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A333446
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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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2
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1, 3, 2, 6, 14, 6, 10, 44, 126, 24, 15, 100, 630, 1704, 120, 21, 190, 1950, 13584, 30360, 720, 28, 322, 4680, 57264, 390720, 666000, 5040, 36, 504, 9576, 173544, 2251200, 14032080, 17302320, 40320, 45, 744, 17556, 428568, 8626800, 110941200, 603353520, 518958720, 362880
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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Seiichi Manyama, Antidiagonals n = 1..140, flattened
Chai Wah Wu, On rearrangement inequalities for multiple sequences, arXiv:2002.10514 [math.CO], 2020.
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FORMULA
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T(n,k) = Sum_{i=1..n} Gamma(ik+1)/Gamma((i-1)k+1).
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EXAMPLE
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From Seiichi Manyama, Jul 23 2020: (Start)
T(3,2) = Sum_{i=1..3} Product_{j=1..2} (i-1)*2+j = 1*2 + 3*4 + 5*6 = 44.
Square array begins:
1, 2, 6, 24, 120, 720, ...
3, 14, 126, 1704, 30360, 666000, ...
6, 44, 630, 13584, 390720, 14032080, ...
10, 100, 1950, 57264, 2251200, 110941200, ...
15, 190, 4680, 173544, 8626800, 538459200, ...
21, 322, 9576, 428568, 25727520, 1940869440, ... (End)
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PROG
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(Python)
def T(n, k): # T(n, k) for A333446
c, l = 0, list(range(1, k*n+1, k))
lt = list(l)
for i in range(n):
for j in range(1, k):
lt[i] *= l[i]+j
c += lt[i]
return c
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CROSSREFS
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Column k=1-3 give A000217, A268684, A268685(n-1).
Main diagonal gives A336513.
Cf. A323663, A331889,
Sequence in context: A078091 A073883 A248982 * A289069 A074718 A285457
Adjacent sequences: A333443 A333444 A333445 * A333447 A333448 A333449
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KEYWORD
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nonn,tabl
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AUTHOR
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Chai Wah Wu, Mar 23 2020
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STATUS
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approved
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