A251938 Number of length 4+2 0..n arrays with the sum of the maximum minus the median of adjacent triples multiplied by some arrangement of +-1 equal to zero

34, 339, 1760, 6293, 17598, 41677, 87328, 166677, 295898, 495745, 791734, 1215373, 1804320, 2602645, 3662110, 5042499, 6811262, 9045701, 11832586, 15267973, 19459520, 24526335, 30597824, 37817343, 46341088, 56337177, 67989404, 81496455

Offset

1,1

Comments

Row 4 of A251935

Links

R. H. Hardin, Table of n, a(n) for n = 1..94

Formula

Empirical: a(n) = a(n-1) +3*a(n-3) -a(n-4) -2*a(n-5) -3*a(n-6) -3*a(n-7) +5*a(n-8) +2*a(n-9) +5*a(n-10) -3*a(n-11) -3*a(n-12) -2*a(n-13) -a(n-14) +3*a(n-15) +a(n-17) -a(n-18)

Empirical for n mod 12 = 0: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (1631/144)*n^2 - (73/20)*n + 1

Empirical for n mod 12 = 1: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (15575/1296)*n^2 - (146743/25920)*n + (3073/576)

Empirical for n mod 12 = 2: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (15127/1296)*n^2 - (7913/1620)*n + (169/54)

Empirical for n mod 12 = 3: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (1631/144)*n^2 - (1423/320)*n + (209/64)

Empirical for n mod 12 = 4: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (15575/1296)*n^2 - (7273/1620)*n + (31/9)

Empirical for n mod 12 = 5: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (15127/1296)*n^2 - (156983/25920)*n + (4355/1728)

Empirical for n mod 12 = 6: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (1631/144)*n^2 - (73/20)*n + (7/2)

Empirical for n mod 12 = 7: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (15575/1296)*n^2 - (137023/25920)*n + (3289/576)

Empirical for n mod 12 = 8: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (15127/1296)*n^2 - (7913/1620)*n + (17/27)

Empirical for n mod 12 = 9: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (1631/144)*n^2 - (1543/320)*n + (185/64)

Empirical for n mod 12 = 10: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (15575/1296)*n^2 - (7273/1620)*n + (107/18)

Empirical for n mod 12 = 11: a(n) = (121223/25920)*n^5 + (5339/5184)*n^4 + (1793/108)*n^3 + (15127/1296)*n^2 - (147263/25920)*n + (5003/1728)

Example

Some solutions for n=6
..2....2....3....5....6....1....1....4....3....5....4....3....1....4....6....6
..4....4....5....1....2....5....2....5....0....2....2....2....5....0....5....6
..5....6....6....0....0....3....2....2....5....1....5....0....6....4....5....2
..3....1....3....5....0....0....2....6....2....4....6....6....1....3....4....3
..4....3....2....4....4....6....6....6....1....0....5....4....5....4....5....6
..0....6....0....3....6....3....2....4....6....0....3....5....4....5....3....6

Crossrefs

Sequence in context: A233061 A281805 A034978 * A059338 A301954 A368719

Adjacent sequences: A251935 A251936 A251937 * A251939 A251940 A251941

Keyword

nonn

Author

R. H. Hardin, Dec 11 2014

Status

approved