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T(n,k)=Number of length n+2 0..k arrays with no three consecutive terms having the sum of any two elements equal to twice the third
13

%I #6 Dec 12 2014 20:43:38

%S 6,18,10,48,36,16,96,148,72,26,174,380,460,144,42,282,862,1512,1436,

%T 288,68,432,1652,4272,6040,4488,576,110,624,2956,9684,21182,24160,

%U 14040,1152,178,870,4860,20236,56782,105026,96736,43940,2304,288,1170,7642,37868

%N T(n,k)=Number of length n+2 0..k arrays with no three consecutive terms having the sum of any two elements equal to twice the third

%C Table starts

%C ...6...18......48.......96.......174........282.........432.........624

%C ..10...36.....148......380.......862.......1652........2956........4860

%C ..16...72.....460.....1512......4272.......9684.......20236.......37868

%C ..26..144....1436.....6040.....21182......56782......138534......295078

%C ..42..288....4488....24160....105026.....332940......948412.....2299356

%C ..68..576...14040....96736....520788....1952254.....6493036....17917712

%C .110.1152...43940...387488...2582406...11447368....44452660...139623544

%C .178.2304..137532..1552448..12805334...67123652...304332258..1088015294

%C .288.4608..430508..6220480..63497776..393591402..2083523194..8478351478

%C .466.9216.1347652.24926080.314866606.2307892826.14264241960.66067495706

%H R. H. Hardin, <a href="/A248461/b248461.txt">Table of n, a(n) for n = 1..9999</a>

%F Empirical for column k:

%F k=1: a(n) = a(n-1) +a(n-2)

%F k=2: a(n) = 2*a(n-1)

%F k=3: a(n) = 2*a(n-1) +3*a(n-2) +4*a(n-3) -3*a(n-4) -12*a(n-5) -4*a(n-6)

%F k=4: a(n) = 3*a(n-1) +5*a(n-2) +2*a(n-3) -16*a(n-4) -28*a(n-5) -8*a(n-6)

%F k=5: [order 12]

%F k=6: [order 16]

%F k=7: [order 22]

%F Empirical for row n:

%F n=1: a(n) = 3*a(n-1) -2*a(n-2) -2*a(n-3) +3*a(n-4) -a(n-5); also a cubic polynomial plus a constant quasipolynomial with period 2

%F n=2: a(n) = 2*a(n-1) -a(n-3) -2*a(n-5) +2*a(n-6) +a(n-8) -2*a(n-10) +a(n-11); also a quartic polynomial plus a linear quasipolynomial with period 12

%F n=3: [order 27; also a degree 5 polynomial plus a quadratic quasipolynomial with period 840]

%F n=4: [order 61]

%e Some solutions for n=5 k=4

%e ..3....4....4....0....1....4....3....1....0....0....1....1....1....1....3....4

%e ..4....4....0....1....3....1....2....1....2....3....1....0....3....3....0....1

%e ..4....1....4....1....0....2....3....4....3....1....0....1....4....4....3....1

%e ..3....4....4....0....1....1....3....1....2....0....1....0....0....0....3....4

%e ..0....0....1....4....1....2....2....4....3....1....4....3....0....0....4....1

%e ..3....4....2....0....2....1....3....1....3....4....4....4....3....1....1....4

%e ..3....3....4....0....2....1....2....2....4....2....0....4....1....3....1....4

%Y Column 1 is A006355(n+4)

%Y Column 2 is A005010

%K nonn,tabl

%O 1,1

%A _R. H. Hardin_, Oct 06 2014