A250167 T(n,k)=Number of length n+1 0..k arrays with the sum of adjacent differences multiplied by some arrangement of +-1 equal to zero.
2, 3, 4, 4, 11, 8, 5, 20, 37, 16, 6, 33, 96, 119, 32, 7, 48, 211, 436, 373, 64, 8, 67, 380, 1269, 1880, 1151, 128, 9, 88, 639, 2860, 7109, 7836, 3517, 256, 10, 113, 976, 5831, 19896, 37881, 32032, 10679, 512, 11, 140, 1437, 10460, 49037, 129648, 195927
Offset: 1
Examples
Some solutions for n=5 k=4 ..3....0....3....4....0....3....4....4....2....4....4....2....0....4....3....1 ..2....0....4....2....0....4....1....4....4....1....3....1....1....2....1....1 ..4....4....0....4....4....2....2....2....4....3....4....3....1....2....3....3 ..0....2....0....1....2....1....3....2....1....0....3....2....3....1....0....4 ..1....2....4....1....3....1....3....3....3....0....0....2....0....4....3....3 ..1....0....3....2....0....1....2....4....0....4....0....2....0....2....3....1
Links
- R. H. Hardin, Table of n, a(n) for n = 1..263
Formula
Empirical for column k:
k=1: a(n) = 2*a(n-1)
k=2: a(n) = 5*a(n-1) -6*a(n-2)
k=3: a(n) = 8*a(n-1) -21*a(n-2) +22*a(n-3) -8*a(n-4)
k=4: [order 8]
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
n=3: a(n) = 2*a(n-1) +a(n-2) -4*a(n-3) +a(n-4) +2*a(n-5) -a(n-6); also a cubic polynomial plus a linear quasipolynomial with period 2
n=4: [order 12; also a quartic polynomial plus a quadratic quasipolynomial with period 12]
n=5: [order 24; also a polynomial of degree 5 plus a cubic quasipolynomialwith period 60]
Comments