A250646 T(n,k)=Number of length n+1 0..k arrays with the sum of the maximum of each adjacent pair multiplied by some arrangement of +-1 equal to zero.
1, 1, 6, 1, 17, 6, 1, 36, 23, 20, 1, 65, 44, 125, 28, 1, 106, 89, 476, 280, 72, 1, 161, 134, 1293, 1424, 1061, 120, 1, 232, 219, 2954, 4853, 7696, 2870, 272, 1, 321, 296, 5901, 12473, 34441, 28238, 9495, 496, 1, 430, 433, 10766, 28379, 120114, 163043, 126482
Offset: 1
Examples
Some solutions for n=5 k=4 ..4....2....1....4....0....1....1....2....0....2....1....2....2....4....4....3 ..1....1....1....1....2....2....0....2....0....1....0....1....2....0....1....3 ..1....0....1....1....2....0....0....1....1....0....2....0....2....2....0....1 ..0....1....0....0....0....0....1....1....2....4....3....4....1....3....2....2 ..1....1....1....3....1....4....3....0....1....3....3....3....1....0....1....4 ..3....0....4....3....3....3....3....0....0....2....0....3....3....2....0....4
Links
- R. H. Hardin, Table of n, a(n) for n = 1..267
Formula
Empirical for column k:
k=1: a(n) = 2*a(n-1) +2*a(n-2) -4*a(n-3)
k=2: [order 10]
k=3: [order 24] for n>25
Empirical for row n:
n=1: a(n) = a(n-1)
n=2: a(n) = (1/3)*n^3 + 2*n^2 + (8/3)*n + 1
n=3: a(n) = a(n-1) +3*a(n-2) -3*a(n-3) -3*a(n-4) +3*a(n-5) +a(n-6) -a(n-7); also a polynomial of degree 3 plus a quasipolynomial of degree 2 with period 2
n=4: [order 14; also a polynomial of degree 5 plus a quasipolynomial of degree 2 with period 6]
n=5: [order 25; also a polynomial of degree 5 plus a quasipolynomial of degree 4 with period 12]
Comments