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Find an asymptotically tight estimate for the sum $$ A_n^{k}(\lambda)= \sum_{ \substack{a_i\geq \lambda_i \\ a_1+a_2+\dots a_k=n }} \prod_{i=1}^k a_i! $$

Is the leading term going to be $$|\textrm{Number of Maximal Lambda}|(j-\lambda_1-\lambda_2-\lambda_3- \lambda_4+ \lambda_{max})!\frac{1}{\lambda_{\max}!}\prod_{i=1}^4 \lambda_i! $$

Edit: As of right now there is some discrepency as to weather this conjecture is correct or not.

This question has been asked before in the binomial situation HereHere

Find an asymptotically tight estimate for the sum $$ A_n^{k}(\lambda)= \sum_{ \substack{a_i\geq \lambda_i \\ a_1+a_2+\dots a_k=n }} \prod_{i=1}^k a_i! $$

Is the leading term going to be $$|\textrm{Number of Maximal Lambda}|(j-\lambda_1-\lambda_2-\lambda_3- \lambda_4+ \lambda_{max})!\frac{1}{\lambda_{\max}!}\prod_{i=1}^4 \lambda_i! $$

Edit: As of right now there is some discrepency as to weather this conjecture is correct or not.

This question has been asked before in the binomial situation Here

Find an asymptotically tight estimate for the sum $$ A_n^{k}(\lambda)= \sum_{ \substack{a_i\geq \lambda_i \\ a_1+a_2+\dots a_k=n }} \prod_{i=1}^k a_i! $$

Is the leading term going to be $$|\textrm{Number of Maximal Lambda}|(j-\lambda_1-\lambda_2-\lambda_3- \lambda_4+ \lambda_{max})!\frac{1}{\lambda_{\max}!}\prod_{i=1}^4 \lambda_i! $$

Edit: As of right now there is some discrepency as to weather this conjecture is correct or not.

This question has been asked before in the binomial situation Here

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Daniel Parry
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Find an asymptotically tight estimate for the sum $$ A_b^{k}(\lambda)= \sum_{ \substack{a_i\geq \lambda_i \\ a_1+a_2+\dots a_k=n }} \prod_{i=1}^k a_i! $$$$ A_n^{k}(\lambda)= \sum_{ \substack{a_i\geq \lambda_i \\ a_1+a_2+\dots a_k=n }} \prod_{i=1}^k a_i! $$

Is the leading term going to be $$|\textrm{Number of Maximal Lambda}|(j-\lambda_1-\lambda_2-\lambda_3- \lambda_4+ \lambda_{max})!\frac{1}{\lambda_{\max}!}\prod_{i=1}^4 \lambda_i! $$

Edit: As of right now there is some discrepency as to weather this conjecture is correct or not.

This question has been asked before in the binomial situation Here

Find an asymptotically tight estimate for the sum $$ A_b^{k}(\lambda)= \sum_{ \substack{a_i\geq \lambda_i \\ a_1+a_2+\dots a_k=n }} \prod_{i=1}^k a_i! $$

Is the leading term going to be $$|\textrm{Number of Maximal Lambda}|(j-\lambda_1-\lambda_2-\lambda_3- \lambda_4+ \lambda_{max})!\frac{1}{\lambda_{\max}!}\prod_{i=1}^4 \lambda_i! $$

Edit: As of right now there is some discrepency as to weather this conjecture is correct or not.

This question has been asked before in the binomial situation Here

Find an asymptotically tight estimate for the sum $$ A_n^{k}(\lambda)= \sum_{ \substack{a_i\geq \lambda_i \\ a_1+a_2+\dots a_k=n }} \prod_{i=1}^k a_i! $$

Is the leading term going to be $$|\textrm{Number of Maximal Lambda}|(j-\lambda_1-\lambda_2-\lambda_3- \lambda_4+ \lambda_{max})!\frac{1}{\lambda_{\max}!}\prod_{i=1}^4 \lambda_i! $$

Edit: As of right now there is some discrepency as to weather this conjecture is correct or not.

This question has been asked before in the binomial situation Here

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Daniel Parry
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Find an asymptotically tight estimate for the sum $$ A_b^{k}(\lambda)= \sum_{ \substack{a_i\geq \lambda_i \\ a_1+a_2+\dots a_k=n }} \prod_{i=1}^k a_i $$$$ A_b^{k}(\lambda)= \sum_{ \substack{a_i\geq \lambda_i \\ a_1+a_2+\dots a_k=n }} \prod_{i=1}^k a_i! $$

Is the leading term going to be $$|\textrm{Number of Maximal Lambda}|(j-\lambda_1-\lambda_2-\lambda_3- \lambda_4+ \lambda_{max})!\frac{1}{\lambda_{\max}!}\prod_{i=1}^4 \lambda_i! $$

Edit: As of right now there is some discrepency as to weather this conjecture is correct or not.

This question has been asked before in the binomial situation Here

Find an asymptotically tight estimate for the sum $$ A_b^{k}(\lambda)= \sum_{ \substack{a_i\geq \lambda_i \\ a_1+a_2+\dots a_k=n }} \prod_{i=1}^k a_i $$

Is the leading term going to be $$|\textrm{Number of Maximal Lambda}|(j-\lambda_1-\lambda_2-\lambda_3- \lambda_4+ \lambda_{max})!\frac{1}{\lambda_{\max}!}\prod_{i=1}^4 \lambda_i! $$

Edit: As of right now there is some discrepency as to weather this conjecture is correct or not.

This question has been asked before in the binomial situation Here

Find an asymptotically tight estimate for the sum $$ A_b^{k}(\lambda)= \sum_{ \substack{a_i\geq \lambda_i \\ a_1+a_2+\dots a_k=n }} \prod_{i=1}^k a_i! $$

Is the leading term going to be $$|\textrm{Number of Maximal Lambda}|(j-\lambda_1-\lambda_2-\lambda_3- \lambda_4+ \lambda_{max})!\frac{1}{\lambda_{\max}!}\prod_{i=1}^4 \lambda_i! $$

Edit: As of right now there is some discrepency as to weather this conjecture is correct or not.

This question has been asked before in the binomial situation Here

edited body
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Daniel Parry
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