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I just answered this on mathstackexchange, but I duplicate it here (later edit: But include here a proof for the more general formula): You haven't told the reader what the formula is. I believe I know the formula you mean, which is $$\left( \frac{|G|^{n-1}}{\prod_{j=1}^{n} |C_{G}(\sigma_{j})|} \right) \sum_{i=1}^{k} \frac{\prod_{j = 1}^{n} \chi_{i}(\sigma_{j})}{\chi_{i}(1)^{n-2}},$$ where the $\chi_{i}$ are all the complex irreducible characters of $G$. This formula was probably known to Burnside- I know no special name for it- it is a special case of a general formula for the product of class sums in the group algebra. The formula can be derived by writing the class sums as linear combinations of the primitive idempotents of $Z(\mathbb{C}G).$ Such linear combinations are easy to multiply since these idempotents are mutually orthogonal. Then one recovers the coefficient of a particular element $g$ in the product by using the fact that $g$ occurs with coefficient $\frac{\chi(1)\overline{\chi(g)}}{|G|}$ in the primitive central idempotent corresponding to the irreducible character $\chi$. The formula for the product of two class sums appears explicitly in Burnside's book, but some of the exercises in the book make it clear that he was aware of the general formula. As far as I know, there is no obvious connection with Burnside's orbit counting formula.

Later edit in response: I outlined above how to obtain the general formula, which eally indicates the proof: Let $e_{i}$ be the pirmitive central idempotent corresponding to the irreducible character $\chi_{i}.$ Then $e_{i} = \frac{\chi_{i}(1)}{|G|} \sum_{g \in G}\chi_{i}(g^{-1})g.$ Let $x_{j}$ be a representative for the $j$-the conjugacy class $C_{j}.$ Let $C_{j}^{+}$ denote the class sum of $C_{j}$ in the center of the group algebra. Then evaluating the central characters on both sides shows that $C_{j}^{+} = \sum_{i=1}^{k} \frac{[G:C_{G}(x_{j})]\chi_{i}(x_{j})}{\chi_{i}(1)} e_{i}.$

Now if we take any $n$ class sums, which we may as well label $C_{1}^{+},\ldots, C_{n}^{+},$ we multiply them using the orthogonality of the central primitive idempotents, and then recover the coefficient of an element $x_{n+1}$ in the product, to see that the coefficient of $x_{n+1}$ in $\prod_{j=1}C_{j}^{+}$ is $$\left( \frac{|G|^{n-1}}{\prod_{j=1}^{n} |C_{G}(x_{j})|} \right) \sum_{i=1}^{k} \frac{\chi_{i}(x_{n+1}^{-1})\prod_{j = 1}^{n} \chi_{i}(x_{j})}{\chi_{i}(1)^{n-1}}.$$

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I just answered this on mathstackexchange, but I duplicate it here: You haven't told the reader what the formula is. I believe I know the formula you mean, which is $$\left( \frac{|G|^{n-1}}{\prod_{j=1}^{n} |C_{G}(\sigma_{j})|} \right) \sum_{i=1}^{k} \frac{\prod_{j = 1}^{n} \chi_{i}(\sigma_{j})}{\chi_{i}(1)^{n-2}},$$ where the $\chi_{i}$ are all the complex irreducible characters of $G$. This formula was probably known to Burnside- I know no special name for it- it is a special case of a general formula for the product of class sums in the group algebra. The formula can be derived by writing the classY class sums as linear combinations of the primitive idempotents of $Z(\mathbb{C}G).$ Such linear combinations are easy to multiply since these idempotents are mutually orthogonal. Then one recovers the coefficient of a particular element $g$ in the product by using the fact that $g$ occurs with coefficient $\frac{\chi(1)\overline{\chi(g)}}{|G|}$ in the primitive central idempotent corresponding to the irreducible character $\chi$. The formula for the product of two class sums appears explicitly in Burnside's book, but some of the exercises in the book make it clear that he was aware of the general formula. As far as I know, there is no obvious connection with Burnside's orbit counting formula.

Later edit in response: I outlined above how to obtain the general formula, which eally indicates the proof: Let $e_{i}$ be the pirmitive central idempotent corresponding to the irreducible character $\chi_{i}.$ Then $e_{i} = \frac{\chi_{i}(1)}{|G|} \sum_{g \in G}\chi_{i}(g^{-1})g.$ Let $x_{j}$ be a representative for the $j$-the conjugacy class $C_{j}.$ Let $C_{j}^{+}$ denote the class sum of $C_{j}$ in the center of the group algebra. Then evaluating the central characters on both sides shows that $C_{j}^{+} = \sum_{i=1}^{k} \frac{[G:C_{G}(x_{j})]\chi_{i}(x_{j})}{\chi_{i}(1)} e_{i}.$

Now if we take any $n$ class sums, which we may as well label $C_{1}^{+},\ldots, C_{n}^{+},$ we multiply them using the orthogonality of the central primitive idempotents, and then recover the coefficient of an element $x_{n+1}$ in the product, to see that the coefficient of $x_{n+1}$ in $\prod_{j=1}C_{j}^{+}$ is $$\left( \frac{|G|^{n-1}}{\prod_{j=1}^{n} |C_{G}(x_{j})|} \right) \sum_{i=1}^{k} \frac{\chi_{i}(x_{n+1}^{-1})\prod_{j = 1}^{n} \chi_{i}(x_{j})}{\chi_{i}(1)^{n-1}}.$$

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I just answered this on mathstackexchange, but I duplicate it here: You haven't told the reader what the formula is. I believe I know the formula you mean, which is $$\left( \frac{|G|^{n-1}}{\prod_{j=1}^{n} |C_{G}(\sigma_{j})|} \right) \sum_{i=1}^{k} \frac{\prod_{j = 1}^{n} \chi_{i}(\sigma_{j})}{\chi_{i}(1)^{n-2}},$$ where the $\chi_{i}$ are all the complex irreducible characters of $G$. This formula was probably known to Burnside- I know no special name for it- it is a special case of a general formula for the product of class sums in the group algebra. The formula can be derived by writing the classY sums as linear combinations of the primitive idempotents of $Z(\mathbb{C}G).$ Such linear combinations are easy to multiply since these idempotents are mutually orthogonal. Then one recovers the coefficient of a particular element $g$ in the product by using the fact that $g$ occurs with coefficient $\frac{\chi(1)\overline{\chi(g)}}{|G|}$ in the primitive central idempotent corresponding to the irreducible character $\chi$. The formula for the product of two class sums appears explicitly in Burnside's book, but some of the exercises in the book make it clear that he was aware of the general formula. As far as I know, there is no obvious connection with Burnside's orbit counting formula.

Later edit in response: I outlined above how to obtain the general formula, which eally indicates the proof: Let $e_{i}$ be the pirmitive central idempotent corresponding to the irreducible character $\chi_{i}.$ Then $e_{i} = \frac{\chi_{i}(1)}{|G|} \sum_{g \in G}\chi_{i}(g^{-1})g.$ Let $x_{j}$ be a representative for the $j$-the conjugacy class $C_{j}.$ Let $C_{j}^{+}$ denote the class sum of $C_{j}$ in the center of the group algebra. Then evaluating the central characters on both sides shows that $C_{j}^{+} = \sum_{i=1}^{k} \frac{[G:C_{G}(x_{j})]\chi_{i}(x_{j})}{\chi_{i}(1)} e_{i}.$

Now if we take any $n$ class sums, which we may as well label $C_{1}^{+},\ldots, C_{n}^{+},$ we multiply them using the orthogonality of the central primitive idempotents, and then recover the coefficient of an element $x_{n+1}$ in the product, to see that the coefficient of $x_{n+1}$ in $\prod_{j=1}C_{j}^{+}$ is $$\left( \frac{|G|^{n-1}}{\prod_{j=1}^{n} |C_{G}(x_{j})|} \right) \sum_{i=1}^{k} \frac{\chi_{i}(x_{n+1}^{-1})\prod_{j = 1}^{n} \chi_{i}(x_{j})}{\chi_{i}(1)^{n-1}}.$$

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