A333420 Table T(n,k) read by upward antidiagonals. T(n,k) is the maximum value of Product_{i=1..n} Sum_{j=1..k} r[(i-1)*k+j] among all permutations r of {1..kn}.

1, 2, 3, 6, 25, 6, 24, 343, 110, 10, 120, 6561, 3375, 324, 15, 720, 161051, 144400, 17576, 756, 21, 5040, 4826809, 7962624, 1336336, 64000, 1521, 28, 40320, 170859375, 535387328, 130691232, 7595536, 185193, 2756, 36, 3628800, 6975757441

Offset

1,2

Comments

A dual sequence to A331889.

k 1 2 3 4 5 6 7 8 9

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

n 1| 1 3 6 10 15 21 28 36 45

2| 2 25 110 324 756 1521 2756 4624 7310

3| 6 343 3375 17576 64000 185193 456533 1000000 2000376

4| 24 6561 144400 1336336 7595536 31640625 106131204

5| 120 161051 7962624 130691232

6| 720 4826809 535387328

7| 5040 170859375

8| 40320 6975757441

9| 3628800

10| 39916800

Links

Table of n, a(n) for n=1..38.

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

Formula

T(n,k) <= floor((k*(k*n+1)/2)^n) with equality if k = 2*t+n*u for nonnegative integers t and u.

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

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

T(n,2) = (2*n+1)^n = A085527(n).

If n is even, k is odd and k >= n-1, then T(n,k) = ((k^2*(k*n+1)^2-1)/4)^(n/2).

Prog

(Python)
from itertools import combinations, permutations
from sympy import factorial
def T(n, k): # T(n, k) for A333420
    if k == 1:
        return int(factorial(n))
    if n == 1:
        return k*(k+1)//2
    if k % 2 == 0 or (k >= n-1 and n % 2 == 1):
        return (k*(k*n+1)//2)**n
    if k >= n-1 and n % 2 == 0 and k % 2 == 1:
        return ((k**2*(k*n+1)**2-1)//4)**(n//2)
    nk = n*k
    nktuple = tuple(range(1, nk+1))
    nkset = set(nktuple)
    count = 0
    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()
            w = 1
            for i in range(n):
                w *= llist[i]+t[i]
            if w > count:
                count = w
    return count

Crossrefs

Cf. A000142, A000217, A085527, A331889.

Sequence in context: A099000 A032540 A063728 * A296259 A344935 A000341

Adjacent sequences: A333417 A333418 A333419 * A333421 A333422 A333423

Keyword

nonn,more,tabl

Author

Chai Wah Wu, Mar 23 2020

Status

approved