A248461 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.
6, 18, 10, 48, 36, 16, 96, 148, 72, 26, 174, 380, 460, 144, 42, 282, 862, 1512, 1436, 288, 68, 432, 1652, 4272, 6040, 4488, 576, 110, 624, 2956, 9684, 21182, 24160, 14040, 1152, 178, 870, 4860, 20236, 56782, 105026, 96736, 43940, 2304, 288, 1170, 7642, 37868
Offset: 1
Examples
Some solutions for n=5 k=4 ..3....4....4....0....1....4....3....1....0....0....1....1....1....1....3....4 ..4....4....0....1....3....1....2....1....2....3....1....0....3....3....0....1 ..4....1....4....1....0....2....3....4....3....1....0....1....4....4....3....1 ..3....4....4....0....1....1....3....1....2....0....1....0....0....0....3....4 ..0....0....1....4....1....2....2....4....3....1....4....3....0....0....4....1 ..3....4....2....0....2....1....3....1....3....4....4....4....3....1....1....4 ..3....3....4....0....2....1....2....2....4....2....0....4....1....3....1....4
Links
- R. H. Hardin, Table of n, a(n) for n = 1..9999
Formula
Empirical for column k:
k=1: a(n) = a(n-1) +a(n-2)
k=2: a(n) = 2*a(n-1)
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)
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)
k=5: [order 12]
k=6: [order 16]
k=7: [order 22]
Empirical for row n:
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
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
n=3: [order 27; also a degree 5 polynomial plus a quadratic quasipolynomial with period 840]
n=4: [order 61]
Comments