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How to find the end of a series representation of the product $$ \prod_{\substack{i=1...\infty\\\ j=0...i\\\ k=0...j}}\frac{1}{1-x^{i-j}y^{j-k}z^{k}}? $$

For example for product $$ \prod_{\substack{i=1...\infty\\\ j=0...i}}\frac{1}{1-x^{i-j}y^j} $$ the ends of series is $$ ...+7x^5 + 12x^4y + 16x^3y^2 + 16x^2y^3 + 12xy^4 + 7y^5 + 5x^4 +\\\ +7x^3y + 9x^2y^2 + 7xy^3 + 5y^4 + 3x^3 + 4x^2y + 4xy^2+\\\ + 3y^3 + 2x^2 + 2xy + 2y^2 + x + y + 1 $$

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I think there's an error in your indexing. Consider the example product. Because the power of y always starts with j=1, there should be no pure x power in the series. (I posted this by accident as an answer. I'm reposting it here in case someone with the appropriate MO clout can delete its appearance as an answer. I couldn't figure out how to delete the answer myself.) – Barry Cipra Jun 28 '12 at 14:19
What technique to find the end of a series representation of the product from example? – Alexander Jun 28 '12 at 14:22
@Alexander, I don't understand your question. My point is, if you let $y=0$, then every term in your product is 1, so its series cannot end $\cdots + 3x^3 + 2x^2 + x + 1$. – Barry Cipra Jun 28 '12 at 14:37
If y=0 this product $$ \prod_{n=1}^\infty\frac{1}{1-x^n} $$ will be generating function of number of partitions of $n$ $$ 1+x+2x^2+3x^3+5x^4+7x^5... $$ ( – Alexander Jun 28 '12 at 16:33
@Alexander: that would be for $\prod_{i=1}^\infty \prod_{j={\bf 0}}^i \dfrac{1}{1-x^{i-j} y^j}$, not $\prod_{i=1}^\infty \prod_{j={\bf 1}}^i \dfrac{1}{1-x^{i-j} y^j}$ – Robert Israel Jun 28 '12 at 16:41
up vote 1 down vote accepted

Your example should be $$ \prod_{i=1}^\infty \prod_{j=1}^i \dfrac{1}{1-x^{i-j} y^j} = 1+y+2{y}^{2}+xy+3{y}^{3}+2x{y}^{2}+{x}^{2}y+{x}^{3}y+3{x}^{2}{ y}^{2}+4x{y}^{3}+5{y}^{4} + \ldots$$ The triple product is $$ \eqalign{\prod _{i=1}^{\infty } & \prod _{j=1}^{i} \prod _{k=1}^{j} \dfrac{1}{ 1-{x}^{i-j}{y}^{j-k}{z}^{k}} \cr & =1+ z+2{z}^{2}+xz+yz+3{z}^{3}+2x{z}^{2}+2y{z}^{2}+{x}^{2}z+xyz+{y} ^{2}z\cr&+{x}^{3}z+3{x}^{2}{z}^{2}+{x}^{2}yz+4x{z}^{3}+x{y}^{2}z+3xy {z}^{2}+{y}^{3}z+3{y}^{2}{z}^{2}+4y{z}^{3}+5{z}^{4}+\ldots\cr} $$ EDIT: If you meant the $j$ and $k$ indexing to start from $0$ instead of $1$, $$ \eqalign{\prod _{i=1}^{\infty } & \prod _{j=0}^{i} \prod _{k=0}^{j} \dfrac{1}{ 1-{x}^{i-j}{y}^{j-k}{z}^{k}} \cr & = 1+x+y+z+2\,{x}^{2}+2xy+2xz+2{y}^{2}+2yz+2{z}^{2}+3{x}^{3}+ 4{x}^{2}y+4{x}^{2}z\cr&+4x{y}^{2}+5xyz+4x{z}^{2}+3{y}^{3}+4{ y}^{2}z+4y{z}^{2}+3{z}^{3}+5{x}^{4}+7{x}^{3}y+7{x}^{3}z+9{ x}^{2}{y}^{2}\cr&+11{x}^{2}yz+9{x}^{2}{z}^{2}+7x{y}^{3}+11x{y}^{2} z+11xy{z}^{2}+7x{z}^{3}+5{y}^{4}+7{y}^{3}z\cr&+9{y}^{2}{z}^{2}+7 y{z}^{3}+5{z}^{4}+\ldots\cr}$$

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Thank you. How to get this? Or where I can see more members of the series? – Alexander Jun 28 '12 at 16:47
First of all, if you only want terms up to total degree $d$ you only need to take $i=1..d$. I used Maple's mtaylor command. – Robert Israel Jun 28 '12 at 16:54
Can you write youre Maple's commands? – Alexander Jun 28 '12 at 16:58
P4:= product(product(product(1/(1-x^(i-j)* y^(j-k)*z^k),k=0..j),j=0..i),i=1..4); mtaylor(P4,[x,y,z],5); – Robert Israel Jun 28 '12 at 17:41
Thank you, I already figured out. – Alexander Jun 28 '12 at 17:54

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