A247535 Number of length 3+3 0..n arrays with some disjoint pairs in every consecutive four terms having the same sum.

8, 61, 220, 561, 1124, 2009, 3220, 4901, 7016, 9737, 13000, 17025, 21688, 27245, 33572, 40929, 49140, 58553, 68924, 80613, 93400, 107641, 123056, 140113, 158440, 178509, 200012, 223393, 248260, 275209, 303748, 334453, 366920, 401689, 438264

Offset: 1

Author

R. H. Hardin, Sep 18 2014

Keywords

• nonn

Examples

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

Links

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

Crossrefs

Row 3 of A247533.

Formula

Empirical: a(n) = a(n-2) + 2*a(n-3) + a(n-4) - 2*a(n-5) - 2*a(n-6) - 2*a(n-7) + a(n-8) + 2*a(n-9) + a(n-10) - a(n-12).

Also as a cubic plus a linear quasipolynomial with period 12:

Empirical for n mod 12 = 0: a(n) = (563/54)*n^3 - (127/18)*n^2 + 2*n + 1

Empirical for n mod 12 = 1: a(n) = (563/54)*n^3 - (127/18)*n^2 - (5/2)*n + (385/54)

Empirical for n mod 12 = 2: a(n) = (563/54)*n^3 - (127/18)*n^2 + (10/9)*n + (97/27)

Empirical for n mod 12 = 3: a(n) = (563/54)*n^3 - (127/18)*n^2 - (5/2)*n + (19/2)

Empirical for n mod 12 = 4: a(n) = (563/54)*n^3 - (127/18)*n^2 + 2*n - (37/27)

Empirical for n mod 12 = 5: a(n) = (563/54)*n^3 - (127/18)*n^2 - (61/18)*n + (761/54)

Empirical for n mod 12 = 6: a(n) = (563/54)*n^3 - (127/18)*n^2 + 2*n - 1

Empirical for n mod 12 = 7: a(n) = (563/54)*n^3 - (127/18)*n^2 - (5/2)*n + (385/54)

Empirical for n mod 12 = 8: a(n) = (563/54)*n^3 - (127/18)*n^2 + (10/9)*n + (151/27)

Empirical for n mod 12 = 9: a(n) = (563/54)*n^3 - (127/18)*n^2 - (5/2)*n + (19/2)

Empirical for n mod 12 = 10: a(n) = (563/54)*n^3 - (127/18)*n^2 + 2*n - (91/27)

Empirical for n mod 12 = 11: a(n) = (563/54)*n^3 - (127/18)*n^2 - (61/18)*n + (761/54).

Empirical g.f.: x*(8 + 61*x + 212*x^2 + 484*x^3 + 774*x^4 + 963*x^5 + 892*x^6 + 661*x^7 + 330*x^8 + 120*x^9 - x^11) / ((1 - x)^4*(1 + x)^2*(1 + x^2)*(1 + x + x^2)^2). - Colin Barker, Nov 07 2018