is computing Euler characteristics of the Chevalley-Eilenberg chain complex in two ways). However, the sequence of squares does not seem to correspond to a basis of any sensible (not too Abelian) Lie algebra; this is another reason to not expect good formulas $\endgroup$ Dec 9, 2014 at 10:34

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