A331889 - OEIS

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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

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

1,2

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

Table of n, a(n) for n=1..31.

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

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).

PROG

(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

CROSSREFS

Cf. A000142, A000217, A006331, A072368.

Sequence in context: A072765 A210756 A210748 * A369247 A367544 A340612

Adjacent sequences: A331886 A331887 A331888 * A331890 A331891 A331892

KEYWORD

nonn,more,tabl

AUTHOR

Chai Wah Wu, Mar 20 2020

STATUS

approved