A248464 Number of length 3+2 0..n arrays with no three consecutive terms having the sum of any two elements equal to twice the third.

16, 72, 460, 1512, 4272, 9684, 20236, 37868, 67140, 111104, 177024, 269892, 400040, 574140, 806808, 1107132, 1493952, 1978984, 2586340, 3330852, 4242444, 5338788, 6656352, 8217136, 10064140, 12222784, 14744480, 17659176, 21025952, 24879392

Offset: 1

Author

R. H. Hardin, Oct 06 2014

Keywords

• nonn

Comments

Row 3 of A248461

Examples

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

Links

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

Formula

Empirical: a(n) = a(n-1) +a(n-3) -2*a(n-7) +a(n-8) -2*a(n-9) +a(n-10) -a(n-11) +2*a(n-12) +2*a(n-15) -a(n-16) +a(n-17) -2*a(n-18) +a(n-19) -2*a(n-20) +a(n-24) +a(n-26) -a(n-27)

Also a polynomial of degree 5 plus a quadratic quasipolynomial with period 840, the first 12 being:

Empirical for n mod 840 = 0: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (197/30)*n^2 + (51/5)*n

Empirical for n mod 840 = 1: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (31/15)*n^2 + (66/5)*n - (33/10)

Empirical for n mod 840 = 2: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (197/30)*n^2 + (113/15)*n - (152/15)

Empirical for n mod 840 = 3: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (31/15)*n^2 + (26/5)*n - (1/2)

Empirical for n mod 840 = 4: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (197/30)*n^2 + (51/5)*n - (12/5)

Empirical for n mod 840 = 5: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (31/15)*n^2 + (158/15)*n + (1/6)

Empirical for n mod 840 = 6: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (197/30)*n^2 + (51/5)*n - (24/5)

Empirical for n mod 840 = 7: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (31/15)*n^2 + (26/5)*n + (67/10)

Empirical for n mod 840 = 8: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (197/30)*n^2 + (113/15)*n - (4/3)

Empirical for n mod 840 = 9: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (31/15)*n^2 + (66/5)*n - (9/10)

Empirical for n mod 840 = 10: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (197/30)*n^2 + (51/5)*n - 8

Empirical for n mod 840 = 11: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (31/15)*n^2 + (38/15)*n + (41/30)