A248433 T(n,k)=Number of length n+2 0..k arrays with every three consecutive terms having the sum of some two elements equal to twice the third.
2, 9, 2, 16, 9, 2, 29, 20, 9, 2, 42, 45, 24, 9, 2, 61, 70, 69, 28, 9, 2, 80, 105, 118, 101, 36, 9, 2, 105, 140, 185, 198, 165, 44, 9, 2, 130, 189, 252, 327, 342, 261, 52, 9, 2, 161, 242, 357, 462, 601, 590, 389, 68, 9, 2, 192, 301, 470, 691, 884, 1105, 1014, 645, 84, 9, 2, 229, 360
Offset: 1
Examples
Some solutions for n=6 k=4 ..2....3....3....0....4....3....1....4....0....2....3....0....2....0....2....1 ..4....3....4....2....2....4....3....2....2....4....2....2....0....2....1....0 ..0....3....2....4....0....2....2....3....1....3....1....4....4....1....0....2 ..2....3....3....3....1....3....4....4....0....2....0....0....2....3....2....4 ..1....3....4....2....2....4....3....2....2....1....2....2....0....2....4....0 ..0....3....2....4....0....2....2....3....1....0....4....1....1....4....3....2 ..2....3....0....3....1....0....4....4....0....2....0....3....2....3....2....4 ..1....3....1....2....2....4....0....2....2....4....2....2....0....2....1....3
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)
k=2: a(n) = a(n-1)
k=3: a(n) = a(n-1) +2*a(n-3) -2*a(n-4)
k=4: a(n) = a(n-1) +4*a(n-3) -4*a(n-4)
k=5: a(n) = a(n-1) +6*a(n-3) -6*a(n-4) -4*a(n-6) +4*a(n-7)
k=6: a(n) = 8*a(n-3) -11*a(n-6) +4*a(n-9)
k=7: [order 13]
Empirical for row n:
n=1: a(n) = 2*a(n-1) -2*a(n-3) +a(n-4); also quadratic polynomial plus a constant quasipolynomial with period 2
n=2: a(n) = a(n-1) +a(n-3) -a(n-5) -a(n-7) +a(n-8); also a quadratic polynomial plus a constant quasipolynomial with period 12
n=3: [order 18; also a quadratic polynomial plus a constant quasipolynomial with period 840]
n=4: [order 36]
n=5: [order 70]
Comments