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Let $k_1, k_2, k_3$ be natural non-negative numbers such that $k_1+k_2+k_3=K$. Let $n_1, n_2, n_3 \in \{0, \ldots, N\}$ and such that $n_1+n_2+n_3=N$.

Calculate

$$ S=\sum_{(k_1, k_2, k_3): k_1+k_2+k_3=K, \,\, n_1+n_2+n_3=N}k_1^{n_1}\times k_2^{n_2} \times k_3^{n_3} $$

My attempt: I am thinking on representing this sum as a chain of sums over each summand $k_j$. For example, the interior sum would be: $ \sum_{k_3=0}^{K-k_1-k_2}k_3^{n_3}. $ Using Sums of p-th powers formula we can get $$\sum_{k_3=0}^{K-k_1-k_2}k_3^{n_3}=\frac{B_{n_3+1}(K-k_1-k_2+1)-B_{n_3+1}}{n_3}.$$ So, the sum $S$ would be represented as a product of these ratios with Bernoulli numbers $B_n$.

Is there a better way on computing/estimating from above sum $S$?

user4164
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    I may not be following the notation in the first expression. What are the variables that change within the summation? For example in $\sum_j (\sum_i {k_j}^{n_i})$ the inner sum goes from $i=$ initial value to final value in $i$, while the second has analogous limits for $j$. However, I am unable to read the expression in the statement. – Basco Jul 19 '20 at 23:09
  • By Bell numbers do you really mean Bernoulli?? – Phicar Jul 20 '20 at 01:33
  • The first formula under "My attempt" doesn't make any sense to me. Have you tried calculating it for some small values of $K$ and $N$? If you can calculate it at all, does it give you the right answers? – Gerry Myerson Jul 20 '20 at 03:11
  • @Phicar: yes, I meant Bernoulli numbers. Thank you for catching it. – user4164 Jul 20 '20 at 04:05
  • I think you should try out if your approach works for a simpler version of you r problem, e.g. k1+k2=5, n1+n2=5 – miracle173 Jul 20 '20 at 10:13
  • I expect there's a multivariate generalization of the Stirling number version of Faulhaber's formula that would handle your expression. I rediscovered the univariate version yet again when making a homework problem on finite differences. See part (f) for the formula; the curly braces indicate Stirling numbers of the second kind. The rest of the problem is essentially an outline of a proof. There are other approaches, but I didn't see anybody else write it this (very simple...) way in a brief search. – Joshua P. Swanson Jul 22 '20 at 08:06
  • @Joshua P.Swanson: yes, the Faulhaber's formula is connected to Bernoulli polynomials. The whole problem is to generalize for sum $S$. – user4164 Jul 22 '20 at 15:23
  • @user4164: well, that's not what your question asks, which is for "a better way on computing/estimating from above sum". I'm saying in the univariate case there are other families of formulas than Faulhaber type, in particular using the falling factorial basis instead of the monomial basis. If you really want to use the monomial basis, you should ask for that specifically. I am not sufficiently invested to churn through the falling factorial calculations myself. – Joshua P. Swanson Jul 22 '20 at 20:03

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