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

1,1

Links

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

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

Example

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

Crossrefs

Row 3 of A248433.

Sequence in context: A079997 A351252 A275260 * A354016 A222667 A122006

Adjacent sequences: A248433 A248434 A248435 * A248437 A248438 A248439

Keyword

nonn

Author

R. H. Hardin, Oct 06 2014

Status

approved