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Number of -n..n arrays x(0..3) of 4 elements with zero sum and no two consecutive zero elements.
Row 4 of A199530.
Empirical: a(n) = (16/3)*n^3 + 8*n^2 - (4/3)*n.
Conjectures from Colin Barker, May 15 2018: (Start)
G.f.: 4*x*(3 + 6*x - x^2) / (1 - x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>4.
(End)
Some solutions for n=5:
Cf. A199530.
R. H. Hardin , Nov 07 2011
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_R. H. Hardin (rhhardin(AT)att.net) _ Nov 07 2011
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R. H. Hardin, <a href="/A199531/b199531.txt">Table of n, a(n) for n = 1..200</a>
allocated for Ron HardinNumber of -n..n arrays x(0..3) of 4 elements with zero sum and no two consecutive zero elements
12, 72, 212, 464, 860, 1432, 2212, 3232, 4524, 6120, 8052, 10352, 13052, 16184, 19780, 23872, 28492, 33672, 39444, 45840, 52892, 60632, 69092, 78304, 88300, 99112, 110772, 123312, 136764, 151160, 166532, 182912, 200332, 218824, 238420, 259152
1,1
Row 4 of A199530
Empirical: a(n) = (16/3)*n^3 + 8*n^2 - (4/3)*n
Some solutions for n=5
.-4...-4....4....0...-5....5....2....0...-1...-3....2....1....4...-1....4...-1
..4....5...-4....2....3...-2...-2...-2....0...-2...-2...-3...-1....2...-2....5
..3....1...-2...-5...-3...-1....1....1...-2....2....3....1....0....4....1....1
.-3...-2....2....3....5...-2...-1....1....3....3...-3....1...-3...-5...-3...-5
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nonn
R. H. Hardin (rhhardin(AT)att.net) Nov 07 2011
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allocated for Ron Hardin
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