A250277 T(n,k)=Number of length n+1 0..k arrays with the sum of the squares of adjacent differences multiplied by some arrangement of +-1 equal to zero.
2, 3, 4, 4, 11, 8, 5, 20, 27, 16, 6, 33, 52, 79, 32, 7, 48, 89, 240, 255, 64, 8, 67, 140, 581, 984, 843, 128, 9, 88, 207, 1132, 2909, 4412, 2763, 256, 10, 113, 288, 1991, 6732, 17885, 20252, 8903, 512, 11, 140, 389, 3156, 14003, 51884, 107387, 91808, 28215, 1024, 12
Offset: 1
Examples
Some solutions for n=6 k=4 ..4....0....0....1....3....3....4....3....3....1....2....0....4....1....3....3 ..2....2....0....1....1....3....2....3....2....4....3....3....4....3....2....4 ..4....2....2....4....2....3....0....3....4....4....0....2....4....2....2....4 ..2....1....4....4....1....3....4....1....2....1....2....4....2....3....1....2 ..2....2....4....2....1....1....4....0....1....2....4....3....1....1....1....4 ..2....3....2....1....1....3....2....1....2....1....3....4....2....1....0....0 ..4....2....0....3....3....3....0....3....3....1....0....2....4....1....1....3
Links
- R. H. Hardin, Table of n, a(n) for n = 1..181
Formula
Empirical for column k:
k=1: a(n) = 2*a(n-1)
k=2: a(n) = 9*a(n-1) -31*a(n-2) +51*a(n-3) -40*a(n-4) +12*a(n-5) for n>6
k=3: [order 15] for n>18
Empirical for row n:
n=1: a(n) = n + 1
n=2: a(n) = 2*a(n-1) -2*a(n-3) +a(n-4); also a quadratic polynomial plus a constant quasipolynomial with period 2
Comments