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Table T(n,k) read by upward antidiagonals. T(n,k) = Sum_{i=1..n} Product_{j=1..k} (i-1)*k+j.
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%I #35 Jul 24 2020 06:17:17

%S 1,3,2,6,14,6,10,44,126,24,15,100,630,1704,120,21,190,1950,13584,

%T 30360,720,28,322,4680,57264,390720,666000,5040,36,504,9576,173544,

%U 2251200,14032080,17302320,40320,45,744,17556,428568,8626800,110941200,603353520,518958720,362880

%N Table T(n,k) read by upward antidiagonals. T(n,k) = Sum_{i=1..n} Product_{j=1..k} (i-1)*k+j.

%C T(n,k) is the maximum value of Sum_{i=1..n} Product_{j=1..k} r[(i-1)*k+j] among all permutations r of {1..kn}. For the minimum value see A331889.

%H Seiichi Manyama, <a href="/A333446/b333446.txt">Antidiagonals n = 1..140, flattened</a>

%H Chai Wah Wu, <a href="https://arxiv.org/abs/2002.10514">On rearrangement inequalities for multiple sequences</a>, arXiv:2002.10514 [math.CO], 2020.

%F T(n,k) = Sum_{i=1..n} Gamma(ik+1)/Gamma((i-1)k+1).

%e From _Seiichi Manyama_, Jul 23 2020: (Start)

%e T(3,2) = Sum_{i=1..3} Product_{j=1..2} (i-1)*2+j = 1*2 + 3*4 + 5*6 = 44.

%e Square array begins:

%e 1, 2, 6, 24, 120, 720, ...

%e 3, 14, 126, 1704, 30360, 666000, ...

%e 6, 44, 630, 13584, 390720, 14032080, ...

%e 10, 100, 1950, 57264, 2251200, 110941200, ...

%e 15, 190, 4680, 173544, 8626800, 538459200, ...

%e 21, 322, 9576, 428568, 25727520, 1940869440, ... (End)

%o (Python)

%o def T(n,k): # T(n,k) for A333446

%o c, l = 0, list(range(1,k*n+1,k))

%o lt = list(l)

%o for i in range(n):

%o for j in range(1,k):

%o lt[i] *= l[i]+j

%o c += lt[i]

%o return c

%Y Column k=1-3 give A000217, A268684, A268685(n-1).

%Y Main diagonal gives A336513.

%Y Cf. A323663, A331889,

%K nonn,tabl

%O 1,2

%A _Chai Wah Wu_, Mar 23 2020