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A199530
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T(n,k)=Number of -k..k arrays x(0..n-1) of n elements with zero sum and no two consecutive zero elements
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11
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1, 1, 2, 1, 4, 6, 1, 6, 18, 12, 1, 8, 36, 72, 32, 1, 10, 60, 212, 320, 80, 1, 12, 90, 464, 1324, 1414, 200, 1, 14, 126, 860, 3734, 8342, 6346, 520, 1, 16, 168, 1432, 8470, 30484, 53302, 28766, 1336, 1, 18, 216, 2212, 16682, 84852, 252154, 343710, 131246, 3472, 1, 20, 270
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OFFSET
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1,3
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COMMENTS
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Table starts
....1......1........1.........1.........1..........1...........1...........1
....2......4........6.........8........10.........12..........14..........16
....6.....18.......36........60........90........126.........168.........216
...12.....72......212.......464.......860.......1432........2212........3232
...32....320.....1324......3734......8470......16682.......29750.......49284
...80...1414.....8342.....30484.....84852.....197962......407946......766664
..200...6346....53302....252154....860854....2378412.....5662636....12071420
..520..28766...343710...2105064...8815392...28844590....79345982...191873280
.1336.131246..2232322..17701326..90927530..352355640..1119873360..3071898666
.3472.602390.14582218.149708146.943302430.4329146404.15897133212.49465959068
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LINKS
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FORMULA
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Empirical for rows:
T(1,k) = 1
T(2,k) = 2*k
T(3,k) = 3*k^2 + 3*k
T(4,k) = (16/3)*k^3 + 8*k^2 - (4/3)*k
T(5,k) = (115/12)*k^4 + (115/6)*k^3 + (41/12)*k^2 - (1/6)*k
T(6,k) = (88/5)*k^5 + 44*k^4 + (58/3)*k^3 - 3*k^2 + (31/15)*k
T(7,k) = (5887/180)*k^6 + (5887/60)*k^5 + (620/9)*k^4 + (11/12)*k^3 + (433/180)*k^2 - (91/30)*k
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EXAMPLE
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Some solutions for n=6 k=5
..1...-1....0...-4...-1....4....1...-5....1....2...-1....1....3....3....1....5
..0...-1...-1...-3....0....5...-3....0....1...-3...-3...-4....2...-5....1....2
.-4....5....3...-1...-2...-3....5....1...-1....5....0....2...-3....4...-4...-5
..3...-2...-3....2....2...-4...-1....2...-4....4...-2....2....2...-5....5...-1
..1...-3...-2....4...-4...-4....0....5....2...-5....5....4...-4....0...-2....1
.-1....2....3....2....5....2...-2...-3....1...-3....1...-5....0....3...-1...-2
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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