A251429 Number of length 2+2 0..n arrays with the sum of the maximum minus twice the median plus the minimum of adjacent triples multiplied by some arrangement of +-1 equal to zero.

12, 41, 116, 237, 432, 725, 1128, 1641, 2316, 3145, 4148, 5357, 6776, 8413, 10328, 12489, 14924, 17689, 20764, 24149, 27928, 32061, 36568, 41513, 46868, 52641, 58924, 65653, 72856, 80621, 88896, 97681, 107092, 117057, 127596, 138805, 150624, 163061

Offset: 1

Author

R. H. Hardin, Dec 02 2014

Keywords

• nonn

Examples

			Some solutions for n=10:
..1....1....5....1....6....1....1....0....5....4....8....2....7....8....0....6
..7....8...10....7....4....6....3....3....6....6....3....6....2....6....6....7
.10....3....7....3....3...10....5....1....3....7....9....5...10....6....2....1
..7....7....8....9....6....1....1....4....5....4....1....9....7....8....5....9
		

Links

• R. H. Hardin, Table of n, a(n) for n = 1..210

Crossrefs

Row 2 of A251428.

Formula

Empirical: a(n) = a(n-1) + 2*a(n-3) - a(n-4) - a(n-5) - a(n-6) - a(n-7) + 2*a(n-8) + a(n-10) - a(n-11).

Empirical for n mod 12 = 0: a(n) = (79/27)*n^3 + (29/18)*n^2 + (17/3)*n + 1

Empirical for n mod 12 = 1: a(n) = (79/27)*n^3 + (29/18)*n^2 + (17/3)*n + (97/54)

Empirical for n mod 12 = 2: a(n) = (79/27)*n^3 + (29/18)*n^2 + (43/9)*n + (43/27)

Empirical for n mod 12 = 3: a(n) = (79/27)*n^3 + (29/18)*n^2 + (17/3)*n + (11/2)

Empirical for n mod 12 = 4: a(n) = (79/27)*n^3 + (29/18)*n^2 + (17/3)*n + (35/27)

Empirical for n mod 12 = 5: a(n) = (79/27)*n^3 + (29/18)*n^2 + (43/9)*n + (113/54)

Empirical for n mod 12 = 6: a(n) = (79/27)*n^3 + (29/18)*n^2 + (17/3)*n + 1

Empirical for n mod 12 = 7: a(n) = (79/27)*n^3 + (29/18)*n^2 + (17/3)*n + (313/54)

Empirical for n mod 12 = 8: a(n) = (79/27)*n^3 + (29/18)*n^2 + (43/9)*n + (43/27)

Empirical for n mod 12 = 9: a(n) = (79/27)*n^3 + (29/18)*n^2 + (17/3)*n + (3/2)

Empirical for n mod 12 = 10: a(n) = (79/27)*n^3 + (29/18)*n^2 + (17/3)*n + (35/27)

Empirical for n mod 12 = 11: a(n) = (79/27)*n^3 + (29/18)*n^2 + (43/9)*n + (329/54)

Empirical g.f.: x*(12 + 29*x + 75*x^2 + 97*x^3 + 125*x^4 + 114*x^5 + 98*x^6 + 55*x^7 + 27*x^8 + x^9 - x^10) / ((1 - x)^4*(1 + x)*(1 + x^2)*(1 + x + x^2)^2). - Colin Barker, Nov 29 2018