A248436 Number of length 3+2 0..n arrays with every three consecutive terms having the sum of some two elements equal to twice the third.

2, 9, 24, 69, 118, 185, 252, 357, 470, 593, 716, 881, 1046, 1217, 1400, 1621, 1842, 2081, 2320, 2593, 2874, 3161, 3448, 3785, 4126, 4473, 4828, 5213, 5598, 6005, 6412, 6857, 7310, 7769, 8232, 8737, 9242, 9753, 10272, 10833, 11394, 11973, 12552, 13161, 13782

Offset: 1

Author

R. H. Hardin, Oct 06 2014

Keywords

• nonn

Examples

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

Links

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

Crossrefs

Row 3 of A248433.

Formula

Empirical: a(n) = a(n-1) - a(n-2) + a(n-3) - a(n-4) + 2*a(n-5) - a(n-6) + a(n-7) - a(n-11) + a(n-12) - 2*a(n-13) + a(n-14) - a(n-15) + a(n-16) - a(n-17) + a(n-18).

Also a quadratic polynomial plus a constant quasipolynomial with period 840, the first 12 being:

Empirical for n mod 840 = 0: a(n) = (106/15)*n^2 - (178/15)*n + 1

Empirical for n mod 840 = 1: a(n) = (106/15)*n^2 - (178/15)*n + (34/5)

Empirical for n mod 840 = 2: a(n) = (106/15)*n^2 - (178/15)*n + (67/15)

Empirical for n mod 840 = 3: a(n) = (106/15)*n^2 - (178/15)*n - 4

Empirical for n mod 840 = 4: a(n) = (106/15)*n^2 - (178/15)*n + (17/5)

Empirical for n mod 840 = 5: a(n) = (106/15)*n^2 - (178/15)*n + (2/3)

Empirical for n mod 840 = 6: a(n) = (106/15)*n^2 - (178/15)*n + (9/5)

Empirical for n mod 840 = 7: a(n) = (106/15)*n^2 - (178/15)*n - (56/5)

Empirical for n mod 840 = 8: a(n) = (106/15)*n^2 - (178/15)*n - (1/3)

Empirical for n mod 840 = 9: a(n) = (106/15)*n^2 - (178/15)*n + (22/5)

Empirical for n mod 840 = 10: a(n) = (106/15)*n^2 - (178/15)*n + 5

Empirical for n mod 840 = 11: a(n) = (106/15)*n^2 - (178/15)*n - (128/15).

Empirical g.f.: x*(2 + 7*x + 17*x^2 + 52*x^3 + 66*x^4 + 117*x^5 + 124*x^6 + 200*x^7 + 175*x^8 + 222*x^9 + 167*x^10 + 210*x^11 + 118*x^12 + 113*x^13 + 52*x^14 + 54*x^15 - x^16 + x^17) / ((1 - x)^3*(1 + x)*(1 - x + x^2)*(1 + x^2)*(1 + x + x^2)*(1 + x^4)*(1 + x + x^2 + x^3 + x^4)). - Colin Barker, Nov 08 2018