A250169 Number of length 4+1 0..n arrays with the sum of adjacent differences multiplied by some arrangement of +-1 equal to zero.

16, 119, 436, 1269, 2860, 5831, 10460, 17765, 27984, 42611, 61752, 87477, 119672, 161051, 211212, 273601, 347428, 436963, 540948, 664553, 805976, 971367, 1158288, 1373969, 1615248, 1890503, 2195780, 2540693, 2920396, 3345831, 3811180, 4328789

Offset: 1

Author

R. H. Hardin, Nov 13 2014

Keywords

• nonn

Examples

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

Links

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

Crossrefs

Row 4 of A250167.

Formula

Empirical: a(n) = a(n-1) + 2*a(n-2) - a(n-3) - a(n-4) - 2*a(n-5) + 2*a(n-7) + a(n-8) + a(n-9) - 2*a(n-10) - a(n-11) + a(n-12).

Empirical for n mod 12 = 0: a(n) = (197/48)*n^4 + (5/12)*n^3 + (45/4)*n^2 + (8/3)*n + 1

Empirical for n mod 12 = 1: a(n) = (197/48)*n^4 + (5/12)*n^3 + (69/8)*n^2 + (77/12)*n - (57/16)

Empirical for n mod 12 = 2: a(n) = (197/48)*n^4 + (5/12)*n^3 + (45/4)*n^2 + (8/3)*n - (1/3)

Empirical for n mod 12 = 3: a(n) = (197/48)*n^4 + (5/12)*n^3 + (69/8)*n^2 + (77/12)*n - (73/16)

Empirical for n mod 12 = 4: a(n) = (197/48)*n^4 + (5/12)*n^3 + (45/4)*n^2 + (8/3)*n + 1

Empirical for n mod 12 = 5: a(n) = (197/48)*n^4 + (5/12)*n^3 + (69/8)*n^2 + (77/12)*n - (235/48)

Empirical for n mod 12 = 6: a(n) = (197/48)*n^4 + (5/12)*n^3 + (45/4)*n^2 + (8/3)*n + 1

Empirical for n mod 12 = 7: a(n) = (197/48)*n^4 + (5/12)*n^3 + (69/8)*n^2 + (77/12)*n - (73/16)

Empirical for n mod 12 = 8: a(n) = (197/48)*n^4 + (5/12)*n^3 + (45/4)*n^2 + (8/3)*n - (1/3)

Empirical for n mod 12 = 9: a(n) = (197/48)*n^4 + (5/12)*n^3 + (69/8)*n^2 + (77/12)*n - (57/16)

Empirical for n mod 12 = 10: a(n) = (197/48)*n^4 + (5/12)*n^3 + (45/4)*n^2 + (8/3)*n + 1

Empirical for n mod 12 = 11: a(n) = (197/48)*n^4 + (5/12)*n^3 + (69/8)*n^2 + (77/12)*n - (283/48).

Empirical g.f.: x*(16 + 103*x + 285*x^2 + 611*x^3 + 854*x^4 + 1020*x^5 + 852*x^6 + 612*x^7 + 274*x^8 + 101*x^9 - x^10 + x^11) / ((1 - x)^5*(1 + x)^3*(1 + x^2)*(1 + x + x^2)). - Colin Barker, Nov 12 2018