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A251937
Number of length 3+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.
1
18, 115, 431, 1191, 2695, 5340, 9615, 16098, 25474, 38538, 56176, 79403, 109350, 147253, 194487, 252561, 323091, 407854, 508767, 627868, 767366, 929628, 1117144, 1332597, 1578834, 1858839, 2175799, 2533083, 2934199, 3382880, 3883047, 4438774
OFFSET
1,1
LINKS
FORMULA
Empirical: a(n) = 3*a(n-1) - 2*a(n-2) - 3*a(n-4) + 3*a(n-5) + 3*a(n-6) - 3*a(n-7) - 2*a(n-9) + 3*a(n-10) - a(n-11).
Empirical for n mod 6 = 0: a(n) = (1/30)*n^5 + (653/216)*n^4 + (239/54)*n^3 + 5*n^2 + (43/10)*n + 1
Empirical for n mod 6 = 1: a(n) = (1/30)*n^5 + (653/216)*n^4 + (239/54)*n^3 + 5*n^2 + (1241/270)*n + (199/216)
Empirical for n mod 6 = 2: a(n) = (1/30)*n^5 + (653/216)*n^4 + (239/54)*n^3 + 5*n^2 + (1201/270)*n + (34/27)
Empirical for n mod 6 = 3: a(n) = (1/30)*n^5 + (653/216)*n^4 + (239/54)*n^3 + 5*n^2 + (43/10)*n + (5/8)
Empirical for n mod 6 = 4: a(n) = (1/30)*n^5 + (653/216)*n^4 + (239/54)*n^3 + 5*n^2 + (1241/270)*n + (35/27)
Empirical for n mod 6 = 5: a(n) = (1/30)*n^5 + (653/216)*n^4 + (239/54)*n^3 + 5*n^2 + (1201/270)*n + (191/216).
Empirical g.f.: x*(18 + 61*x + 122*x^2 + 128*x^3 + 38*x^4 - 72*x^5 - 121*x^6 - 78*x^7 - 26*x^8 + 3*x^9 - x^10) / ((1 - x)^6*(1 + x)*(1 + x + x^2)^2). - Colin Barker, Dec 01 2018
EXAMPLE
Some solutions for n=6:
..2....1....2....5....6....5....1....1....4....5....3....0....4....3....2....3
..4....0....1....4....2....5....2....6....3....2....5....0....4....1....0....6
..6....3....4....2....5....1....0....2....6....2....3....1....1....5....1....2
..0....4....3....3....3....3....1....6....5....0....0....3....2....4....3....4
..6....1....3....3....4....0....3....1....3....5....3....4....4....0....2....5
CROSSREFS
Row 3 of A251935.
Sequence in context: A244866 A125328 A126486 * A061803 A207103 A101089
KEYWORD
nonn
AUTHOR
R. H. Hardin, Dec 11 2014
STATUS
approved