A331889 Table T(n,k) read by upward antidiagonals. T(n,k) is the minimum value of Sum_{i=1..n} Product_{j=1..k} r[(i-1)*k+j] among all permutations r of {1..kn}.

1, 3, 2, 6, 10, 6, 10, 28, 54, 24, 15, 60, 214, 402, 120, 21, 110, 594, 2348, 3810, 720, 28, 182, 1334, 8556, 32808, 43776, 5040, 36, 280, 2614

Offset: 1

Author

Chai Wah Wu, Mar 20 2020

Keywords

• nonn
• more
• tabl

Comments

k 1 2 3 4 5 6 7 8 9 10 11 12

---------------------------------------------------------------------------------

n 1| 1 2 6 24 120 720 5040 40320 362880 3628800 39916800 479001600

2| 3 10 54 402 3810 43776

3| 6 28 214 2348 32808

4| 10 60 594 8556

5| 15 110 1334

6| 21 182 2614

7| 28 280

8| 36 408

9| 45 570

10| 55 770

Links

• Chai Wah Wu, On rearrangement inequalities for multiple sequences, arXiv:2002.10514 [math.CO], 2020-2022.

Crossrefs

Cf. A000142, A000217, A006331, A072368.

Programs

Python

from itertools import combinations, permutations
from sympy import factorial
def T(n,k): # T(n,k) for A331889
    if k == 1:
        return n*(n+1)//2
    if n == 1:
        return int(factorial(k))
    if k == 2:
        return n*(n+1)*(2*n+1)//3
    nk = n*k
    nktuple = tuple(range(1,nk+1))
    nkset = set(nktuple)
    count = int(factorial(nk))
    for firsttuple in combinations(nktuple,n):
        nexttupleset = nkset-set(firsttuple)
        for s in permutations(sorted(nexttupleset),nk-2*n):
            llist = sorted(nexttupleset-set(s),reverse=True)
            t = list(firsttuple)
            for i in range(0,k-2):
                itn = i*n
                for j in range(n):
                        t[j] *= s[itn+j]
            t.sort()
            v = 0
            for i in range(n):
                v += llist[i]*t[i]
            if v < count:
                count = v
    return count

Formula

T(n,k) >= ceiling(n*((kn)!)^(1/n)).

T(n,1) = n*(n+1)/2 = A000217(n).

T(1,k) = k! = A000142(k).

T(n,3) = A072368(n).

T(n,2) = n*(n+1)*(2*n+1)/3 = A006331(n).