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A333446 Table T(n,k) read by upward antidiagonals. T(n,k) = Sum_{i=1..n} Product_{j=1..k} (i-1)*k+j.

%I A333446 #35 Jul 24 2020 06:17:17
%S A333446 1,3,2,6,14,6,10,44,126,24,15,100,630,1704,120,21,190,1950,13584,
%T A333446 30360,720,28,322,4680,57264,390720,666000,5040,36,504,9576,173544,
%U A333446 2251200,14032080,17302320,40320,45,744,17556,428568,8626800,110941200,603353520,518958720,362880
%N A333446 Table T(n,k) read by upward antidiagonals. T(n,k) = Sum_{i=1..n} Product_{j=1..k} (i-1)*k+j.
%C A333446 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 A333446 Seiichi Manyama, <a href="/A333446/b333446.txt">Antidiagonals n = 1..140, flattened</a>
%H A333446 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 A333446 T(n,k) = Sum_{i=1..n} Gamma(ik+1)/Gamma((i-1)k+1).
%e A333446 From _Seiichi Manyama_, Jul 23 2020: (Start)
%e A333446 T(3,2) = Sum_{i=1..3} Product_{j=1..2} (i-1)*2+j = 1*2 + 3*4 + 5*6 = 44.
%e A333446 Square array begins:
%e A333446    1,   2,    6,     24,      120,        720, ...
%e A333446    3,  14,  126,   1704,    30360,     666000, ...
%e A333446    6,  44,  630,  13584,   390720,   14032080, ...
%e A333446   10, 100, 1950,  57264,  2251200,  110941200, ...
%e A333446   15, 190, 4680, 173544,  8626800,  538459200, ...
%e A333446   21, 322, 9576, 428568, 25727520, 1940869440, ... (End)
%o A333446 (Python)
%o A333446 def T(n,k): # T(n,k) for A333446
%o A333446     c, l = 0, list(range(1,k*n+1,k))
%o A333446     lt = list(l)
%o A333446     for i in range(n):
%o A333446         for j in range(1,k):
%o A333446             lt[i] *= l[i]+j
%o A333446         c += lt[i]
%o A333446     return c
%Y A333446 Column k=1-3 give A000217, A268684, A268685(n-1).
%Y A333446 Main diagonal gives A336513.
%Y A333446 Cf. A323663, A331889,
%K A333446 nonn,tabl
%O A333446 1,2
%A A333446 _Chai Wah Wu_, Mar 23 2020