Prove or disprove:

$$m(n,k,s)=\sum_{a_1=1}^n \sum_{a_2=1}^n \cdots \sum_{a_k=1}^n \min(a_1, a_2,\cdots, a_k)^s =$$

$$ \sum _{i=0}^{k-1} \frac{(-1)^i}{i!} F(n,i+s) \sum _{j=0}^{k-1} \frac{\partial ^i \left(n^j \binom{k}{j}\right)}{\partial n^i}$$

Where $$F(n,s)=\sum_{t=1}^{n}{t^s}$$ is Faulhaber's formula

  • 3
    $\begingroup$ Some context please? $\endgroup$ – Per Alexandersson Jun 20 '15 at 22:02
  • $\begingroup$ This is a generalization from matrices: $$f(n,k)=\sum _{i=1}^n {\sum_{j=1}^n {\min(i,j)^k}} = \sum_{s=1}^{n}{(2n-2s-1)\cdot s^k} = (2n+1)\cdot F(n,k)-2F(n,k+1)$$ Sums related to min matrix entrywise norms $\endgroup$ – Enrique Pérez Herrero Jun 20 '15 at 22:06
  • $\begingroup$ $$\sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n \min(i,j,k) = \sum_{i=1}^n \sum_{j=1}^n \max(i\cdot j,j) = \frac{1}{4} n^2 (1+n)^2 $$ A000537(n) in OEIS related to the Jordan functions $\endgroup$ – Enrique Pérez Herrero Jun 20 '15 at 22:10
  • $\begingroup$ Obviously $$m(n,k,s) = \sum_{t=1}^n t^s((n-t+1)^k-(n-t)^k).$$ What is the point of the more complicated answer? $\endgroup$ – Brendan McKay Jun 21 '15 at 9:27
  • $\begingroup$ Another formula is $$m(n,k,s) = \sum_{t=1}^n t^k((n-t+1)^s-(n-t)^s),$$ which shows a pleasing symmetry $m(n,k,s)=m(n,s,k)$. $\endgroup$ – Brendan McKay Jun 21 '15 at 9:41

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